Two small spheres of putty, A and B, of equal mass m, hang from theceiling on massless strings of equal length.

Sphere A is raised to a height h0 as shown below and released. Itcollides with sphere B (which is initially at rest). The two spheres stick and swing together to a maximum height hf. (assume a perfectly inelastic collision, where

there is no internal energy lost to deformation, heating, etc.)

Find the height hf in terms of h0.

y

xh0 hf

A

B

BA

Lowest point in path is the point of zero gravitational potential

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

1. Which of the following physics principles should we use to solve this

problem.

A) Conservation of Total Mechanical Energy

B) Conservation of Momentum

C) Newton’s 2nd Law

Choice ACorrect, but this is not the only principle that we need to apply.

We will need to apply this principle to this problem twice in order to find relations between

the initial height and the speed of ball A, and between the speed of the merged balls and the

maximum height that the two balls reach together.

We will also need to think about the conservation of momentum.

We will need to apply this principle to this problem in order to find a relation between the speed of ball A right before collision and the speed of the two balls

together.

We will also need to think about the conservation of energy.

Choice B Correct, but this is not the only principle that we need to apply.

Newton’s 2nd Law will be of no use to us here.

Remember, we are concerned with the speeds of the putty balls, and their initial and final heights. Newton’s 2nd Law will

not lead us to useful relations.

Choice CIncorrect

2. We want to find the speed at which putty ball A strikes putty ball B. We can find this using the law of conservation of energy.

What types of mechanical energy does ball A have initially and just before colliding with ball B (final)?

A) Kinetic Gravitational potential

B) Gravitational potential Kinetic

C) Kinetic & Gravitational potential Gravitational potential

Initial Energy Final Energy

Putty ball A starts from rest, so it will not have kinetic energy initially.

Also, the lowest point of the ball’s motion is considered to be the point of zero

gravitational potential (h=0).

Choice AIncorrect

This is true because ball A starts from rest and reaches a point of zero gravitational

potential with a speed vA.

Choice BCorrect

Choice CIncorrect

Putty ball A starts from rest, so it will not have kinetic energy initially.

Also, the lowest point of the balls’ motion is considered to be the point of zero gravitational potential (h=0).

3. Applying the law of conservation of energy to ball A gives which of the following

expressions for the speed of ball A at the moment right before it collides with ball B?

A)

B)

C)

€

vA = gh0

€

vA =gh02

€

vA = 2gh0

Initial

Final

Choice ACorrect

€

mAgh0 =12mAvA

2

gh0 =12vA

2

2gh0 =vA2

vA = 2gh0

Reasoning:

Choice BIncorrect

€

mAgh0 =12mAvA

2

gh0 =12vA

2

2gh0 =vA2

vA = 2gh0

kinetic energy

kinetic energy

gravitational potential energy

gravitational potential energy

Choice CIncorrect

€

mAgh0 =12mAvA

2

gh0 =12vA

2

2gh0 =vA2

vA = 2gh0

kinetic energy

kinetic energy

gravitational potential energy

gravitational potential energy

4. In order to find an expression for the speed of the merged balls (vAB), in terms of the speed of ball A immediately before

the collision (vA), we need to use which conservation principle?

A) Conservation of Momentum

B) Conservation of Total Energy

C) Conservation of Angular Momentum

When we set the momentum before the collision equal to the momentum of the

system after the collision, there will not be other variables in our relation between the speeds, because we know that mA=mB=m.

Choice ACorrect

Note:

In general, this problem can be solved for spheres with different masses. This problem is a more specific case of colliding pendulums.

Note:

In general, this problem can be solved for spheres with different masses. This problem is a more specific case of colliding pendulums.

We can relate the two speeds this way, but this expression will involve

other variables.

Choice BIncorrect

This does not help us with the problem at hand.

Choice CIncorrect

5. Which of the following expressions correctly relates the speed of ball A immediately before the

collision and the speed of both balls moving together?

A)

B)

C)

€

vAB =vA2

€

vAB =2vA

€

vAB =vA4

Choice ACorrect

€

P0 =PfmAvA =mA +mB( )vABmvA =2mvAB

vA =2vAB

vA2

=vAB

Reasoning:

mA=mB=m

Choice BIncorrect

P=total linear momentum

0,f subscripts represent before and after collision

P=total linear momentum

0,f subscripts represent before and after collision Since

mA=mB=m

Since

mA=mB=m

€

P0 =PfmAvA =mA +mB( )vABmvA =2mvAB

vA =2vAB

vA2

=vAB

Choice CIncorrect

Since

mA=mB=m

Since

mA=mB=m

P=total linear momentum

0,f subscripts represent before and after collision

P=total linear momentum

0,f subscripts represent before and after collision

€

P0 =PfmAvA =mA +mB( )vABmvA =2mvAB

vA =2vAB

vA2

=vAB

6. Once again, use the law of conservation of energy for the initial moment right before the collision to the final moment

where balls A & B reach their maximum height.

Which one of the following expressions is correct for hf?

A)

B)

C)

€

hf =vA

2 −2vAB2

4g

€

hf =vA

2 −vAB2

2g

€

hf =2vA

2 −vAB2

4g

Choice AIncorrect

€

12mAvA

2 =12mA +mB( )vAB

2 + mA +mB( )ghf

12m( )vA

2 =12

2m( )vAB2 + 2m( )ghf

12vA

2 −122vAB

2 =2ghf

vA2 −2vAB

2

2=2ghf

vA2 −2vAB

2

4g=hf

mA=mB=mmA=mB=m

Choice BIncorrect

€

12mAvA

2 =12mA +mB( )vAB

2 + mA +mB( )ghf

12m( )vA

2 =12

2m( )vAB2 + 2m( )ghf

12vA

2 −122vAB

2 =2ghf

vA2 −2vAB

2

2=2ghf

vA2 −2vAB

2

4g=hf

mA=mB=mmA=mB=m

Choice CCorrect

€

12mAvA

2 =12mA +mB( )vAB

2 + mA +mB( )ghf

12m( )vA

2 =12

2m( )vAB2 + 2m( )ghf

12vA

2 −122vAB

2 =2ghf

vA2 −2vAB

2

2=2ghf

vA2 −2vAB

2

4g=hf

Reasoning:

7. Use the relations that we found in previous questions to find an expression for the

maximum height of the two balls together (hf) in terms of the initial height of ball A (h0).

Which of the following expressions is correct?

A)

B)

C)

€

hf =2h0

€

hf =h0

2

€

hf =h0

4

Choice AIncorrect

€

hf =vA

2 −2vAB2

4g

hf =2gh0 −2

2gh04

⎛ ⎝ ⎜

⎞ ⎠ ⎟

4g

hf =2h0 −h0

4

hf =h0

4

€

vAB2 =

vA2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟2

=2gh04

€

vA2 =2gh0

€

vA = 2gh0 =2vAB

From our previous expressions:

We see that:

€

hf =vA

2 −2vAB2

4g

hf =2gh0 −2

2gh04

⎛ ⎝ ⎜

⎞ ⎠ ⎟

4g

hf =2h0 −h0

4

hf =h0

4

Choice BCorrect

Reasoning:

The two putty balls will reach a maximum height that is 1/4th of ball A’s initial height.

The two putty balls will reach a maximum height that is 1/4th of ball A’s initial height.

€

hf =vA

2 −2vAB2

4g

hf =2gh0 −2

2gh04

⎛ ⎝ ⎜

⎞ ⎠ ⎟

4g

hf =2h0 −h0

4

hf =h0

4

Choice CIncorrect

From our previous expressions:

We see that:

€

vAB2 =

vA2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟2

=2gh04

€

vA2 =2gh0

€

vA = 2gh0 =2vAB

Reflection Questions:

• Answer this question twice more for balls of differing mass: –

–

€

mA =mB

2

€

mA =2mB