Conservation of momentum/energy of stacked balls

In summary: What is the answer to the problem?In summary, the problem involves dropping a tennis ball and basketball from a height of 1m, with the tennis ball on top of the basketball. When dropped separately, the tennis ball bounces to a height of 0.5m and the basketball to a height of 0.8m. The task is to find the maximum height of the tennis ball when the balls are dropped together, given that the basketball's velocity after the bounce is 1m/s. The equations used are conservation of momentum, conservation of energy, and the formula for kinetic energy. However, the problem is not entirely clear as it is missing information about the radius of the basketball and the coefficient of restitution for the collision
  • #1
GamrCorps
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Homework Statement



A tennis ball and basketball are dropped from a height of 1m (the tennis ball on top of the basketball). The tennis ball has a mass of 75g and the basketball has a mass of 1kg. When dropped separately, the tennis ball bounces to a height of 0.5m and the basketball to a height of 0.8. Find the max height of the tennis ball when the balls are dropped together if the basketball's velocity after the bounce is 1m/s.

Use g=10m/s/s

Homework Equations



p=mv
U=mgh
K=0.5mv^2
p1=p2
E1=E2
v^2=v0^2+2gh

The Attempt at a Solution



I tried to do standard conservation of momentum and conservation of energy, but realized I needed to account for energy loss. That is where I got lost. How do you account for energy loss and/or what is the answer so I can check my work when I'm done?
 
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  • #2
Hello Gamr, :welcome:

Wel, if the tennis ball bounces back to 0.5 m when dropped from 1 m, you can calculate the energy loss...
 
  • #3
GamrCorps said:
A tennis ball and basketball are dropped from a height of 1m (the tennis ball on top of the basketball). The tennis ball has a mass of 75g and the basketball has a mass of 1kg. When dropped separately, the tennis ball bounces to a height of 0.5m and the basketball to a height of 0.8. Find the max height of the tennis ball when the balls are dropped together if the basketball's velocity after the bounce is 1m/s.
I have a bad feeling about this question. There are three things unclear about the interaction between the balls when dropped together.

We are not given the radius of the basketball, so we do not know the height or speed of the tennis ball when it collides with the basketball. Will this matter? I think it would, so we have to pretend the basketball is minute.

Should we treat it as though the basketball bounces first, then immediately collides with the tennis ball; or consider both balls undergoing compression then decompression simultaneously (a somewhat complex interaction since one compression might complete before the other)?

Secondly, we can compute the coefficients of restitution for tennis ball/floor and basket ball/floor, but how do we deduce the coefficient of restitution for the collision between the two balls? I tried to come up with a law by considering that each object acts like a spring with a lower constant on decompression than on compression. This did not lead to any way to find the coefficient of restitution for an impact between the two bodies given their separate coefficients on the rigid floor - it depends on the details of the spring constants.

However, if we could figure out the answers to the above then we would not need to be told the basketball's velocity after the bounce. So maybe there is a way.
E.g., if we treat it as the basketball bouncing first then we know its velocity at that instant (from the bounce height with no tennis ball), then from momentum conservation we can deduce the change in velocity of the tennis ball when they collide, reducing the basketball's speed to 1m/s. But if that is the way we would not need the information about the tennis ball bouncing alone.

Bringing in my experience of the two types of ball, I would say the basketball is much stiffer, i.e. a much higher "spring constant". That means the bounce between basketball and ground will complete before that between the balls progresses very far. Moreover, the coefficient of restitution of the two balls in collision will be much closer to that of the tennis ball/floor than to that of basketball/floor.

If I take it as basketball bouncing first, ignore the radius of the basketball and the information that the tennis ball alone bounces .5m, I get the tennis ball bounces to 81- √320 ≈ 63m. This is clearly impossible since it implies a net gain in energy when the two balls collide.
 
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  • #4
Yeah, I had the same feeling. It seemed like there is information missing and everytime I approached the problem I got answers over 100m. No clue what to do.
 
  • #5
GamrCorps said:
Yeah, I had the same feeling. It seemed like there is information missing and everytime I approached the problem I got answers over 100m. No clue what to do.

My suggestion is to assume that the basketball loses 20% of its energy in the collision and the tennis ball loses 50% of its energy and work it out from there.

You shouldn't get 100m because that would be more potential energy than the system has at the outset.
 
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  • #6
PeroK said:
assume that the basketball loses 20% of its energy in the collision and the tennis ball loses 50% of its energy
That would not conserve momentum.
Seems to me that no matter in what reasonable way we handle the collision between the two balls, conserved momentum is going to imply gained energy. The basketball would bounce at 3m/s. If slowed to 1m/s by collision with an object 3/40 of its mass, that smaller object is going to gain a lot of KE.
 
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  • #7
haruspex said:
That would not conserve momentum.
Seems to me that no matter in what reasonable way we handle the collision between the two balls, conserved momentum is going to imply gained energy. The basketball would bounce at 3m/s. If slowed to 1m/s by collision with an object 3/40 of its mass, that smaller object is going to gain a lot of KE.

My guess is that whoever set the question simply used energy and didn't think about how a basketball could lose so much energy to a tennis ball.
 

FAQ: Conservation of momentum/energy of stacked balls

How does conservation of momentum/energy apply to stacked balls?

The principle of conservation of momentum and energy states that in a closed system, the total momentum and energy remains constant. This means that within a system of stacked balls, the total momentum and energy of the balls will remain the same throughout any interactions or movements.

What factors affect the conservation of momentum/energy in stacked balls?

Several factors can affect the conservation of momentum and energy in stacked balls, including the mass and velocity of each ball, the angle at which they are stacked, and any external forces acting on the system.

Can the conservation of momentum/energy be violated in stacked balls?

No, the conservation of momentum and energy is a fundamental law of physics and cannot be violated. It applies to all systems, including stacked balls.

How does friction impact the conservation of momentum/energy in stacked balls?

Friction can cause a loss of kinetic energy in the system of stacked balls, as some energy is converted into heat. However, the total momentum and energy of the system will still remain constant.

Is conservation of momentum/energy only applicable to stacked balls moving in a straight line?

No, conservation of momentum and energy applies to all types of motion, including stacked balls moving in curved paths or undergoing collisions. As long as the system is closed, the total momentum and energy will remain constant.

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