Elastic Collision of a ball and an elephant

[kaka2007]

A charging elephant with mass 5230kg comes toward you with speed of 4.45m/s. You toss a 0.15 rubber ball at the elephant with a speed of 7.91 m/s. (a) When the ball bounces back toward you, what is the speed?

Answer= 16.8

I don't know if the answer is written wrong in the back of the book or if I'm oversimplifying it.

Since the mass of the elephant is so big compared to the ball shouldn't it just be 2 x 7.91?.

I just used v2,f = (2m1/(m1+m2))v0

since m2 approaches zero I would expect (2m1/m1)v0 --> 2v0

 

[alphysicist]

A charging elephant with mass 5230kg comes toward you with speed of 4.45m/s. You toss a 0.15 rubber ball at the elephant with a speed of 7.91 m/s. (a) When the ball bounces back toward you, what is the speed?

Answer= 16.8

I don't know if the answer is written wrong in the back of the book or if I'm oversimplifying it.

Since the mass of the elephant is so big compared to the ball shouldn't it just be 2 x 7.91?.

If the ball was at rest and collided with the elephant, the ball would rebound with approximately twice the elephants speed.

If the elephant was at rest, after the collision, the ball would rebound with approximately the same speed it had before the collision, but moving in the opposite direction.

I just used v2,f = (2m1/(m1+m2))v0

This is not the full equation; this is what's left of the equation after you have set $$v_{2i}\to 0$$. What is the full equation for $v_{2f}$? That should give you the correct answer.

 

[Chrisas]

I think you need to go ahead and apply the conservation of momentum and conservation of energy to get two equations for two unknowns, i.e. the final velocities of the elephant and ball. The equation you quoted doesn't fit this situation. It only works if the initial speed of the elephant is zero (i.e. the ball bounces against a wall) It is true that the velocity of the elephant will hardly change. However, his mass is so large that this amounts to a significant transfer of momentum to the small ball.

Think about swinging a sledge hammer at a ping pong ball. You won't even know that you hit it. You can bet that the ping pong ball will go flying back at more than twice what it came in at.

 

[alphysicist]

Hi Chrisas,

I think you need to go ahead and apply the conservation of momentum and conservation of energy to get two equations for two unknowns, i.e. the final velocities of the elephant and ball. The equation you quoted doesn't fit this situation. It only works if the initial speed of the elephant is zero

I don't believe that is correct. The equation in the original post only works if the ball's initial speed is zero, not the elephant's.


There is a standard set of equations for the final velocities that are derived by using the momentum and kinetic energy conservation equations, and the equation kaka2007 quoted was just that part that remains after you set the ball's initial velocity to zero.

 

[Chrisas]

Your probably right. I didn't work through the math myself. I was looking at the equations as derived by Serway and trying to match to what he wrote. I was attempting to juggle Serway's i's and f's with balls and elephants and the OP's 1's and 2's in my head and probably got them swapped at some point.