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Momentum

Slide 2 / 90 Momentum

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· Momentum · Conservation of Momentum · The Momentum of a System of Objects · Momentum Change and Impulse · Conservation of Momentum in Collisions and Explosions · Collisions in Two and Three Dimensions

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Momentum

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

• ne with most mass is hardest to stop.

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

Momentum Defined Slide 5 / 90

· 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 and: · momentum = mass × velocity Momentum is a vector quantity

Momentum is a Vector Quantity Slide 7 / 90

There no specially named units for momentum. We just use the product of the units of mass and velocity...

mass x velocity

kg⋅m/s SI Unit for Momentum Slide 8 / 90

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. Answer

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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 5.0 m/s? Answer

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2 What is the momentum of a 20 kg object moving to the right with a velocity of 5.0 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? Answer

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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 5.0kg object whose momentum is −15.0 kg# m/s? Answer

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4 What is the velocity of a 5.0kg object whose momentum is −15.0 kg# m/s?

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Answer m = 5 kg p = −15 kg ⋅m/s p = mv divide both sides by m v = p/m = (−15 kg ⋅m/s)/(5 kg) = − 3 m/s

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5 What is the mass of an object whose momentum is 35 kg⋅m/s when its velocity is 7.0 m/s? Answer

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5 What is the mass of an object whose momentum is 35 kg⋅m/s when its velocity is 7.0 m/s?

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Answer v = 7 m/s p = 35 kg ⋅m/s p = mv divide both sides by v m = p/v = (35 kg ⋅m/s)/(7 m/s ) = 5 kg

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Momentum Change & Impulse

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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 15 / 90

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

• bject won't change unless the object is

affected by an external force. When an outside force F acts on the object for a time Δt, it delivers an impulse I to the

• bject that changes its momentum:

Where the impulse is:

Momentum Change = Impulse Slide 16 / 90

There no specially named unit for impulse. We just 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/s

2 x s

= kg⋅m/s

• the same as momentum!

SI Unit for Impulse Slide 17 / 90 Effect of Collision Time on Force

∆t (seconds) F (newtons)

Changing the duration ∆t of an impulse by a small amount can greatly change the force exerted on an

• bject!

Impulse = F(∆t) = F(∆t)

Slide 18 / 90 Real World Applications

· Car Design / Accidents · airbags · collisions head-on vs walls · crush zones · Jumping / Landing · Boxing / Martial Arts · Hitting Balls - Golf, Baseball... · Catching Balls

Impulse = F(∆t) = F(∆t)

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6 An external force of 25 N acts on a system for 10 s. How big is the impulse delivered to the system? Answer

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6 An external force of 25 N acts on a system for 10 s. How big is 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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7 In the previous problem, an external force of 25 N acted on a system for 10 s. We found that the impulse delivered was 250 N-s. What was the change in momentum of the system? Answer

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7 In the previous problem, an external force of 25 N acted on a system for 10 s. We found that 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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8 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 Answer

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8 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 22 (Answer) / 90

9 Air bags are use 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

Answer

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9 Air bags are use 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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10 One car crashes into a concrete barrier. Another car crashes into a collapsible barrier at the same speed. What is the difference between the 2 crashes? A change in momentum B force on the car C impact time D both B & C Answer

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10 One car crashes into a concrete barrier. Another car crashes into a collapsible barrier at the same speed. What is the difference between the 2 crashes? A change in momentum B force on the car C impact time D both B & C

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Answer

C

Whether the wall is padded or not, the car experiences the same change in momentum, and so the same impulse. The

• nly difference is that the impact time has

increased (which in turn reduces the force experienced by the car).

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11 In order to increase the final momentum of a golf ball: A maintain the speed of the golf club after the collision (called "following through") B increase the force hitting the ball C increase the time of contact between the club and ball D any or all of the above Answer

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11 In order to increase the final momentum of a golf ball: A maintain the speed of the golf club after the collision (called "following through") B increase the force hitting the ball C increase the time of contact between the club and ball D any or all of the above

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Answer

D

Following through, hitting the golf ball harder, and/or increasing the impact time will all result in an increase in the final momentum of the golf ball.

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12 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? Answer

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12 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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13 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?

*

Answer

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13 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) ⨉ (0.03 s) = 600 m/s

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14 A 1200 kg car slows from 40 m/s to 5 m/s in 2 seconds. What force was applied by the brakes to decelerate the car?

*

Answer

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14 A 1200 kg car slows from 40 m/s to 5 m/s in 2 seconds. What force was applied by the brakes to decelerate 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

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15 Using the F-t graph shown, what is the change in momentum during the time interval from 0 to 6 seconds? Answer

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15 Using the F-t graph shown, what is the change in momentum during the time interval from 0 to 6 seconds?

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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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16 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 seconds? Answer

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16 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 seconds?

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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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The Momentum of a System of Objects

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If a system contains more than one object, it's 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.

+

• Slide 35 / 90

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 Answer

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

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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 (+)

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Answer 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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17 Determine the magnitude of the momentum of a system of two

• bjects: m1, has a mass of 6.0 kg and a velocity of 20 m/s north

and m2, has a mass of 3 kg and a velocity 20 m/s south. Answer

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17 Determine the magnitude of the momentum of a system of two

• bjects: m1, has a mass of 6.0 kg and a velocity of 20 m/s north

and m2, has a mass of 3 kg and a velocity 20 m/s south.

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Answer 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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18 Determine the momentum of a system of two objects: the first has a mass of 8 kg and a velocity of 8 m/s to the east while the second has a mass of 5 kg and a velocity of 15 m/s to the west. Answer

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18 Determine the momentum of a system of two objects: the first has a mass of 8 kg and a velocity of 8 m/s to the east while the second has a mass of 5 kg and a velocity of 15 m/s to the west.

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Answer 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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19 Determine the momentum of a system of 3 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. Answer

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19 Determine the momentum of a system of 3 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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Answer psystem = m1v1 + m2v2 + m3v3

= (7kg)(23m/s) + (9kg)(7m/s) +(5 kg)(−42m/s) = (161 kg⋅m/s) + (63 kg⋅m/s) + (−210 kg⋅m/s) = 14 kg⋅m/s (North) m1 = 7 kg v1 = +23 m/s (North) m2 = 9 kg v2 = 7 m/s (North) m3 = 5 kg v3 = −42 m/s (South)

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Conservation of Momentum

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Some of the most powerful concepts in science are called "conservation laws". 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 41 / 90

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 42 / 90

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 43 / 90

Conservation of Momentum and Impulse

Recall that momentum of an objet changes when it experiences an impulse, I: This impulse arises when a non-zero external force acts on the

• bject.

This is exactly the same for a system of objects. If there is no net external force on the system, the momentum of the system is conserved, that is, it does not change.

Slide 44 / 90 Conservation of Momentum Proof

Both the Conservation of Momentum and the concept of Impulse follow directly from Newton's Second Law:

*

F = ma F=m(Δv/Δt) FΔt =mΔv FΔt = Δ(mv) I = Δp I = pf - p0 p0 + I = pf p0 = pf start with Newton's Second Law where F is the net external force... since a = Δv/Δt... after multiplying both sides by Δt... since m is constant... recognizing that FΔt = I and mv = p, this becomes... since Δp = pf - p0

• r, adding p0 to both sides..

when there no net external force (F=0), I=0 and... momentum is conserved!

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Conservation of Momentum in Collisions and Explosions

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Conservation Laws, Collisions and Explosions

Objects in an isolated system can interact with each other in a number of 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 happened or what will happen in one of these events.

Slide 47 / 90 Collisions and Explosions

We differentiate collisions and explosions by the way the energy changes or does not change form. · explosions: an object or objects break apart because potential energy stored in one or more of the objects is transformed into kinetic energy · inelastic collisions: two objects collide and stick together converting some kinetic energy into bonding energy, heat, sound... · elastic collisions: two objects collide and bounce off each

• ther while conserving kinetic energy

Slide 48 / 90 Collisions and Explosions - Summarized

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

• ff each other

Yes

• No. Some kinetic

energy is converted to heat, sound... energy Inelastic Collision Objects stick together Yes

• No. Kinetic energy is

converted to potential energy, bonding, or heat, sound...energy Elastic Collision Objects bounce

• ff each other

Yes Yes Explosion Object or

• bjects break

apart Yes

• No. Release of

potential energy increases kintetic energy

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20 In _______ collisions momentum is conserved. A Elastic B Inelastic C All Answer

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20 In _______ collisions momentum is conserved. A Elastic B Inelastic C All [This object is a pull tab] Answer C

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21 In ______ collisions kinetic energy is conserved. A Elastic B Inelastic C All Answer

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21 In ______ collisions kinetic energy is conserved. A Elastic B Inelastic C All [This object is a pull tab] Answer A

Slide 51 (Answer) / 90 Conservation of Momentum

During a collision or an explosion, measurements show that the total momentum does not change: A B

mAVA mBVB

A B

A

B mAVA'

mAVB' +x

the prime means "after"

Slide 52 / 90 Explosions

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

• bjects.

We will assume: · the object (or a coupled pair of objects) breaks into two pieces · explosion is along the same line as the initial velocity A B Before (moving together) pA+pB=(mA+ mB)v

A

B

pA'=mAvA' After (moving apart) pB'=mBvB'

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22 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 balls's velocity after the explosion?

Answer

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22 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 balls's velocity after the explosion?

[This object is a pull tab] Answer (m1+m2) v = m1v1'+m2v2' 0 = m1v1'+m2v2' m1v1' = -m2v2' v1' = -m2v2'/m1 = -(300kg)(5m/s) / (5kg) = -300m/s m1 = 5kg m2 = 300kg v1 = v2 = v = 0 v2' = 5m/s

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23 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 6 m/s. How fast is the 6000 kg car going?

Answer

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23 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 6 m/s. How fast is the 6000 kg car going?

[This object is a pull tab] Answer (m1+m2) v = m1v1'+m2v2' 0 = m1v1'+m2v2' m1v1' = -m2v2' v1' = -m2v2'/m1 = -(4000kg)(6m/s) / (6000kg) = -4m/s m1 = 6000kg m2 = 4000kg v1 = v2 = v = 0 v2' = 6m/s

Slide 55 (Answer) / 90 Inelastic Collisions

In an inelastic collision, two objects collide and stick together, moving afterwards as one object.

A B

After (moving together) pA'+pB'=(mA+ mB)v'

A

B

pA=mAvA Before (moving towards the other) pB=mBvB

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24 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?

Answer

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24 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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Answer

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 same direction as first car's initial velocity m1 = 13,500kg m2 = 25,000kg v1 = 4.5m/s v2 = 0 m/s v1' = v2' = v'

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25 A cannon ball with a mass of 100 kg flies in horizontal direction with a speed of 800 m/s and strikes a ship initially at rest. The mass of the ship is 15,000 kg. Find the speed of the ship after the ball becomes embedded in it. Answer

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25 A cannon ball with a mass of 100 kg flies in horizontal direction with a speed of 800 m/s and strikes a ship initially at rest. The mass of the ship is 15,000 kg. Find the speed of the ship after the ball becomes embedded in it.

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Answer

m1v1+m2v2 = m1v1'+m2v2' m1v1+0 = (m1+m2) v' v' = m1v1/(m1+m2) = (100kg)(800m/s) / (100+15,000)kg = 5.3 m/s in same direction as cannon ball's initial velocity m1 = 100kg m2 = 15,000kg v1 = 800m/s v2 = 0 m/s v1' = v2' = v'

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26 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 the speed of the couple? Answer

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26 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 the speed of the couple?

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Answer

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 m1 = 40kg m2 = 70kg v1 = 5.5m/s v2 = 0 m/s v1' = v2' = v'

Slide 59 (Answer) / 90 Elastic Collisions

In a elastic collision, two objects collide and bounce off each other. Both momentum and kinetic energy are conserved. If we know the masses and any two of the velocities, these two conservation laws enable us to calculate the other two velocities.

A

B

pA=m

AvA

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

A

B

pA'=m

AvA'

pB'=mBvB'

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**Derivation of Elastic Collision Condition

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 61 / 90 Properties of Elastic Collisions

1. For all elastic collisions - regardless of mass - the relative velocity of the objects is the same before and after

2. In an elastic collision where one object strikes a much less massive stationary object - the velocity of the light object after the collision is approximately twice that of the heavy object. The heavy

• bject's velocity is almost unchanged.

3. In an elastic collision between two objects of identical masses, the two objects exchange velocities.

v1 - v2 = -(v1' - v'

2)

m1 m1 m2 m2 v1 v2 = 0 v'1 v'2 v1 2v1

v1' = v2 and v2' = v1 v1' = v1 and v2' = 2v1

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27 Two objects have an elastic collision. Before they collide they are approaching each other with a velocity of 4m/s relative to each

• ther. With what velocity do they go apart from one another?

Answer

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27 Two objects have an elastic collision. Before they collide they are approaching each other with a velocity of 4m/s relative to each

• ther. With what velocity do they go apart from one another?

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Answer 4m/s

Initially they are approaching relative to each

• ther at 4m/s, and afterwards they will move

apart relative to each other with a velocity of 4m/s. v1+v1'=v2+v2' The difference in the initial velocities is the same as the difference in the final velocities therefore they will move apart relative to one another at the same velocity

Slide 63 (Answer) / 90

28 Two objects have an elastic collision. One 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? Answer

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28 Two objects have an elastic collision. One 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

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

Slide 64 (Answer) / 90

29 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? Answer

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29 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

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

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30 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? Answer

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30 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

v1+v1' = v2+v2' v2' = v1+v1'-v2 v2' = v + v - (-2v) v2' = 4v (baseball's ball's speed doubled after being hit by the bat) v1 = +v v2 = -2v v1' = +v v2' = ?

Slide 66 (Answer) / 90

31 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +6.0m/s and m2 is -3.0m/s. What is the velocity of m1 after the collision? Answer

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31 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +6.0m/s and m2 is -3.0m/s. What is the velocity of m1 after the collision?

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Answer

In elastic collisions between objects of identical mass, the velocities are swapped on collision: v1' = v2 = -3m/s v2' = v1 = 6m/s v1 = +6m/s v2 = -3m/s v1' = ? v2' = ?

Slide 67 (Answer) / 90

32 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +6.0m/s and m2 is -3.0 m/s. What is the velocity of m2 after the collision? Answer

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32 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +6.0m/s and m2 is -3.0 m/s. What is the velocity of m2 after the collision?

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Answer

In elastic collisions between objects of identical mass, the velocities are swapped on collision: v1' = v2 = -3m/s v2' = v1 = 6m/s v1 = +6m/s v2 = -3m/s v1' = ? v2' = ?

Slide 68 (Answer) / 90

33 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +3.0m/s and m2 is +2.0m/s. What is the velocity of m1 after the collision? Answer

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33 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +3.0m/s and m2 is +2.0m/s. What is the velocity of m1 after the collision?

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Answer

In elastic collisions between objects of identical mass, the velocities are swapped on collision: v1' = v2 = +2m/s v2' = v1 = +3m/s v1 = +3m/s v2 = +2m/s v1' = ? v2' = ?

Slide 69 (Answer) / 90

34 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +3.0m/s and m2 is +2.0m/s. What is the velocity of m2 after the collision? Answer

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34 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +3.0m/s and m2 is +2.0m/s. What is the velocity of m2 after the collision?

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Answer

In elastic collisions between objects of identical mass, the velocities are swapped on collision: v1' = v2 = +2m/s v2' = v1 = +3m/s v1 = +3m/s v2 = +2m/s v1' = ? v2' = ?

Slide 70 (Answer) / 90

Collisions in Two and Three Dimensions

Return to Table of Contents

Slide 71 / 90 Conservation of Momentum in 2 and 3 Dimensions

Momentum vectors (like all others) 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:

Slide 72 / 90 Example: Collision with a Wall

When a ball bounces off a wall... m m θ θ p p' Its momentum perpendicular to the wall reverses Its momentum parallel to the wall is unchanged

Slide 73 / 90

35 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 cos(θ) D -2mv cos(θ) m m θ θ v v Answer

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35 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 cos(θ) D -2mv cos(θ) m m θ θ v v

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Answer

D

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36 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with half the speed as it had initially. What was the change in momentum of the ball? A -2.5 mv cos(θ) B -mv cos(θ) C -1.5 mv cos(θ) D -2 mv cos(θ) m m θ θ v v/2 Answer

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36 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with half the speed as it had initially. What was the change in momentum of the ball? A -2.5 mv cos(θ) B -mv cos(θ) C -1.5 mv cos(θ) D -2 mv cos(θ) m m θ θ v v/2

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Answer

C

Slide 75 (Answer) / 90 Collisions Between Two Objects

· All the collisions we will examine will only involve two objects. · Since there is no absolute reference frame, we can line up the x- axis with the velocity of one of the objects. · And we only need to consider 2 dimensions (x- and y-).

m1 m2

Slide 76 / 90 Collisions Between Two Objects

Consider a system of two objects where one of the objects (the blue one) is initially at rest... What is the momentum of the blue object after the collision?

m1 m2

Before After

m1 m2

p2 = ? p2 = 0

Slide 77 / 90

Conserving momentum along the x-direction

Before After

Add the components for the blue object

Collisions Between Two Objects

Conserving momentum along the y-direction ?

Slide 78 / 90

37 After the collision shown below, Which of the following is the most likely momentum vector for the second ball? A B C D E before after ? Answer

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37 After the collision shown below, Which of the following is the most likely momentum vector for the second ball? A B C D E before after ?

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Answer

D

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38 A bowling ball with a momentum of 20 kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 12 kg- m/s directed 60° to left of its initial direction (as shown). What is the momentum and direction of the pin? before after 60°

12 kg-m/s

θ

20 kg-m/s

Answer

Slide 80 / 90

38 A bowling ball with a momentum of 20 kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 12 kg- m/s directed 60° to left of its initial direction (as shown). What is the momentum and direction of the pin? before after 60°

12 kg-m/s

θ

20 kg-m/s

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Answer

before p1x = 20 kg-m/s p2x = 0 p1y = 0 p2y = 0 x-direction p1x + p2x = p'1x + p'2x 20 + 0 = 12cos(60°) + p'2x p'2x = 20 - 6 = 14 kg-m/s y-direction p1y + p2y = p'1y + p'2y 0 + 0 = 12sin(60°) + p'2y 0 = 10.39 + p'2y p'2y = -10.39 kg-m/s

p2' = √((p'2x)2 + (p'2y)2) = 17.4 kg-m/s θ = tan-1(p'2y/p'2x) = -36.6°

after p'1x = 12cos(60°)kg-m/s p'2x = p'1y = 12sin(60°)kg-m/s p'2y = ?

Slide 80 (Answer) / 90

39 A 5 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 of 18 kg-m/s. What was the velocity of the ball before the collision? before after 30°

18 kg-m/s ?

Answer 53.1° pin ball

Slide 81 / 90

39 A 5 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 of 18 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 = 22.35 kg-m/s v1x = p1x /m1 = 22.35/5 = 4.47 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 81 (Answer) / 90 Inelastic Collisions in Two Dimensions

m1 m2

Before 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. p1 p2

Slide 82 / 90 Inelastic Collisions in Two Dimensions

m1 m2

Before After

m1 m2

p' θ p1 p2 p-conservation in x: p1x = px' in y: p2y = py' final momentum: p' = √[(px')2 + (py')2] = √[(px)2 + (py)2] final velocity:

v' = p'/(m1+m2)

final direction:

θ = tan-1(py / px)

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40 Object A with mass 20 kg travels to the east at 10 m/s and

• bject B with mass 5 kg travels south at 20 m/s. What is the

magnitude of the velocity they have after the collision? A 8.9 m/s B 22.36 m/s C 30 m/s D 27.78 m/s E 223.6 m/s

Answer

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40 Object A with mass 20 kg travels to the east at 10 m/s and

• bject B with mass 5 kg travels south at 20 m/s. What is the

magnitude of the velocity they have after the collision? A 8.9 m/s B 22.36 m/s C 30 m/s D 27.78 m/s E 223.6 m/s

[This object is a pull tab] Answer before p1x = 200 kg-m/s p2x = 0 p1y = 0 p2y = 100 kg-m/s p' = √((p'x)2 + (p'y)2) = 223.6 kg-m/s v' = p'/(m1 + m2) = 223.6 / 25 = 8.9 m/s θ = tan-1(p'y/p'x) = tan-1(100/200) = 26.57° (south of east) after p'x = 200 kg-m/s p'y = 100 kg-m/s A

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41 Object A with mass 20 kg travels to the east at 10 m/s and object B with mass 5 kg travels south at 20 m/s. What is the direction

• f the velocity they have after the collision (use east as 0o, south

as 90o, etc.)? A 12.62o B 28.98o C 26.57o D 29.45o E 31.09o

Answer

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41 Object A with mass 20 kg travels to the east at 10 m/s and object B with mass 5 kg travels south at 20 m/s. What is the direction

• f the velocity they have after the collision (use east as 0o, south

as 90o, etc.)? A 12.62o B 28.98o C 26.57o D 29.45o E 31.09o

[This object is a pull tab] Answer before p1x = 200 kg-m/s p2x = 0 p1y = 0 p2y = 100 kg-m/s p' = √((p'x)2 + (p'y)2) = 223.6 kg-m/s v' = p'/(m1 + m2) = 223.6 / 25 = 8.9 m/s θ = tan-1(p'y/p'x) = tan-1(100/200) = 26.57° (south of east) after p'x = 200 kg-m/s p'y = 100 kg-m/s C

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Explosions in Two Dimensions

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. · if the initial momentum is zero, the final momentum is zero. · the third piece must have equal and opposite momentum to the sum of the other two.

p2 p1 p3

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Explosions in Two Dimensions

p2 p1 p3

θ

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: p1x + p2x + p3x = 0 p1y + p2y + p3y = 0 In this case the blue and red pieces are moving perpendicularly to each

• ther, so:

p1y = 0 and p2x = 0 p3x = -p1x and p3y = -p2y p3 = √((p1x)2 + (p2y)2) θ = tan-1(p2y / p1x)

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42 A stationary cannon ball explodes in 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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43 A stationary 10 kg bomb explodes into 3 pieces. The 2 kg piece moves west at 200 m/s. Another piece with a mass of 3 kg moves north with a velocity of 100 m/s. What is the velocity (speed and direction) of the third piece?

Answer

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43 A stationary 10 kg bomb explodes into 3 pieces. The 2 kg piece moves west at 200 m/s. Another piece with a mass of 3 kg moves north with a velocity of 100 m/s. What is the velocity (speed and direction) of the third piece?

[This object is a pull tab] Answer Since the first two pieces are moving perpendicular to each other p3x = -p1x = -m1v1 = -2kg × (-200m/s) = 400 kg-m/s p3y = -p2y = -m2v2 = -3kg × 100m/s = -300 kg-m/s m3 = M - m1 - m2 =10kg - 2kg - 3kg = 5 kg p' = √((p3x)2 + (p3y)2) = 500 kg-m/s v' = p'/(m3) = 500 / 5 = 100 m/s θ = tan-1(p3y/p3x) = tan-1(-300/400) = 36.87° (south of east)

Slide 89 (Answer) / 90

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