Transcript Tutorial_Putty Collision

Two small spheres of putty, A and B, of equal mass m, hang from the
ceiling on massless strings of equal length.
Sphere A is raised to a height h0 as shown below and released. It
collides 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
A
h0
B
hf
x
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 A
Correct, 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.
Choice B
Correct, but this is not the only principle that we need to apply.
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 C
Incorrect
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.
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)?
Initial Energy
Final Energy
A)
Kinetic
Gravitational potential
B)
Gravitational potential
Kinetic
C)
Kinetic & Gravitational potential
Gravitational potential
Choice A
Incorrect
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 B
Correct
This is true because ball A starts from rest
and reaches a point of zero gravitational
potential with a speed vA.
Choice C
Incorrect
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)
vA  2gh0
B)
vA  gh0
C)
gh0
vA 
2
Initial
Final
Choice A
Correct
Reasoning:
1
2
mAgh0  mA v A
2
1 2
gh0  v A
2
2
2gh0  v A
v A  2gh0
Choice B
Incorrect
gravitational
potential
energy
kinetic energy
1
2
mAgh0  mA v A
2
1 2
gh0  v A
2
2
2gh0  v A
v A  2gh0
Choice C
Incorrect
gravitational
potential
energy
kinetic energy
1
2
mAgh0  mA v A
2
1 2
gh0  v A
2
2
2gh0  v A
v A  2gh0
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
Choice A
Correct
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.
Note:
In general, this problem can be
solved for spheres with different
masses. This problem is a more
specific case of colliding
pendulums.
Choice B
Incorrect
We can relate the two speeds this
way, but this expression will involve
other variables.
Choice C
Incorrect
This does not help us with the
problem at hand.
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)
vA
vAB 
2
vA
vAB 
4
vAB  2vA
Choice A
Correct
Reasoning:
P0  Pf
mA v A  mA  mB v AB
mA=mB=m
mv A  2mv AB
v A  2v AB
vA
 v AB
2
Choice B
Incorrect
P0  Pf
P=total linear
momentum
mA v A  mA  mB v AB
0,f subscripts
represent
before and
after collision
mv A  2mv AB
v A  2v AB
Since
mA=mB=m
vA
 v AB
2
Choice C
Incorrect
P0  Pf
P=total linear
momentum
mA v A  mA  mB v AB
0,f subscripts
represent
before and
after collision
mv A  2mv AB
v A  2v AB
Since
mA=mB=m
vA
 v AB
2
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)
2v A2  v AB2
hf 
4g
B)
v A2  v AB2
hf 
2g

C)

v A2  2v AB2
hf 
4g
Choice A
Incorrect
mA=mB=m
1
1
2
2
mA v A  mA  mB v AB  mA  mB ghf
2
2
1
1
mv A2  2mv AB2  2mghf
2
2
1 2 1
2
v A  2v AB  2ghf
2
2
v A2  2v AB2
 2ghf
2
v A2  2v AB2
 hf
4g
Choice B
Incorrect
1
1
2
2
mA v A  mA  mB v AB  mA  mB ghf
2
2
1
1
mv A2  2mv AB2  2mghf
2
2
mA=mB=m
1 2 1
2
v A  2v AB  2ghf
2
2
v A2  2v AB2
 2ghf
2
v A2  2v AB2
 hf
4g

Choice C
Correct
Reasoning:
1
1
2
2
mA v A  mA  mB v AB  mA  mB ghf
2
2
1
1
mv A2  2mv AB2  2mghf
2
2
1 2 1
2
v A  2v AB  2ghf
2
2
v A2  2v AB2
 2ghf
2
v A2  2v AB2
 hf
4g
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
h0
hf 
4
h0
hf 
2
Choice A
Incorrect
v A  2v AB
hf 
4g
2gh0 
2gh0  2

 4 
hf 
4g
2
From our previous expressions:
vA  2gh0  2vAB
We see that:
v A  2gh0
2
hf 
v A  2gh0
   
 2 
4
2h0  h0
4
2
v
2
AB
h0
hf 
4
2
Choice B
Correct
Reasoning:
v A  2v AB
hf 
4g
2gh0 
2gh0  2

 4 
hf 
4g
2
hf 
2h0  h0
4
h0
hf 
4

2
The two putty
balls will reach a
maximum height
that is 1/4th of ball
A’s initial height.
Choice C
Incorrect
From our previous expressions:
vA  2gh0  2vAB
v A  2v AB
hf 
4g
2gh0 
2gh0  2

 4 
hf 
4g
2
We see that:
v A  2gh0
2
hf 
v A  2gh0
   
 2 
4
2h0  h0
4
2
v
2
AB
h0
hf 
4
2
Reflection Questions:
• Answer this question twice more for
balls of differing mass:


–
mB
mA 
2
–
mA  2mB