# Momentum of golf ball and basketball

1. Oct 16, 2012

### Delta Sheets

1. The problem statement, all variables and given/known data
Place a golf ball on top of a basketball, and drop the pair from rest so they fall to the ground.The golf ball stays on top of the basketball until the basketball hits the floor. The mass of the golf ball is 0.0459 kg, and the mass of the basketball is 0.587 kg.

a) If the balls are released from a height where the bottom of the basketball is at 1.021 m above the ground, what is the absolute value of the basketball’s momentum just before it hits the ground? Answer calculated: 2.627 kg*m/s

b) What is the absolute value of the momentum of the golf ball at this instant?

c) Treat the collision of the basketball with the floor and the collision of the golf ball with the basketball as totally elastic collisions in one dimension. What is the absolute magnitude of the momentum of the golf ball after these collisions?

d) Now comes the interesting question: How high, measured from the ground, will the golf ball bounce up after its collision with the basketball? (Hint: do not forget to add the diameter of the basketball, 23.87 cm).

Would like to find how to solve parts c) and d)

2. Relevant equations
m1v1 + m2v2(initial)=m1v1 + m2v2(final)

3. The attempt at a solution
In these equations since momentum is conserved I thought that all of the momentum would just be emitted to the golf ball in the end, with the momentum of the basketball(initial) added to the momentum of the golf ball(initial) giving a total momentum of 2.832 kg*m/s, but that was not the case. Would like a little help on how the momentum would be shown for the gold ball after the collision, with this I could solve part d).

2. Oct 16, 2012

### cepheid

Staff Emeritus
With conservation of momentum alone, you have only one equation and two unknowns. So you are not going to arrive at a unique solution for v1 and v2: there are infinitely many. In other words, after the collision, conservation of momentum only demands that the total momentum be the same as before. The momentum can be divided up between the two balls any way you like, subject to this constraint.

So, to get a unique solution, you need another constraint (i.e. another equation). Where do you think you might get this from? Hint: do you know what an elastic collision IS?

3. Oct 16, 2012

### Delta Sheets

Elastic collision is conservation of kinetic energy and momentum correct?

4. Oct 16, 2012

### cepheid

Staff Emeritus
An elastic collision is a collision in which kinetic energy is conserved. Unlike momentum, kinetic energy doesn't *have* to be conserved, but in the case of an elastic collision, it is.

5. Oct 16, 2012

### Delta Sheets

Figured it out thank you, found velocity of basketball after first collision, which is the same as before, and then velocity of the golf ball after hitting the basketball. Momentum of the golf ball was 0.557 kg*m/s, and height after being hit is 7.737m

6. Oct 16, 2012

### cepheid

Staff Emeritus
Cool. How does this compare to the height reach by the basketball? (I've done this demo in tutorials, so I know that the smaller ball goes crazy. Accidently hit someone in the first row one time, although to be fair, I did warn them that it was tough control things such that the ball bounced exactly straight up. Good thing we used a tennis ball and not a golf ball).

7. Oct 16, 2012

### Delta Sheets

Height reached by the basketball is much smaller as the velocity is 3.18 m/s in the upward direction, making it only go 0.515m up

8. Oct 17, 2012

### cepheid

Staff Emeritus
I don't get the same numbers as you. For one thing, the speed of the balls at the instant they hit the ground is given by $\sqrt{2gh}$ = 4.48 m/s. Now, the assumption is that the basketball collides elastically with a stationary object of infinite mass. You can work out using conservation of momentum and conservation of kinetic energy that the basketball would then be travelling upwards after the collision at the same speed: 4.48 m/s (as opposed to 3.18 m/s -- no energy is lost, so I have no idea how you got that number). So, in the instant just after the collision with the ground, the basketball is moving upwards at 4.48 m/s, whereas the golf ball is still moving downwards towards it at -4.48 m/s, so their *relative* velocity is 2*4.48 m/s. So, applying the result of the elastic collision again, this time to the golf ball, assuming that it collides at 8.96 m/s with a stationary object much more massive than it (i.e. the basketball), it will be moving upwards at 8.96 m/s *relative to the basketball* after the collision. That's 8.96 m/s + 4.48 m/s relative to the ground. So it's travelling upwards at 3 times its original speed, which means it will reach nine times its original height.

Last edited: Oct 17, 2012
9. Oct 17, 2012

### cepheid

Staff Emeritus
Now that I think about it, maybe you didn't assume that m2/m1 --> infinity (2 = basketball) but instead used the proper full expressions. So that's v2 = 2m1u1/(m1+m2)

u = vels before collision
v = after

V2 = 1.3 m/s downward in the basketball reference frame.

4.48 - 1.3 = 3.18 m/s upward relative to the ground after the collision. My mistake.

10. Oct 17, 2012

### cepheid

Staff Emeritus
v1 = u1[ (m1-m2)/(m1+m2) ]

= 7.66 m/s upward relative to basketball frame. I think you still need to add 4.48 to this, not 3.18, because we are measuring speeds in the original frame of reference of the basketball. I get a height of 7.51 m from that, but I did this on my phone in a rush.