Bouncing balls
A tennis ball rests directly on top of a cannon ball, which has a much larger mass, i.e. M cannonball ≫ M tennis .
When the balls are dropped, to what height (in c m ) will the bottom of the tennis ball bounce?
Details and Assumptions:
- The bottom of the cannon ball starts at a height of h = 6 0 c m from the ground, and the cannon ball has diameter d = 3 0 c m .
- All collisions are perfectly elastic, and wind resistance is ignored.
The answer is 570.
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3 solutions
The interesting point to note here is that the ball bounces off to a much greater height compared to where it was dropped. It looks like a super-elastic collision.
How will you explain the conservation of energy here with regards to the initial and final heights of the balls?
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@Rohit Gupta ...well , do not consider mass of smaller ball to be much less as compared to the larger ball....use the exact equations and you can explain energy conservation very easily...
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Agreed, the cannon ball will not bounce as high as it would without the Cosco ball on its top. That loss of energy of the cannonball, when received by the Cosco ball, will rise a lot higher as given in the solution.
And if I don't understand physics not the equations or anything , I try to use my logic because I can't be a robot and do an equation that I don't know , how can I solve it ?
I said the hight of the cannonball is 30 so when it hits the ground the tennis ball will hit the maximum hight so the hight must be greater than 30
And then I am stuck Also why do I have to use an equation that others have proven without proving it
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Can you elaborate on which part are you stuck and which equation are you referring?
So then, a three-stack would then be (3+2)^2 = 25 times as high, and a four-stack would bounce almost (5+2)^2 = 50 times as high!
I was misled by the cannon ball, which intuitively would not bounce. However, as stated, the problem is well posed and the answer is correct. You got me!
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Wait a minute! The solution given does not seem to conserve momentum or kinetic energy. Some of the cannonball's upward velocity must have been imparted to the smaller ball, and I am thinking it will not rise quite as high. I am not sure, will look at the equations later.
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Yes, you are correct. However, the mass of the cannon ball is given to be much larger than the mass of the tennis ball, therefore, its velocity will not change significantly and the ball will rise nearly to the height given in the answer. So, it is a fair approximation.
Forgot to add 30 lol.
i don't have any to add to @Denton Young 's solution, but here is a video u might find interesting :)
The speed of the basketball decreases due to its collision with the Cosco ball and the equivalent energy is gained by the latter. Hence, the Cosco ball bounces a lot more.
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I prefer this to her compression description
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Well yes, however, I am assuming elastic collisions, in that case, the change in potential energy is zero and the balls will regain their deformations completely.
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@Rohit Gupta – This may assume the cannonball is moving upward at speed v and not slowing down as conservation of linear momentum would require.
The model in the proposed solution is interesting, as the instant of collision is critical to the analysis. It assumes separation of the two balls prior to impact so that a collision of one moving downward and one moving upward actually occurs.
If they remained in contact during their descent, why would they necessarily separate after the bounce? Also, if you assume there is a collision at the moment the cannonball hits the ground, at which time it's velocity is momentarily 0, the outcome is the same, as it would rebound upward at speed v.
This is a great "Gedank" exercise, and one that would be interesting to try to perform physically to see if the different "mathematical" outcomes might be replicable with enough accuracy to see them clearly.
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@Will Heierman – While in the air they do not both are falling at the same acceleration and thus they won't apply any force to each other, so it is safe to imagine a microscopic separation between the balls. During the collision of the cannonball with the ground, its velocity changes from v downwards to v upwards in a fraction of a second. Also, during this fraction of a second, the cannonball also passes through the state of momentary rest. You may try a similar problem .
That was cool:)
When the impact happens, both tennisball and cannonball have downward speed v .
Because the cannonball is so much more massive, its motion is not really affected by the tennisball. After the bounce it has speed v in upward direction.
That is when it hits the tennisball. Since the tennisball is still moving down at speed v but the cannonball is moving up at speed v , the relative speed at impact is 2 v . This is also the speed of the tennisball to the center of mass (which essentially is the center of the cannonball).
The collision is elastic, so the tennisball bounces back with upward speed 2 v relative to the cannonball; i.e. with upward speed 3 v relative to the ground.
Had the tennisball bounced directly off a stationary surface with speed v , it would have reached a height of 60 cm above that surface. Now it is launched three times faster and will therefore travel 3 2 = 9 times higher, i.e. 540 cm. Add the 30 cm of the height of the cannonball, and you find the answer 5 7 0 cm.
At first sight, it looks counterintuitive that the tennis ball would go so much higher than its initial height. This is probably because we don't notice the small change in the speed of the cannon ball.
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Exactly. If we account for the finite mass of the cannonball, we find for the speeds after impact v T ′ = M + m 3 M − m v 0 ; v C ′ = M + m M − 3 m v 0 .
The total momentum of the system is M + m m ( 3 M − m ) + M ( M − 3 m ) v 0 = M + m M 2 − m 2 v 0 = ( M − m ) v 0 , which makes sense because immediately after the cannonball bounces upward, it has a momentum of M v 0 and the tennisball (still going down) has a momentum of − m v 0 . Thus the collision between cannonball and tennisball conserves the momentum.
Zeroeth order approximation (as we did in this problem), with m / M ≈ 0 : v T ′ ≈ 3 v 0 , v C ′ ≈ v 0 ;
First order approximation (as we did in this problem), with α = m / M and α 2 ≈ 0 : v T ′ ≈ ( 3 − 2 α ) v 0 , v C ′ ≈ ( 1 − 4 α ) v 0 .
Relevant wiki: Analyzing Elastic Collisions
For simplicity, assume that the balls are separated by a very small distance, so that the relevant bounces happen a short time apart. This assumption isn’t necessary, but it makes for a slightly cleaner solution.
Just before the cannon ball hits the ground, both balls are moving downward with speed (using m v 2 / 2 = m g h )
v = 2 g h
Just after the cannon ball bounces off the ground, it moves upward with speed v , while the tennis ball still moves downward with speed v . The relative speed is therefore 2 v . After the balls bounce off each other, the relative speed is still 2 v .
Since the upward speed of the cannon ball essentially stays equal to v , the upward speed of the tennis ball is 2 v + v = 3 v . By conservation of energy, it will therefore rise to a height of H = d + ( ( 3 v ) 2 ) / ( 2 g )
But v 2 = 2 g h , so we have H = d + 9 h = 3 0 + 5 4 0 = 5 7 0 c m