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Momentum

Slide 3 / 133 Table of Contents

· Momentum · Impulse-Momentum Equation · The Momentum of a System of Objects · Types of Collisions

Click on the topic to go to that section

· Conservation of Momentum · Collisions in Two Dimensions

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Return to Table of Contents

Momentum

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Newton’s First Law tells us that objects remain in motion with a constant velocity unless acted upon by an external force (Law of Inertia). In our experience: · When objects of different mass travel with the same velocity, the

• ne with more mass is harder to stop.

· When objects of equal mass travel with different velocities, the faster one is harder to stop.

Momentum Defined Slide 6 / 133

Define a new quantity, momentum (p), that takes these

• bservations into account:

momentum = mass × velocity

Momentum Defined p =mv

click here for a introductory video on momentum from Bill Nye!

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Since: · mass is a scalar quantity · velocity is a vector quantity · the product of a scalar and a vector is a vector and: · momentum = mass × velocity therefore: Momentum is a vector quantity - it has magnitude and direction

Momentum is a Vector Quantity Slide 8 / 133

There is no specially named unit for momentum - so there is an

• pportunity for it to be named after a renowned physicist!

We use the product of the units of mass and velocity.

mass x velocity

kg⋅m/s SI Unit for Momentum Slide 9 / 133

1 Which has more momentum? A A large truck moving at 30 m/s B A small car moving at 30 m/s C Both have the same momentum.

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1 Which has more momentum? A A large truck moving at 30 m/s B A small car moving at 30 m/s C Both have the same momentum.

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Answer

A

Both have the same speed, but the truck has the larger mass

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2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 m/s?

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2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 m/s?

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Answer m= 20 kg v = 5 m/s p = mv = (20 kg)(5 m/s) = 100 kg⋅m/s

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3 What is the momentum of a 20 kg object with a velocity of 5.0 m/s to the left?

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3 What is the momentum of a 20 kg object with a velocity of 5.0 m/s to the left?

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Answer m= 20 kg v = −5 m/s p = mv = (20 kg)(−5 m/s) = −100 kg- m/s

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4 What is the velocity of a 5 kg object whose momentum is −15 kg-m/s?

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4 What is the velocity of a 5 kg object whose momentum is −15 kg-m/s?

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Answer m= 5 kg p = −15 m/s p = mv v = p/m v= (-15 kg-m/s)/5 kg v = -3 m/s

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5 What is the mass of an object that has a momentum of 35 kg-m/s with a velocity of 7 m/s?

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5 What is the mass of an object that has a momentum of 35 kg-m/s with a velocity of 7 m/s?

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Answer v= 7 m/s p = 35 m/s p = mv m = p/v m= (35 kg-m/s)/(7 m/s) m = 5 kg

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Impulse-Momentum Equation

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Suppose that there is an event that changes an object's momentum. · from p0 - the initial momentum (just before the event) · by Δp - the change in momentum · to pf - the final momentum (just after the event) The equation for momentum change is:

Change in Momentum Slide 16 / 133

Momentum change equation: Newton's First Law tells us that the velocity (and so the momentum)

• f an object won't change unless the object is affected by an external

force.

Momentum Change = Impulse

Look at the above equation. Can you relate Newton's First Law to the Δp term? Δp would represent the external force.

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There no special unit for impulse. We use the product of the units of force and time.

force x time

N⋅s

Recall that N=kg⋅m/s2, so

N⋅s=kg⋅m/s2 x s = kg⋅m/s SI Unit for Impulse

This is also the unit for momentum, which is a good thing since Impulse is the change in momentum.

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6 There is a battery powered wheeled cart moving towards you at a constant velocity. You want to apply a force to the cart to move it in the opposite direction. Which combination of the following variables will result in the greatest change of momentum for the cart? Select two answers. A Increase the applied force. B Increase the time that the force is applied. C Maintain the same applied force. D Decrease the time that the force is applied.

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6 There is a battery powered wheeled cart moving towards you at a constant velocity. You want to apply a force to the cart to move it in the opposite direction. Which combination of the following variables will result in the greatest change of momentum for the cart? Select two answers. A Increase the applied force. B Increase the time that the force is applied. C Maintain the same applied force. D Decrease the time that the force is applied.

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Answer A, B

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7 From which law or principle is the Impulse-Momentum equation derived from? A Conservation of Energy. B Newton's First Law. C Newton's Second Law. D Conservation of Momentum.

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7 From which law or principle is the Impulse-Momentum equation derived from? A Conservation of Energy. B Newton's First Law. C Newton's Second Law. D Conservation of Momentum.

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Answer C

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8 Can the impulse applied to an object be negative? Why or why not? Give an example to explain your answer.

Students type their answers here

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8 Can the impulse applied to an object be negative? Why or why not? Give an example to explain your answer.

Students type their answers here

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Answer Yes, since Impulse is a vector quantity, it has magnitude and

• direction. Direction can be described

as negative. An example would be if a baseball is thrown in the positive direction, and you are catching it with a mitt. The mitt changes the baseball's momentum from positive to zero - hence applying a negative impulse to it.

Slide 23 / 133 Effect of Collision Time on Force

∆t (seconds) F (newtons)

Since force is inversely proportional to Δt, changing the ∆t of a given impulse by a small amount can greatly change the force exerted on an object!

Impulse = F(#t) = F(#t)

Slide 24 / 133 Every Day Applications

Impulse = F(#t) = F(#t) The inverse relationship of Force and time interval leads to many interesting applications of the Impulse-Momentum equation to everyday experiences such as: · car structural safety design · car air bags · landing after parachuting · martial arts · hitting a baseball · catching a basebal

Slide 25 / 133 Every Day Applications

I=F# t=# p

Let's analyze two specific cases from the previous list: · car air bags · hitting a baseball Whenever you have an equation like the one above, you have to decide which values will be fixed and which will be varied to determine the impact on the third value. For the car air bags, we'll fix Δp, vary Δt and see its impact on

• F. For the bat hitting a ball, we'll fix F, vary

Δt and see the impact on Δp.

Slide 26 / 133 Car Air Bags

I=F# t=# p

In the Dynamics unit of this course, it was shown how during an accident, seat belts protect passengers from the effect of Newton's First Law by stopping the passenger with the car, and preventing them from striking the dashboard and window. They also provide another benefit explained by the Impulse- Momentum equation. But, this benefit is greatly enhanced by the presence of air bags. Can you see what this benefit is?

Slide 27 / 133 Car Air Bags

I=F# t=# p

The seat belt also increases the time interval that it takes the passenger to slow down - there is some play in the seat belt, that allows you to move forward a bit before it stops you. The Air bag will increase that time interval much more than the seat belt by rapidly expanding, letting the passenger strike it, then deflating. Earlier it was stated that for the Air bag example, Δp would be fixed and Δt would be varied. So, we've just increased Δt. Why is Δp fixed?

Slide 28 / 133 Car Air Bags

I=F# t=# p

Δp is fixed, because as long as the passenger remains in the car, the car (and the passengers) started with a certain velocity and finished with a final velocity of zero, independent of seat belts or air bags. Rearranging the equation, we have: F represents the Force delivered to the passenger due to the accident.

Slide 29 / 133 Car Air Bags

I=F# t=# p

Since Δp is fixed, by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag), the force, F delivered to the passenger is smaller. Less force on the passenger means less physical harm. Also, another benefit needs a quick discussion of Pressure. Pressure is Force per unit area. By increasing the area of the body that feels the force (the air bag is big), less pressure is delivered to parts of the body - reducing the chance of puncturing the body. Also good.

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9 If a car is in a front end collision, which of the below factors will help reduce the injury to the driver and passengers? Select two answers. A An absolutely rigid car body that doesn't deform. B Deployment of an air bag for each adult in the car. C Deployment of an air bag only for the driver. D The proper wearing of a seatbelt or child seat for each person in the car.

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9 If a car is in a front end collision, which of the below factors will help reduce the injury to the driver and passengers? Select two answers. A An absolutely rigid car body that doesn't deform. B Deployment of an air bag for each adult in the car. C Deployment of an air bag only for the driver. D The proper wearing of a seatbelt or child seat for each person in the car.

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Answer B, D

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10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest? A Force delivered to the car. B Interval of time for the collision. C Momentum change. D Acceleration of the car.

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10 Which of the following variables are fixed when a moving car strikes a barrier and comes to rest? A Force delivered to the car. B Interval of time for the collision. C Momentum change. D Acceleration of the car.

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Answer C

Slide 32 / 133 Hitting a Baseball

I=F# t=# p

Now, we're going to take a case where there is a given force, and the time interval will be varied to find out what happens to the change of momentum. This is different from the Air Bag example just worked (Δp was constant, Δt was varied, and its impact on F was found). Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat. The ball is approaching the batter with a positive momentum. What is the goal of the batter?

Slide 33 / 133 Hitting a Baseball

I=F# t=# p

The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing. The greater the Δp, the faster the ball will fly off his bat, which will result in it going further, hopefully to the seats. Hitting a baseball is way more complex than the analysis that will follow. If you're interested in more information, please check out the book, The Physics of Baseball, written by Robert Adair, a Yale University physicist.

Slide 34 / 133 Hitting a Baseball

I=F# t=# p

In this case, the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball. This means he should follow through with his swing. The batter needs to apply a large impulse to reverse the ball's large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers).

Slide 35 / 133 Every Day Applications

I=F# t=# p

Now, discuss the other examples. Make sure you decide which

• bject in the collision is more "affected" by the force or the

change in momentum, and which variables are capable of being varied. Consider the Air Bag example - the car experiences the same impulse as the passenger during an accident, but a car is less valuable than a human being - so it is more important for the passenger that less force is delivered to his body - and more force is absorbed by the car. · car structural safety design · landing after parachuting · martial arts · catching a basebal

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11 An external force of 25 N acts on a system for 10.0 s. What is the

magnitude of the impulse delivered to the system?

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11 An external force of 25 N acts on a system for 10.0 s. What is the

magnitude of the impulse delivered to the system?

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Answer F = 25 N Δt = 10 s I = F Δt = (25N) (10 s) = 250 N•s

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12 In the previous problem, an external force of 25 N acted on a

system for 10.0 s. The impulse delivered was 250 N-s. What was the change in momentum of the system?

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12 In the previous problem, an external force of 25 N acted on a

system for 10.0 s. The impulse delivered was 250 N-s. What was the change in momentum of the system?

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Answer I = 250 N•s I = Δp Δp = 250 N•s

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13 The momentum change of an object is equal to the

______. A force acting on it B impulse acting on it C velocity change of the object D object's mass times the force acting on it

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13 The momentum change of an object is equal to the

______. A force acting on it B impulse acting on it C velocity change of the object D object's mass times the force acting on it

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Answer

B Slide 39 / 133

14 Air bags are used in cars because they:

A increase the force with which a passenger hits the dashboard B increase the duration (time) of the passenger's impact C decrease the momentum of a collision D decrease the impulse in a collision

B

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14 Air bags are used in cars because they:

A increase the force with which a passenger hits the dashboard B increase the duration (time) of the passenger's impact C decrease the momentum of a collision D decrease the impulse in a collision

B

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Answer

B

By increasing the amount of time during the collision, the force required to reduce the passenger's momentum is reduced. This in turn reduces or prevents injury.

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15 One car crashes into a concrete barrier. Another car crashes into a

collapsible barrier at the same speed. What is the difference between the two crashes? Select two answers. A change in momentum B force on the car C impact time D final momentum

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15 One car crashes into a concrete barrier. Another car crashes into a

collapsible barrier at the same speed. What is the difference between the two crashes? Select two answers. A change in momentum B force on the car C impact time D final momentum

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Answer

B, C Slide 41 / 133

16 In order to increase the final momentum of a golf ball, the golfer should: Select two answers: A maintain the speed of the golf club after the impact (follow through). B Hit the ball with a greater force. C Decrease the time of contact between the club and the ball. D Decrease the initial momentum of the golf club.

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16 In order to increase the final momentum of a golf ball, the golfer should: Select two answers: A maintain the speed of the golf club after the impact (follow through). B Hit the ball with a greater force. C Decrease the time of contact between the club and the ball. D Decrease the initial momentum of the golf club.

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Answer

A, B

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17 An external force acts on an object for 0.0020 s. During that time

the object's momentum increases by 400 kg-m/s. What was the magnitude of the force?

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17 An external force acts on an object for 0.0020 s. During that time

the object's momentum increases by 400 kg-m/s. What was the magnitude of the force?

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Answer

Δt = 0.002 s Δp = 400 kg•m/s I = FΔt = Δp F = Δp/Δt = (400 kg•m/s)/(0.002 s) = 200,000 N

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18 A 50,000 N force acts for 0.030 s on a 2.5 kg object that was

initially at rest. What is its final velocity?

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18 A 50,000 N force acts for 0.030 s on a 2.5 kg object that was

initially at rest. What is its final velocity?

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Answer

F = 50,000 N Δt = 0.03 s m = 2.5 kg v0 = 0 FΔt = Δp ; Δp = mΔv FΔt = mΔv ; Δv = (vf-v0) FΔt = m(vf-v0) = mvf (since v0 = 0) vf = (F/m) Δt = (50,000 N)/(2.5 kg) x (0.03 s) = 600 m/s

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19 A 1200 kg car slows from 40.0 m/s to 5.0 m/s in 2.0 s. What

force was applied by the brakes to slow the car?

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19 A 1200 kg car slows from 40.0 m/s to 5.0 m/s in 2.0 s. What

force was applied by the brakes to slow the car?

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Answer

Δt = 2 s m = 1200 kg v0 = 40 m/s vf = 5 m/s FΔt = Δp ; Δp = mΔv FΔt = m(vf-v0) F = m(vf-v0) /Δt = (1200 kg)(5 m/s -40 m/s)/(2 s) = 21,000 N

Slide 45 / 133 Graphical Interpretation of Impulse

Graphs are a very valuable method to solve physics problems. So far, we've dealt with a constant force exerted over a given time

• interval. But forces are not always constant - most of the time,

they are changing over time. In the baseball example, the force of the bat starts out very small as it makes initial contact with the ball. It then rapidly rises to its maximum value, then decreases as the ball leaves the bat. This would be a very difficult problem to handle without a graph. Start with representing a constant force over a time interval, Δt, and plot Force vs. time on a graph.

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20 Using the F-t graph shown, what is the change in momentum

during the time interval from 0 to 6 s?

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20 Using the F-t graph shown, what is the change in momentum

during the time interval from 0 to 6 s?

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Answer

∆p = area under the graph from 0 to 6 seconds. = area of the triangle from 0 to 3 seconds + area of the rectangle from 3 to 6 seconds = ½ (30 N) (3 s) + (30 N)(3 s) =135 N•s or 135 kg•m/s

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21 A 5 kg object with an initial velocity of 3 m/s experiences the force

shown in the graph. What is its velocity at 6 s?

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21 A 5 kg object with an initial velocity of 3 m/s experiences the force

shown in the graph. What is its velocity at 6 s?

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Answer

From the last question ∆p = 135 kg•m/s m = 5 kg, v0 = 3 m/s p0 = mv0 = (5 kg)(3 m/s) =15 kg•m/s pf = p0 + ∆p = (15 +135) kg•m/s = 150 kg•m/s pf = mvf or vf=pf/m = (150 kg•m/s) / (5kg) = 30 m/s

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22 Using the F-t graph shown, what is the change in momentum

during the time interval from 6 to 10 s?

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22 Using the F-t graph shown, what is the change in momentum

during the time interval from 6 to 10 s?

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Answer

∆p = area under the graph from 6-10 s = area of the triangle from 6 to 7 s + area of the rectangle from 6 to 7 s + area of the rectangle from 7 to 10 s = ½ (20 N) (1 s)+(10 N)(1 s)+(10 N)(3 s) =50 N•s or 50 kg•m/s

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The Momentum of a System of Objects

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If a system contains more than one object, its total momentum is the vector sum of the momenta of those

• bjects.

psystem = # p psystem = p1 + p2 + p3 +... psystem = m1v1 + m2v2 + m3v3 +...

The Momentum of a System of Objects

It's critically important to note that momenta add as vectors, not as scalars.

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psystem = m1v1 + m2v2 + m3v3 +...

In order to determine the total momentum of a system: First: · Determine a direction to be considered positive. · Assign positive values to momenta in that direction. · Assign negative values to momenta in the opposite direction.

The Momentum of a System of Objects

Then: Add the momenta to get a total momentum of the system.

+

• Slide 54 / 133

Determine the momentum of a system of two objects: m1, has a mass

• f 6 kg and a velocity of 13 m/s towards the east and m2, has a mass
• f 14 kg and a velocity of 7 m/s towards the west.

psystem = p1 + p2 psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kg v1 = 13 m/s v2 = −7 m/s

6 kg 14 kg 13 m/s 7 m/s East (+)

psystem = m1v1 + m2v2 = (6 kg)(13 m/s) + (14 kg)(−7 m/s) = (78 kg⋅m/s) + (−98 kg⋅m/s) = −20 kg⋅m/s

Slide 54 (Answer) / 133

Determine the momentum of a system of two objects: m1, has a mass

• f 6 kg and a velocity of 13 m/s towards the east and m2, has a mass
• f 14 kg and a velocity of 7 m/s towards the west.

psystem = p1 + p2 psystem = m1v1 + m2v2

Example

m1 = 6 kg m2 = 14 kg v1 = 13 m/s v2 = −7 m/s

6 kg 14 kg 13 m/s 7 m/s East (+)

psystem = m1v1 + m2v2 = (6 kg)(13 m/s) + (14 kg)(−7 m/s) = (78 kg⋅m/s) + (−98 kg⋅m/s) = −20 kg⋅m/s

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23 Determine the momentum of a system of two objects: m1, has a

mass of 6.0 kg and a velocity of 20 m/s north and m2, has a mass

• f 3 kg and a velocity 20 m/s south.

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23 Determine the momentum of a system of two objects: m1, has a

mass of 6.0 kg and a velocity of 20 m/s north and m2, has a mass

• f 3 kg and a velocity 20 m/s south.

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psystem = m1v1 + m2v2 = (6 kg)(20 m/s) + (3 kg)(−20 m/s) = (120 kg⋅m/s) + (−60 kg⋅m/s) = 60 kg⋅m/s (magnitude) direction is North m1 = 6 kg v1 = +20 m/s (North) m2 = 3 kg v2 = −20 m/s (South)

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24 Determine the momentum of a system of two objects: the first has

a mass of 8.0 kg and a velocity of 8.0 m/s to the east while the second has a mass of 5.0 kg and a velocity of 15 m/s to the west.

Slide 56 (Answer) / 133

24 Determine the momentum of a system of two objects: the first has

a mass of 8.0 kg and a velocity of 8.0 m/s to the east while the second has a mass of 5.0 kg and a velocity of 15 m/s to the west.

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psystem = m1v1 + m2v2 = (8 kg)(8 m/s) + (5 kg)(−15 m/s) = (64 kg⋅m/s) + (−75 kg⋅m/s) = −11 kg⋅m/s (West) m1 = 8 kg v1 = +8 m/s (East) m2 = 5 kg v2 = −15 m/s (West)

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25 Determine the momentum of a system of three objects:

The first has a mass of 7.0 kg and a velocity of 23 m/s north; the second has a mass of 9.0 kg and a velocity of 7 m/s north; and the third has a mass of 5.0 kg and a velocity of 42 m/s south.

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25 Determine the momentum of a system of three objects:

The first has a mass of 7.0 kg and a velocity of 23 m/s north; the second has a mass of 9.0 kg and a velocity of 7 m/s north; and the third has a mass of 5.0 kg and a velocity of 42 m/s south.

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Conservation of Momentum

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Slide 59 / 133

Some of the most powerful concepts in science are called "conservation laws". This was covered in detail in the Energy unit - please refer back to that for more detail. Here is a summary. Conservation laws: · apply to closed systems - where the objects only interact with each other and nothing else. · enable us to solve problems without worrying about the details

• f an event.

Conservation Laws Slide 60 / 133

In the last unit we learned that energy is conserved. Like energy, momentum is a conserved property of nature. This means: · Momentum is not created or destroyed. · The total momentum in a closed system is always the same. · The only way the momentum of a system can change is if momentum is added or taken away by an outside force.

Momentum is Conserved Slide 61 / 133

We will use the Conservation of Momentum to explain and predict the motion of a system of objects. As with energy, it will only be necessary to compare the system at two times: just before and just after an event.

Conservation of Momentum Slide 62 / 133 Slide 63 / 133 Slide 64 / 133

26 Why don't the internal forces on a system change the momentum

• f the system?

Students type their answers here

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26 Why don't the internal forces on a system change the momentum

• f the system?

Students type their answers here

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Answer The internal forces result from the interaction of the objects within the

• system. By Newton's Third Law,

these are all action-reaction pairs, and viewed from outside the system, all cancel out. Thus there is no net force acting on the system and the momentum is conserved.

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27 An external, positive force acts on a system of objects. Which of

the following are true? Select two answers. A The velocity of the system remains the same. B The velocity of the system increases. C The momentum of the system decreases. D The momentum of the system increases.

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27 An external, positive force acts on a system of objects. Which of

the following are true? Select two answers. A The velocity of the system remains the same. B The velocity of the system increases. C The momentum of the system decreases. D The momentum of the system increases.

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Answer

B, D Slide 66 / 133

Types of Collisions

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Slide 67 / 133 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.

Slide 68 / 133 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 69 / 133 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 70 / 133 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.

Slide 71 / 133 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 72 / 133 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 73 / 133 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

Slide 74 / 133

28 In _______ collisions momentum is conserved.

A Elastic B Inelastic C All

Slide 74 (Answer) / 133

28 In _______ collisions momentum is conserved.

A Elastic B Inelastic C All

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Answer

C Slide 75 / 133

29 In ______ collisions kinetic energy is conserved.

A Elastic B Inelastic C All

Slide 75 (Answer) / 133

29 In ______ collisions kinetic energy is conserved.

A Elastic B Inelastic C All

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Answer

A Slide 76 / 133

30 In an inelastic collision, kinetic energy is transformed into what

after the collision? Select two answers. A Nothing, kinetic energy is conserved. B More kinetic energy. C Thermal energy. D Light energy.

Slide 76 (Answer) / 133

30 In an inelastic collision, kinetic energy is transformed into what

after the collision? Select two answers. A Nothing, kinetic energy is conserved. B More kinetic energy. C Thermal energy. D Light energy.

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Answer

C, D Slide 77 / 133 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!

Slide 78 / 133 Slide 79 / 133

31 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?

Slide 79 (Answer) / 133

31 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

Slide 80 / 133

32 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.

Slide 80 (Answer) / 133

32 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

Slide 81 / 133

33 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?

Slide 81 (Answer) / 133

33 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

Slide 82 / 133 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.

Slide 83 / 133 Slide 84 / 133

34 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?

Slide 84 (Answer) / 133

34 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

Slide 85 / 133

35 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?

Slide 85 (Answer) / 133

35 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

Slide 86 / 133 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'

Slide 87 / 133 Slide 88 / 133 Slide 89 / 133 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 90 / 133 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.

Slide 91 / 133

36 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?

Slide 91 (Answer) / 133

36 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.

Slide 92 / 133

37 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?

Slide 92 (Answer) / 133

37 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')

Slide 93 / 133

38 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?

Slide 93 (Answer) / 133

38 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')

Slide 94 / 133 Slide 95 / 133 Slide 96 / 133 Slide 97 / 133 Slide 98 / 133 Slide 99 / 133

Slide 100 / 133 Slide 101 / 133 Slide 102 / 133 Slide 103 / 133 Slide 104 / 133

39 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?

Slide 104 (Answer) / 133

39 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 105 / 133

40 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?

Slide 105 (Answer) / 133

40 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 106 / 133

41 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?

Slide 106 (Answer) / 133

41 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 107 / 133

42 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 107 (Answer) / 133

42 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 108 / 133

43 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 108 (Answer) / 133

43 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 109 / 133

Collisions in Two Dimensions

Return to Table of Contents

Slide 110 / 133 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 111 / 133 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 112 / 133 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 113 / 133 Slide 114 / 133 Slide 115 / 133

44 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 m m θ θ p p' px' py' px' py'

Slide 115 (Answer) / 133

44 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 m m θ θ p p' px' py' px' py'

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Answer D

Slide 116 / 133

45 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 in the y direction? A -mv B 0 C mv D 2mv m m θ θ p p' px' py' px' py'

Slide 116 (Answer) / 133

45 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 in the y direction? A -mv B 0 C mv D 2mv m m θ θ p p' px' py' px' py'

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Answer B

Slide 117 / 133 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 118 / 133 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 reduce the amount of variables that must be specified, we will also assume 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 119 / 133 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 120 / 133 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 121 / 133

46 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 121 (Answer) / 133

46 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 122 / 133 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 123 / 133 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 124 / 133 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 125 / 133 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 126 / 133

47 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 126 (Answer) / 133

47 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 127 / 133 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 128 / 133 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 129 / 133

48 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 129 (Answer) / 133

48 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 130 / 133

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.

Slide 131 / 133 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 132 / 133

49 A stationary cannon ball explodes in three pieces. The momenta

• f two of the pieces is shown below. What is the direction of the

momentum of the third piece? A B C D Answer

Slide 133 / 133

50 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 133 (Answer) / 133

50 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