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Momentum

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· Impulse - Momentum Equation · Collisions in One Dimension · Collisions in Two Dimensions · It is Rocket Science · Center of Mass · Ballistic Pendulum · Conservation of Linear Momentum

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Conservation of Linear Momentum

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The most powerful concepts in science are called "conservation principles". Without worrying about the details of a process, conservation principles can be used to solve problems. If we were to take a snapshot of the initial and final system, comparing the two would provide a lot of information. The last unit presented the Conservation of Energy, which proved helpful in solving problems where the situation was too complex for Newton's Laws to be effectively used. Conservation of Energy is not enough.

Conservation of Momentum Slide 5 / 140

If two ice skaters are holding hands, but then push away from each other, Conservation of Energy will not be able to determine each skater's velocity after the push. Assume a closed system with no external forces. We need something new. Let's start with Newton's Third Law. Assume the ice is frictionless. Then the force exerted by one skater on the second is equal and opposite to the force exerted by the second skater on the first and there are no external forces.

Conservation of Momentum

Stating Newton's Third Law and doing a little bit of substitution

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Make the assumption that we're not dealing with relativistic conditions (the skaters are moving way slower than the speed of light), and the skaters don't lose any mass in the process. Then, the constant mass terms can be put within the derivative.

Conservation of Momentum

This is good, as it looks like we can now say something about the skaters's velocities. The time rate

• f change of the sum of each

skater's mass times velocity is zero - it doesn't change. The quantity is defined as linear momentum and will be represented as - it is a vector.

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The initial linear momentum is then equal to the final linear momentum of the system. Thus, linear momentum is conserved in a closed system with no external forces. For two particles, , and since the initial and final momentum remain constant, we can write:

Conservation of Momentum Slide 8 / 140

The Energy chapter of this course discussed how potential energy could be calculated for a system that had only conservative forces. The Conservation of Momentum does not put this restriction on forces - as long as the forces are internal, momentum is conserved. We will leave the Conservation of Momentum for now, and apply it to problems involving collisions between objects in an upcoming chapter. Many of the great discoveries in nuclear and particle physics involve smashing atoms and nuclei into each other and seeing what comes out. The Conservation of Momentum is key in these experiments.

Conservation of Momentum Slide 9 / 140

Let's take a brief detour and examine Newton's Second Law as it is taught today, and use the definition of momentum. This is how Newton presented his Second Law in Principia - the sum of forces on an object changes its momentum over time. The more general statement of the Second Law has the benefit that it can be extended to cases where the mass of the objects change - not just the velocity. More on this later.

Newton's Second Law restated

The sum of forces on a particle is equal to ma. Since we assume the mass is constant, we can bring it inside the derivative.

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A key difference between momentum and energy is that energy is a scalar, while momentum is a vector. When there is more than one object in a system, the total momentum of the system is found by the vector addition of the each object's momentum. Another key difference is that momentum comes in only one flavor (there is no kinetic, potential, elastic, etc.). The unit for momentum is kg-m/s. There is no special unit for momentum, which presents an excellent opportunity to honor another physicist.

Momentum is a Vector Quantity Slide 11 / 140

1 What is the momentum of a 20 kg object with a velocity of +5.0 m/s? A -100 kg-m/s B -50 kg-m/s C 0 kg-m/s D 50 kg-m/s E 100 kg-m/s

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1 What is the momentum of a 20 kg object with a velocity of +5.0 m/s? A -100 kg-m/s B -50 kg-m/s C 0 kg-m/s D 50 kg-m/s E 100 kg-m/s

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Answer E

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2 What is the momentum of a 20 kg object with a velocity of

• 5.0 m/s?

A -100 kg-m/s B -50 kg-m/s C 0 kg-m/s D 50 kg-m/s E 100 kg-m/s

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2 What is the momentum of a 20 kg object with a velocity of

• 5.0 m/s?

A -100 kg-m/s B -50 kg-m/s C 0 kg-m/s D 50 kg-m/s E 100 kg-m/s

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Answer A

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If a system contains more than one object, its total momentum is the vector sum of each object's individual momentum:

Momentum of a System of Objects Slide 14 / 140

In order to determine the total momentum of a system : · Select a direction to be positive for each dimension that's being considered. · Assign positive values to each momentum in that direction. · Assign negative values to each momentum in the opposite direction. · Add the momenta together independently for each dimension. · Vectorially add each dimension's momentum together to get the total momentum - the Pythagoras theorem is used to find the magnitude, and trigonometry is used to find its direction.

Momentum of a System of Objects Slide 15 / 140 Momentum of a System of Objects

Let's work an example in one dimension. Determine the momentum

• f a system of two objects: m1, has a mass of 15 kg and a velocity of

16 m/s towards the east and m2,has a mass of 42 kg and a velocity

• f 6.0 m/s towards the west.

Choose East as positive. to the west

• r

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3 Determine the magnitude of the momentum of a system of two objects: m1, which has a mass of 6.0 kg and a velocity of 13 m/s north and m2, which has a mass of 14 kg and a velocity of 7.0 m/s south. Assume north is positive. A 10 kg m/s B 15 kg m/s C 20 kg m/s D -15 kg m/s E -20 kg m/s

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3 Determine the magnitude of the momentum of a system of two objects: m1, which has a mass of 6.0 kg and a velocity of 13 m/s north and m2, which has a mass of 14 kg and a velocity of 7.0 m/s south. Assume north is positive. A 10 kg m/s B 15 kg m/s C 20 kg m/s D -15 kg m/s E -20 kg m/s

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Answer E

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4 Determine the momentum of a system of 3 objects: m1, which has a mass of 7.0 kg and a velocity of 23 m/s north, m2, which has a mass of 9.0 kg and a velocity of 7.0 m/s north and m3, which has a mass of 5.0 kg and a velocity of 42 m/s south. Assume north is positive. A -12 kg m/s B 12 kg m/s C -14 kg m/s D 14 kg m/s E 15 m/s

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4 Determine the momentum of a system of 3 objects: m1, which has a mass of 7.0 kg and a velocity of 23 m/s north, m2, which has a mass of 9.0 kg and a velocity of 7.0 m/s north and m3, which has a mass of 5.0 kg and a velocity of 42 m/s south. Assume north is positive. A -12 kg m/s B 12 kg m/s C -14 kg m/s D 14 kg m/s E 15 m/s

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Answer D

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Impulse - Momentum Equation

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Slide 19 / 140 Impulse-Momentum Theorem

The Conservation of Linear Momentum applies to an isolated system of particles. The overall momentum is conserved, but what about the momentum of each particle? Start with Newton's Second Law, as expressed in Principia, where we look at all the forces on one of the particles. Assume the force acts over a time interval t0 to tf, and integrate this expression. The particle's momentum will change.

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Impulse-Momentum Theorem

We will examine the specific case where a single, constant, very large force acts on the particle for a very short time, so all other forces need not be considered. The equation simplifies to: Define Impulse as: The Impulse-Momentum equation is then: Impulse is a vector, and it is in the same direction as the change of momentum or velocity of the particle acted on by the force.

Slide 21 / 140 Impulse-Momentum Theorem

The force is not always constant - for example when a tennis racquet strikes a tennis ball, the force starts out small, and increases as the ball increases its contact time with the racquet, then decreases as it leaves. The large force at the peak results in a deformation of the ball.

F(N) t (s)

Slide 22 / 140 Impulse-Momentum Theorem

F(N) F(N) t (s) t (s) Favg

The shaded areas are equal in magnitude.

The force - time graph can be used to find the Impulse delivered by the racquet in two ways: · Find the area underneath the curve, either by integration if the force is specified as a function of time, or by numerical methods. · Find the average force delivered by the racquet and multiply if by the time interval.

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Impulse-Momentum Theorem

F(N) F(N) t (s) t (s) Favg

· Force known as a function of time: · Average force known: But what if the force is not easily expressed in terms of time or you don't have a numerical integration capability?

Slide 24 / 140 Impulse-Momentum Theorem

F(N) F(N) t (s) t (s) Favg

The equation also works in reverse (of course). If you have the change in momentum of the object, and the time over which it

• ccurs, the average force can be found.

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Impulse tells us that we can get the same change in momentum with a large average force acting for a short time, or a small average force acting for a longer time. For a given change of momentum (when a person stops moving because of an impact, like a car accident or falling), the force to the person can be minimized by extending the duration of the velocity reducing event. This is why one should bend their knees during a parachute landing, why airbags are used, and why landing on a pillow hurts less than landing on concrete.

Implications of Impulse Slide 26 / 140

5 An external force of 25 N acts on a system for 10 s. What is the magnitude of the Impulse delivered to the system? A 25 N-s B 100 N-s C 150 N-s D 200 N-s E 250 N-s

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5 An external force of 25 N acts on a system for 10 s. What is the magnitude of the Impulse delivered to the system? A 25 N-s B 100 N-s C 150 N-s D 200 N-s E 250 N-s

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Answer E

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6 An external force of 25 N acts on a system for 10 s. What is the change in momentum of the system? A 25 N-s B 100 N-s C 150 N-s D 200 N-s E 250 N-s

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6 An external force of 25 N acts on a system for 10 s. What is the change in momentum of the system? A 25 N-s B 100 N-s C 150 N-s D 200 N-s E 250 N-s

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Answer E

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7 An average force of 5,000 N acts for 0.03 s on a 2.5 kg object that is initially at rest. What is its velocity after the application

• f the force?

A 80 m/s B 70 m/s C 60 m/s D 50 m/s E 40 m/s

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7 An average force of 5,000 N acts for 0.03 s on a 2.5 kg object that is initially at rest. What is its velocity after the application

• f the force?

A 80 m/s B 70 m/s C 60 m/s D 50 m/s E 40 m/s

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Answer C

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8 An object at rest experiences a net horizontal force in the +x direction and begins moving. Use the force - time graph below to find the net impulse delivered by the force after 6 s. A 12 N-s B 14 N-s C 16 N-s D 18 N-s E 20 N-s

F (N) t (s)

2 2 4 4 6 6

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8 An object at rest experiences a net horizontal force in the +x direction and begins moving. Use the force - time graph below to find the net impulse delivered by the force after 6 s. A 12 N-s B 14 N-s C 16 N-s D 18 N-s E 20 N-s

F (N) t (s)

2 2 4 4 6 6

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Answer A

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9 A 2 kg object at rest experiences a net horizontal force in the +x direction and begins moving. Use the force - time graph below to find the net impulse delivered by the force after 6 s. A 4 N-s B 6 N-s C 8 N-s D 10 N-s E 12 N-s

F (N) t (s)

2 2 4 4 6 6

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9 A 2 kg object at rest experiences a net horizontal force in the +x direction and begins moving. Use the force - time graph below to find the net impulse delivered by the force after 6 s. A 4 N-s B 6 N-s C 8 N-s D 10 N-s E 12 N-s

F (N) t (s)

2 2 4 4 6 6

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Answer C

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10 A 2 kg object at rest experiences a net horizontal force in the +x direction and begins moving. Use the force - time graph below to find the object's velocity after 6 s. A 10 m/s B 8 m/s C 6 m/s D 5 m/s E 4 m/s

F (N) t (s)

2 2 4 4 6 6

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10 A 2 kg object at rest experiences a net horizontal force in the +x direction and begins moving. Use the force - time graph below to find the object's velocity after 6 s. A 10 m/s B 8 m/s C 6 m/s D 5 m/s E 4 m/s

F (N) t (s)

2 2 4 4 6 6

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Answer E

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11 A force described by F(t) = 190t - 189t2 is applied by a bat to a 0.145 kg ball. Assume the bat loses contact with the ball when the force decreases to zero. Over what time interval does this force act?

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11 A force described by F(t) = 190t - 189t2 is applied by a bat to a 0.145 kg ball. Assume the bat loses contact with the ball when the force decreases to zero. Over what time interval does this force act?

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Answer The force acts from t = 0 to t = 1.01s

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12 A force described by F(t) = 190t - 189t2 is applied by a bat to a 0.145 kg ball. The force acts over a time interval of 1.01s. What is the magnitude of the maximum force delivered to the ball?

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12 A force described by F(t) = 190t - 189t2 is applied by a bat to a 0.145 kg ball. The force acts over a time interval of 1.01s. What is the magnitude of the maximum force delivered to the ball?

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Answer

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13 A force described by F(t) = 190t - 189t2 is applied by a bat to a 0.145 kg ball. The force acts over a time interval of 1.01s. What is the magnitude of the impulse delivered to the ball?

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13 A force described by F(t) = 190t - 189t2 is applied by a bat to a 0.145 kg ball. The force acts over a time interval of 1.01s. What is the magnitude of the impulse delivered to the ball?

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Answer

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14 A force described by F(t) = 190t - 189t2 is applied by a bat to a 0.145 kg ball. The force acts over a time interval of 1.01s. What is the magnitude of the average force delivered to the ball?

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14 A force described by F(t) = 190t - 189t2 is applied by a bat to a 0.145 kg ball. The force acts over a time interval of 1.01s. What is the magnitude of the average force delivered to the ball?

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Answer

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15 A force described by F(t) = 190t - 189t2 is applied by a bat to a 0.145 kg ball and delivers an impulse of 32 N/s. What is the velocity of the ball at t = 1.01 s, assuming it started from rest?

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15 A force described by F(t) = 190t - 189t2 is applied by a bat to a 0.145 kg ball and delivers an impulse of 32 N/s. What is the velocity of the ball at t = 1.01 s, assuming it started from rest?

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Answer

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Collisions in One Dimension

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Slide 38 / 140 Types of Collisions

Objects in an isolated system can interact with each other in two basic ways: · They can collide. · If they are stuck together, they can explode (push apart). In an isolated system both momentum and total energy are

• conserved. But the energy can change from one form to another.

Conservation of momentum and change in kinetic energy can help predict what will happen in these events.

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Types of Collisions

We differentiate collisions and explosions by the way the energy changes or does not change form. · inelastic collisions: two objects collide, converting some kinetic energy into other forms of energy such as potential energy, heat or sound. · elastic collisions: two objects collide and bounce off each other while conserving kinetic energy - energy is not transformed into any other type. · explosions: an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy.

Slide 40 / 140 Inelastic Collisions

There are two types of Inelastic Collisions. · perfect inelastic collisions: two objects collide, stick together and move as one mass after the collision, transferring kinetic energy into other forms of energy. · general inelastic collisions: two objects collide and bounce off each other, transferring kinetic energy into other forms of energy.

Slide 41 / 140 Elastic Collisions

There is really no such thing as a perfect elastic collision. During all collisions, some kinetic energy is always transformed into

• ther forms of energy.

But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions. In chemistry, the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law. Other examples include a steel ball bearing dropping on a steel plate, a rubber "superball" bouncing on the ground, and billiard balls bouncing off each other.

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Explosions

A firecracker is an example of an explosion. The chemical potential energy inside the firecracker is transformed into kinetic energy, light and sound. A cart with a compressed spring is a good example. When the spring is against a wall, and it is released, the cart starts moving

• converting elastic potential energy into kinetic energy and

sound. Think for a moment - can you see a resemblance between this phenomenon and either an elastic or inelastic collision?

Slide 43 / 140 Explosions

In both an inelastic collision and an explosion, kinetic energy is transformed into other forms of energy - such as potential

• energy. But they are time reversed!

An inelastic collision transforms kinetic energy into other forms of energy, such as potential energy. An explosion changes potential energy into kinetic energy. Thus, the equations to predict their motion will be inverted. The next slide summarizes the four types of collisions and explosions.

Slide 44 / 140 Collisions and Explosions

Event Description Momentum Conserved? Kinetic Energy Conserved? General Inelastic Collision Objects bounce

• ff each other

Yes

• No. Kinetic energy is

converted to other forms of energy Perfect Inelastic Collision Objects stick together Yes

• No. Kinetic energy is

converted to other forms of energy Elastic Collision Objects bounce

• ff each other

Yes Yes Explosion Object or

• bjects break

apart Yes

• No. Release of

potential energy increases kinetic energy

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16 Momentum is conserved in which of the following types of collisions? A Perfect Inelastic B Inelastic C Elastic D Explosions E All of the Above

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16 Momentum is conserved in which of the following types of collisions? A Perfect Inelastic B Inelastic C Elastic D Explosions E All of the Above

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Answer E

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17 Kinetic energy is conserved in which of the following types

• f collisions?

A Perfect Inelastic B Inelastic C Elastic D Explosions E All of the Above

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17 Kinetic energy is conserved in which of the following types

• f collisions?

A Perfect Inelastic B Inelastic C Elastic D Explosions E All of the Above

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Answer C

Slide 47 (Answer) / 140 Conservation of Momentum

During a collision or an explosion, measurements show that the total momentum of a closed system does not change. The diagram below shows the objects approaching, colliding and then separating. A B mAvA mBvB A B

A

B

mAvA' mBvB'

+x

the prime means "after"

If the measurements don't show that the momentum is conserved, then this would not be a valid law. Fortunately they do, and it is!

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18 A 13,500 kg railroad freight car travels on a level track at a speed of

4.5 m/s. It collides and couples with a 25,000 kg second car, initially at rest and with brakes released. No external force acts on the

• system. What is the speed of the two cars after colliding?

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18 A 13,500 kg railroad freight car travels on a level track at a speed of

4.5 m/s. It collides and couples with a 25,000 kg second car, initially at rest and with brakes released. No external force acts on the

• system. What is the speed of the two cars after colliding?

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m1v1+m2v2 = m1v1'+m2v2' m1v1+0 = (m1+m2) v' v' = m1v1/(m1+m2) = (13,500kg)(4.5m/s) / (13,500+25,000)kg = 1.6 m/s in the same direction as the first car's initial velocity

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19 A cannon ball with a mass of 100.0 kg flies in horizontal direction

with a speed of 250 m/s and strikes a ship initially at rest. The mass

• f the ship is 15,000 kg. Find the speed of the ship after the ball

becomes embedded in it.

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19 A cannon ball with a mass of 100.0 kg flies in horizontal direction

with a speed of 250 m/s and strikes a ship initially at rest. The mass

• f the ship is 15,000 kg. Find the speed of the ship after the ball

becomes embedded in it.

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m1v1+m2v2 = m1v1'+m2v2' m1v1+0 = (m1+m2) v' v' = m1v1/(m1+m2) = (100kg)(250m/s) / (100+15,000)kg = 1.7 m/s in the same direction as cannon ball's initial velocity

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20 A 40 kg girl skates at 5.5 m/s on ice toward her 70 kg friend who is

standing still, with open arms. As they collide and hold each other, what is their speed after the collision?

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20 A 40 kg girl skates at 5.5 m/s on ice toward her 70 kg friend who is

standing still, with open arms. As they collide and hold each other, what is their speed after the collision?

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m1v1+m2v2 = m1v1'+m2v2' m1v1+0 = (m1+m2) v' v' = m1v1/(m1+m2) = (40kg)(5.5m/s) / (40+70)kg = 2 m/s in same direction as the 40kg girls's initial velocity

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Explosions

In an explosion, one object breaks apart into two or more pieces (or coupled objects break apart), moving afterwards as separate

• bjects.

To make the problems solvable at this math level, we will assume: · the object (or a coupled pair of objects) breaks into two pieces. · the explosion is along the same line as the initial velocity.

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21 A 5 kg cannon ball is loaded into a 300 kg cannon. When the

cannon is fired, it recoils at 5 m/s. What is the cannon ball's velocity after the explosion?

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21 A 5 kg cannon ball is loaded into a 300 kg cannon. When the

cannon is fired, it recoils at 5 m/s. What is the cannon ball's velocity after the explosion?

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Answer

m1 = 5kg m2 = 300kg v1 = v2 = v = 0 v2' = 5m/s (m1+m2) v = m1v1'+m2v2' 0 = m1v1'+m2v2' m1v1' = -m2v2' v1' = -m2v2'/m1

= -(300kg)(5m/s) / (5kg)

= - 300m/s

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22 Two railcars, one with a mass of 4000 kg and the other with a

mass of 6000 kg, are at rest and stuck together. To separate them a small explosive is set off between them. The 4000 kg car is measured travelling at 6 m/s. How fast is the 6000 kg car going?

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22 Two railcars, one with a mass of 4000 kg and the other with a

mass of 6000 kg, are at rest and stuck together. To separate them a small explosive is set off between them. The 4000 kg car is measured travelling at 6 m/s. How fast is the 6000 kg car going?

[This object is a pull tab] (m1+m2) v = m1v1'+m2v2' 0 = m1v1'+m2v2' m1v1' = -m2v2' v1' = -m2v2'/m1 = -(4000kg)(6m/s) / (6000kg) = -4m/s

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Elastic Collisions

In an elastic collision, two objects collide and bounce off each other, as shown below, and both momentum and kinetic energy are conserved. This will give us two simultaneous equations to solve to predict their motion after the collision.

A

B

pA=mAvA Before (moving towards) After (moving apart) pB=mBvB

A

B

pA'=mAvA' pB'=mBvB'

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Elastic Collision Simultaneous Equations

m1v1 + m2v2 = m1v1' +m2v2' m1v1 - m1v1' = m2v2' - m2v2 m1(v1 - v1') = m2(v2' - v2) ½m1v12 + ½m2v22 = ½m1v1'2 +½m2v2'2 m1v12 + m2v22 = m1v1'2 +m2v2'2 m1v12 - m1v1'2 = m2v2'2 - m2v22 m1(v12 - v1'2) = m2(v2'2 - v22) m1(v1 + v1')(v1 - v1') = m2(v2' + v2)(v2' - v2) m1(v1 + v1')(v1 - v1') = m2(v2' + v2)(v2' - v2) m1(v1 - v1') = m2(v2' - v2)

v1 + v1' = v2' + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 - v2 = -(v1' - v'2)

Slide 60 / 140 Properties of Elastic Collisions

By solving the conservation of momentum and constant kinetic energy equations simultaneously, the following result appeared: Do you recognize the terms on the left and right of the above equation? And, what does it mean?

v1 - v2 = -(v1' - v'2)

The terms are the relative velocities of the two objects before and after the collision. It means that for all elastic collisions - regardless of mass - the relative velocity of the objects is the same before and after the collision.

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23 Two objects have an elastic collision. Before they collide, they are

approaching with a velocity of 4 m/s relative to each other. With what velocity do they move apart from one another after the collision?

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23 Two objects have an elastic collision. Before they collide, they are

approaching with a velocity of 4 m/s relative to each other. With what velocity do they move apart from one another after the collision?

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Answer 4 m/s away from each other

v1-v2 = -(v1'-v2') The difference in the initial velocities is the same as the negative difference of the final

• velocities. Or, the relative velocity between the
• bjects before the collision is equal to the

negative of the relative velocity between the

• bjects after the collision.

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24 Two objects have an elastic collision. Object m1, has an initial

velocity of +4.0 m/s and m2 has a velocity of -3.0 m/s. After the collision, m1 has a velocity of 1.0 m/s. What is the velocity of m2?

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24 Two objects have an elastic collision. Object m1, has an initial

velocity of +4.0 m/s and m2 has a velocity of -3.0 m/s. After the collision, m1 has a velocity of 1.0 m/s. What is the velocity of m2?

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Answer

v2' = v1+v1'-v2 v2' = 4m/s + 1m/s - (-3m/s) v2' = 8 m/s v1 = 4 m/s v2 = -3 m/s v1' = 1 m/s v2' = ? v1-v2 = -(v1'-v2')

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25 Two objects have an elastic collision. Object m1, has an initial

velocity of +6.0 m/s and m2 has a velocity of 2.0 m/s. After the collision, m1 has a velocity of 1.0 m/s. What is the velocity of m2?

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25 Two objects have an elastic collision. Object m1, has an initial

velocity of +6.0 m/s and m2 has a velocity of 2.0 m/s. After the collision, m1 has a velocity of 1.0 m/s. What is the velocity of m2?

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Answer

v2' = v1+v1'-v2 v2' = 6m/s + 1m/s - (2m/s) v2' = 5 m/s v1 = 6 m/s v2 = 2 m/s v1' = 1 m/s v2' = ? v1-v2 = -(v1'-v2')

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26 A bowling ball has a velocity of +v when it collides with a ping

pong ball that is at rest. The velocity of the bowling ball is virtually unaffected by the collision. What will be the speed of the ping pong ball?

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26 A bowling ball has a velocity of +v when it collides with a ping

pong ball that is at rest. The velocity of the bowling ball is virtually unaffected by the collision. What will be the speed of the ping pong ball?

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Answer

v2' = v1+v1'-v2 v2' = v + v - 0 v2' = 2 v (ping pong ball's speed is twice that of the bowling ball) v1 = +v v2 = 0 v1' = +v v2' = ? v1-v2 = -(v1'-v2')

Slide 75 (Answer) / 140

27 A baseball bat has a velocity of +v when it collides with a baseball

that has a velocity of -2v. The bat barely changes velocity during the collision. How fast is the baseball going after it's hit?

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27 A baseball bat has a velocity of +v when it collides with a baseball

that has a velocity of -2v. The bat barely changes velocity during the collision. How fast is the baseball going after it's hit?

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Answer

v2' = v1+v1'-v2 v2' = v + v - (-2v) v2' = 4 v v1 = +v v2 = -2v v1' = +v v2' = ? v1-v2 = -(v1'-v2')

Slide 76 (Answer) / 140

28 Two objects with identical masses have an elastic collision: the

initial velocity of m1 is +6.0 m/s and m2 is -3.0 m/s. What is the velocity of m1 after the collision?

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28 Two objects with identical masses have an elastic collision: the

initial velocity of m1 is +6.0 m/s and m2 is -3.0 m/s. What is the velocity of m1 after the collision?

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Answer

When identical mass objects experience an elastic collision, they swap their initial velocities: v1' = v2 = -3.0 m/s v2' = v1 = 6.0 m/s So the velocity of m1 is -3.0 m/s. v1 = +6m/s v2 = -3m/s v1' = ? v2' = ?

Slide 77 (Answer) / 140

29 Two objects with identical masses have an elastic collision: the

initial velocity of m1 is +6.0 m/s and m2 is -3.0 m/s. What is the velocity of m2 after the collision?

Slide 78 / 140

29 Two objects with identical masses have an elastic collision: the

initial velocity of m1 is +6.0 m/s and m2 is -3.0 m/s. What is the velocity of m2 after the collision?

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Answer

When identical mass objects experience an elastic collision, they swap their initial velocities: v1' = v2 = -3.0 m/s v2' = v1 = 6.0 m/s So the velocity of m1 is 6.0 m/s. v1 = +6m/s v2 = -3m/s v1' = ? v2' = ?

Slide 78 (Answer) / 140

30 A golf ball is hit against a solid cement wall, and experiences an elastic collsion. The golf ball strikes the wall with a velocity of +35 m/s. What velocity does it rebound with?

Slide 79 / 140

30 A golf ball is hit against a solid cement wall, and experiences an elastic collsion. The golf ball strikes the wall with a velocity of +35 m/s. What velocity does it rebound with?

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Answer When a light object strikes a very massive object, the light object rebounds with the opposite velocity - in this case, the golf ball will leave the wall with a velocity of -35 m/s.

Slide 79 (Answer) / 140

Collisions in Two Dimensions

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Slide 80 / 140 Conservation of Momentum in Two Dimensions

Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame This means that the momentum conservation equation p = p' can be solved independently for each component: This, of course also applies to three dimensions, but we'll stick with two for this chapter!

Slide 81 / 140

Example: Collision with a Wall

Consider the case of a golf ball colliding elastically with a hard wall, rebounding with the same velocity, where its angle of incidence equals its angle of reflection. Is momentum conserved in this problem? m m θ θ p p' px' py' px py The solid lines represent the momentum of the ball (blue - prior to collision, red - after the collision). The dashed lines are the x and y components of the momentum vectors.

Slide 82 / 140 Example: Collision with a Wall

Momentum is not conserved! An external force from the wall is being applied to the ball in order to reverse its direction in the x axis. However, since we have an elastic collision, the ball bounces off the wall with the same speed that it struck the wall. Hence, the magnitude of the initial momentum and the final momentum is equal: m m θ θ p p' px' py' px py Now it's time to resolve momentum into components along the x and y axis.

Slide 83 / 140 Slide 84 / 140

Slide 85 / 140

31 A tennis ball of mass m strikes a wall at an angle θ relative to

normal then bounces off with the same speed as it had initially. What was the change in momentum of the ball? A -mv B -2mv C -mv cosθ D -2mv cosθ

Slide 86 / 140

32 A tennis ball of mass m strikes a wall an an angle θ relative to normal and then bounces off with the same speed as it had

• initially. What is the change in momentum of the ball?

A 0 B -mv C -2mv D -mv cosθ E -2mv cosθ

m m θ θ p p' px' py' px' py'

Slide 87 / 140

32 A tennis ball of mass m strikes a wall an an angle θ relative to normal and then bounces off with the same speed as it had

• initially. What is the change in momentum of the ball?

A 0 B -mv C -2mv D -mv cosθ E -2mv cosθ

m m θ θ p p' px' py' px' py'

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Answer E

Slide 87 (Answer) / 140

33 A tennis ball of mass m strikes a wall an an angle θ relative to normal and then bounces off with the same speed as it had

• initially. What is the change in momentum of the ball in the

y direction? A 0 B -mv C -2mv D mv E 2mv

m m θ θ p p' px' py' px' py'

Slide 88 / 140

33 A tennis ball of mass m strikes a wall an an angle θ relative to normal and then bounces off with the same speed as it had

• initially. What is the change in momentum of the ball in the

y direction? A 0 B -mv C -2mv D mv E 2mv

m m θ θ p p' px' py' px' py'

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Answer A

Slide 88 (Answer) / 140

General Two Dimensional Collisions

m1 m2

We'll now consider the more general case of two objects moving in random directions in the x-y plane and colliding. Since there is no absolute reference frame, we'll line up the x-axis with the velocity of one of the objects.

Slide 89 / 140 General Two Dimensional Collisions

This is not a head on collision - note how m1 heads off with a y component of velocity after it strikes m2. Also, did you see how we rotated the coordinate system so the x axis is horizontal? To simplify the problem, we will specify that mass 2 is at rest. The problem now is to find the momentum of m2 after the collision.

m1 m2

Before p2 = 0 After

m1 m2

p2 = ?

Slide 90 / 140 General Two Dimensional Collisions

m1 m2

Before p2 = 0 After

m1 m2

p2 = ? This will be done by looking at the vectors first - momentum must be conserved in both the x and the y directions. Since the momentum in the y direction is zero before the collision, it must be zero after the collision. And, the value that m1 has for momentum in the x direction must be shared between both objects after the collision - and not equally - it will depend on the masses and the separation angle.

Slide 91 / 140

General Two Dimensional Collisions

? m1 m2 Here is the momentum vector breakdown of mass 1 after the collision: m2 needs to have a component in the y direction to sum to zero with m1's final y momentum. And it needs a component in the x direction to add to m1's final x momentum to equal the initial x momentum of m1: m2 and this is the final momentum for mass 2 by vectorially adding the final px and py.

Slide 92 / 140

34 After the collision shown below, which of the following is the most

likely momentum vector for the blue ball? A B C D E before after ? m1 m1 m2 m2

Slide 93 / 140

34 After the collision shown below, which of the following is the most

likely momentum vector for the blue ball? A B C D E before after ? m1 m1 m2 m2

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Answer D

Slide 93 (Answer) / 140

General Two Dimensional Collisions

Now that we've seen the vector analysis, let's run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision. There is a bowling ball with momentum 20.0 kg-m/s that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above. What is the final velocity of the pin? before after

60.0° 12.0 kg-m/s

θ

20.0 kg-m/s

m2 m2 m1 m1

Slide 94 / 140 General Two Dimensional Collisions

Given: before after

60.0° 12.0 kg-m/s

θ

20.0 kg-m/s

m2 m2 m1 m1 Find:

Slide 95 / 140 General Two Dimensional Collisions

Use Conservation of Momentum in the x and y directions. x direction y-direction before after

60.0° 12.0 kg-m/s

θ

20.0 kg-m/s

m2 m2 m1 m1

Slide 96 / 140

General Two Dimensional Collisions

Now that the x and y components of the momentum of mass 2 have been found, the total final momentum is calculated. before after

60.0° 12.0 kg-m/s

θ

20.0 kg-m/s

m2 m2 m1 m1

Slide 97 / 140

35 A 5.0 kg bowling ball strikes a stationary bowling pin. After the

collision, the ball and the pin move in directions as shown and the magnitude of the pin's momentum is 18.0 kg-m/s. What was the velocity of the ball before the collision? before after 30°

18 kg-m/s ?

53.1° pin ball

Slide 98 / 140

35 A 5.0 kg bowling ball strikes a stationary bowling pin. After the

collision, the ball and the pin move in directions as shown and the magnitude of the pin's momentum is 18.0 kg-m/s. What was the velocity of the ball before the collision? before after 30°

18 kg-m/s ?

53.1° pin ball

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Answer

before p1x = ? p2x = 0 p1y = 0 p2y = 0 m1 = 5 kg x-direction p1x + p2x = p'1x + p'2x tan(53.1°) = p1y' / p1x' p1x = p'1y/tan(53.1°) + p'2x p1x = 9/tan(53.1°) + 18cos(30°) p1x = 9/1.33 + 15.59 = 22.4 kg-m/s v1x = p1x /m1 = 22.4/5 = 4.48 m/s y-direction p1y + p2y = p'1y + p'2y 0 + 0 = p'1y +18sin(-30°) 0 = p'1y - 9 p'1y = 9 kg-m/s after p'2x = 18cos(30°)kg-m/s p'1x = ? p'2y = 18sin(30°)kg-m/s p'1y = ?

Slide 98 (Answer) / 140

Perfect Inelastic Collisions in Two Dimensions

m1 m2

Before p1 p2 After

m1 m2

p' θ One common kind of inelastic collision is where two cars collide and stick at an intersection. In this situation the two objects are traveling along paths that are perpendicular just prior to the collision.

Slide 99 / 140 Perfect Inelastic Collisions in Two Dimensions

m1 m2

Before p1 p2 After

m1 m2

p' θ p'x p'y p-conservation in x: in y: final momentum: final velocity: final direction:

Slide 100 / 140

36 Object A with mass 20.0 kg travels to the east at 10.0 m/s and

• bject B with mass 5.00 kg travels south at 20.0 m/s. They collide

and stick together. What is the velocity (magnitude and direction)

• f the objects after the collision?

Slide 101 / 140

36 Object A with mass 20.0 kg travels to the east at 10.0 m/s and

• bject B with mass 5.00 kg travels south at 20.0 m/s. They collide

and stick together. What is the velocity (magnitude and direction)

• f the objects after the collision?

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Answer

south of east

Slide 101 (Answer) / 140

Center of Mass

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Slide 102 / 140 Center of Mass

All the problems we've solved have assumed that we're dealing with point particles - this is shown most clearly when free body diagrams were used to find the acceleration of

• bjects due to multiple forces. The object was actually

represented by a point! Other than making the problem simpler (which is good), why did the computed answers actually match what happened to the larger, more extended object? Simple solutions are no good unless they represent reality. To answer this question, the center of mass will be defined, and its relationship to the conservation of momentum will be shown.

Slide 103 / 140

Center of Mass

There is a collection of particles, with different masses, in space, each with an x and y component (x, y) specifying their location. Define the center of mass as having coordinates (xcm, ycm), where:

Slide 104 / 140 Center of Mass

A particle's position is described by a position vector, r: Therefore rcm is expressed as: This is a "weighted average" of each particle - more massive particles contribute more to the coordinates of the center of mass What's next? Think Kinematics.

Slide 105 / 140 Center of Mass

Let's derive the velocity and the acceleration of the center of mass by taking the time derivatives position. First the velocity.

Slide 106 / 140

Slide 107 / 140 Center of Mass

It shows that the total momentum of a system of objects is equal to the total mass of the system times the velocity of its center of mass. The system acts the same as if all of its mass was concentrated at the center of mass. Rewriting: If the momentum of the system is constant (no net external forces acting), then no matter what the individual objects are doing, the velocity of their center of mass remains constant.

Slide 108 / 140 Center of Mass

One more step to finalize the description of the motion of the center

• f mass. Take the derivative of the momentum vector of the center
• f mass with respect to time.

Newton's Second Law This shows that conservation of momentum applies equally to a solid object as well as a system of particles.

Slide 109 / 140

Center of Mass

In a system of objects, both internal and external forces are present. Internal, from the interaction of the objects, and external coming from outside the system, so we write: This enables us to work with solid, macroscopic bodies - the body accelerates as if all of the external forces are acting on the center of mass of all its component parts. Newton's Third Law - the internal forces are action reaction forces and their net effect on the system is zero, so ΣFint = 0.

Slide 110 / 140 Separating Masses

The below photograph shows 3 different fireworks rockets that were sent up into the air and then exploded. The center of mass of each rocket kept moving up, and the particles symmetrically spread out from each one. For each rocket, the velocity of the center of mass is equal to the weighted average of the velocities of each particle, resulting in the starburst pattern. The acceleration of each particle (after the explosion) and the center

• f mass all equal -g.

http://www.flickr.com/photos/stewart/126122/

Slide 111 / 140

37 A wire is bent into the below shape. What are the coordinates for the center of mass of the wire?

Slide 112 / 140

37 A wire is bent into the below shape. What are the coordinates for the center of mass of the wire?

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Answer

Slide 112 (Answer) / 140

38 A missile is launched with velocity v0, and explodes mid flight into over 1000 fragments. What is the velocity of the center of mass of the system after the explosion? A 0 B v0 C 2v0 D -v0 E -2v0

Slide 113 / 140

38 A missile is launched with velocity v0, and explodes mid flight into over 1000 fragments. What is the velocity of the center of mass of the system after the explosion? A 0 B v0 C 2v0 D -v0 E -2v0

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Answer B

Slide 113 (Answer) / 140

Explosions in Two Dimensions

p'3 p'2 p'1 The Black object explodes into 3 pieces (blue, red and green). We want to determine the momentum of the third piece. During an explosion, the total momentum is unchanged, since no EXTERNAL force acts on the system. · By Newton's Third Law, the forces that occur between the particles within the object will add up to zero, so they don't affect the momentum. · If the initial momentum is zero, the final momentum is zero. · The third piece must have equal and

• pposite momentum to the sum of

the other two. Move the dashed box to see the third piece's momentum.

Slide 114 / 140 Explosions in Two Dimensions

#

p'1 p'2 p'3 The Black object explodes into 3 pieces (blue, red and green). We want to determine the momentum of the third piece. before: px = py = 0 after: p'1x + p'2x + p'3x = 0 p'1y + p'2y + p'3y = 0 In this case the blue and red pieces are moving perpendicularly to each other, so:

Slide 115 / 140

39 A stationary cannon ball explodes into three pieces. The momenta of two of the pieces is shown below. What is the direction of the momentum of the third piece? A B C D E

Slide 116 / 140

39 A stationary cannon ball explodes into three pieces. The momenta of two of the pieces is shown below. What is the direction of the momentum of the third piece? A B C D E

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Answer C

Slide 116 (Answer) / 140

40 A stationary 10.0 kg bomb explodes into three pieces. A

2.00 kg piece moves west at 200.0 m/s. Another piece with a mass of 3.00 kg moves north with a velocity of 100.0 m/s. What is the velocity (speed and direction) of the third piece?

Slide 117 / 140

40 A stationary 10.0 kg bomb explodes into three pieces. A

2.00 kg piece moves west at 200.0 m/s. Another piece with a mass of 3.00 kg moves north with a velocity of 100.0 m/s. What is the velocity (speed and direction) of the third piece?

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Answer

Slide 117 (Answer) / 140

It is Rocket Science

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Slide 118 / 140 It is Rocket Science

http://www.nytimes.com/2001/11/14/news/150th- anniversary-1851-2001-the-facts-that-got-away.html

JULY 17, 1969: On Jan. 13, 1920, Topics of The Times, an editorial-page feature of The New Yo Times, dismissed the notion that a rocket could function in a vacuum and commented on the idea

• f Robert H. Goddard, the rocket pioneer, as follows: ''That Professor Goddard, with his 'chair' in

Clark College and the countenancing of the Smithsonian Institution, does not know the relation of action to reaction, and of the need to have something better than a vacuum against which to reac to say that would be absurd. Of course he only seems to lack the knowledge ladled out daily in hi schools.'' Further investigation and experimentation have confirmed the findings of Isaac Newton in the 17t century and it is now definitely established that a rocket can function in a vacuum as well as in an

• atmosphere. The Times regrets the error.

On July 20, 1969, the American Apollo 11 spacecraft landed on the moon, and six hours later, Neil Armstrong became the first human to walk on the moon. While the spacecraft was enroute to the moon, the New York Times published the following article:

Slide 119 / 140 It is Rocket Science

The 1920 article referenced Newton's Third Law, but how did they misinterpret it? And how is Newton's Third Law related to the Conservation of Momentum?

Slide 120 / 140

It is Rocket Science

A rocket operates by expelling burned fuel through its tail. The fuel has both mass and velocity, so it has momentum. Before launch, the rocket was at rest, so its momentum was zero. Define the system as the rocket and its fuel (the onboard and the expelled fuel), and since there are no external forces acting on it, momentum must be conserved. The rocket is launched by burning the fuel and expelling it. It then has a momentum opposite the expelled fuel and moves forward to keep the total momentum of the system equal to zero.

Slide 121 / 140 It is Rocket Science

The action - reaction forces described by Newton's Third Law and the free body diagrams from his Second Law are: · The rocket exerts a force on the burned fuel as it expels it out the tail. · The burned fuel exerts a force on the rocket ship as it leaves. Rocket Burned fuel And that's how Newton's Third Law relates to the Conservation of Momentum!

Slide 122 / 140 It is Rocket Science

This is a slightly more complex problem than the skaters pushing

• ff of each other or an archer shooting an arrow. In those cases,

the masses of each object remains the same (the arrow's mass is negligible compared to the archer) .

Slide 123 / 140

It is Rocket Science

For the more general case, assume the rocket is already moving at a constant veloicty and has a certain momentum. It then lights its engines and starts burning fuel. There is no external force so the momentum right before the burn must equal the momentum at any time. The rocket and the burned fuel both increase momentum but the momentum at any time during the fuel burn is constant.

Slide 124 / 140 Slide 125 / 140 It is Rocket Science

Some algebraic manipulation: Take the limit of Δv and Δmef as Δt approaches zero: As the mass of the expelled fuel increases, the mass of the rocket plus unburned fuel decreases by the same amount .

Slide 126 / 140

It is Rocket Science

Integrate from v0 to vf and from Mr0 (initial mass of rocket with fuel) to Mrf (final mass of rocket with remaining, unburned fuel). v ef is the velocity of the expelled fuel which is kept constant by the pilot: Rocket equation

Slide 127 / 140

41 A missile is launched from rest. The initial mass of the missile and its fuel is 125 kg. The fuel's exhaust velocity is 2500 m/s. How much fuel is used to accelerate the rocket to a speed of 926 m/s? A 7.8 kg B 20 kg C 39 kg D 50 kg E 86 kg

Slide 128 / 140

41 A missile is launched from rest. The initial mass of the missile and its fuel is 125 kg. The fuel's exhaust velocity is 2500 m/s. How much fuel is used to accelerate the rocket to a speed of 926 m/s? A 7.8 kg B 20 kg C 39 kg D 50 kg E 86 kg

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Answer C

Slide 128 (Answer) / 140

Ballistic Pendulum

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Slide 129 / 140 Ballistic Pendulum

Before they were rendered obsolete by modern sensors, Ballistic Pendulums were used to find the velocities of projectiles (bullets) . But, they still are used in Physics Labs, where a combination of Conservation of Momentum, and Conservation of Total Mechanical Energy can be used to find the approximate velocity of a projectile.

M l vM = 0 m v M+m l v' h

A more exact solution requires the use of rotational dynamics to account for the moment of inertia of the block.

Slide 130 / 140 Ballistic Pendulum

A bullet of mass m is fired into a block of mass M and remains embedded in the block. The block is attached to a stand with a fixed length, l, of string/wire and moves as shown below. What type

• f collision is this? What is conserved?

M l vM = 0 m v M+m l v' h

Slide 131 / 140

Ballistic Pendulum

This is a perfectly inelastic collision, so linear momentum is conserved, but kinetic energy is not conserved - some of the bullet's energy goes into non conservative forces, such as friction and material deformation of the block and bullet. After the collision, assume that there are zero non conservative external forces acting on the system. What conservation law can now be used?

M l vM = 0 m v M+m l v' h

Slide 132 / 140 Ballistic Pendulum

Conservation of Total Mechanical Energy. After the collision, the bullet-block system rises a height h, above its initial position, and the string makes an angle θ with the vertical. That is shown in the picture on the right where one of the strings has been removed to see the angle θ more clearly.

M l vM = 0 m v M+m v' l h

θ

Slide 133 / 140 Ballistic Pendulum

A little trigonometry to find the change in gravitational potential energy due to the block rising after the bullet's impact.

M+m v' l h

θ lcosθ l - lcosθ

Slide 134 / 140

Slide 135 / 140 Ballistic Pendulum

M+m v' l h

θ

lcosθ l - lcosθ

Substitute the first into the second equation:

M l vM = 0 m v

Slide 136 / 140 Slide 137 / 140

42 A 13 g bullet is fired at 460 m/s into a stationary 1.0 kg block and embeds within the block (completely inelastic collision). Find the velocity of the bullet - block system. A 3.2 m/s B 4.7 m/s C 5.0 m/s D 5.9 m/s E 6.3 m/s

v

Slide 138 / 140

42 A 13 g bullet is fired at 460 m/s into a stationary 1.0 kg block and embeds within the block (completely inelastic collision). Find the velocity of the bullet - block system. A 3.2 m/s B 4.7 m/s C 5.0 m/s D 5.9 m/s E 6.3 m/s

v

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Answer D

Slide 138 (Answer) / 140

43 A 13 g bullet is fired at 460 m/s into a stationary 1.0 kg block attached to a 2.0 m long string and embeds within the block (completely inelastic collision). Find the maximum height that the bullet - block system moves to. A 1.0 m B 1.5 m C 1.8 m D 2.2 m E 2.3 m

v Before After 2 m

h

Slide 139 / 140

43 A 13 g bullet is fired at 460 m/s into a stationary 1.0 kg block attached to a 2.0 m long string and embeds within the block (completely inelastic collision). Find the maximum height that the bullet - block system moves to. A 1.0 m B 1.5 m C 1.8 m D 2.2 m E 2.3 m

v Before After 2 m

h

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Answer C

Slide 139 (Answer) / 140

44 A 13 g bullet is fired at 460 m/s into a stationary 1.0 kg block attached to a 2.0 m long string and embeds within the block (completely inelastic collision). Find the angle θ that is subtended by the string at the maximum height of the bullet - block system. A 570 B 620 C 650 D 770 E 840

2 m

h

Slide 140 / 140

44 A 13 g bullet is fired at 460 m/s into a stationary 1.0 kg block attached to a 2.0 m long string and embeds within the block (completely inelastic collision). Find the angle θ that is subtended by the string at the maximum height of the bullet - block system. A 570 B 620 C 650 D 770 E 840

2 m

h

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Answer E

Slide 140 (Answer) / 140