Conservation of Momentum in Collision between a Ball and a Truck

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The discussion centers on the conservation of momentum during a collision between a ball and a truck, specifically a 0.2 kg ball traveling at 20 m/s to the left and a 20,000 kg truck moving at 25 m/s to the right. The correct post-collision speed of the ball is determined to be 70 m/s, assuming an elastic collision where the truck's velocity change is negligible. The conversation emphasizes the importance of correctly applying momentum conservation principles and the role of the coefficient of restitution in such scenarios.

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A ball weighing .2kg is traveling 20ms-1 to the left, a 20 000kg truck is traveling 25ms-1 to the right. What is the speed of the ball after the collision
(M1V1) + (M2V2)before=(M1V1) + (M2V2)after
This seemed so obvious, but i keep getting it wrong. The answer is 70ms-1. The only thing that i could be doing wrong is assuming the change in speed of the truck is negligible. Thanks for the help in advance.
 
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The question is quite unclear.They are implying a completely elastic collision here as the theoretical maximum velocity of ball is 70 m/s,assuming there is no change in velocity of the truck.
Your basic assumption was correct but the manner of application was wrong.If there is no change in momentum of truck then no change in momentum of ball too,isn't it? Then the ball would probably go through the truck! Hence momentum conservation is not justified.
In such cases the coefficient of restitution is defined (which is arbitratily 1 in this case).
 
Try looking at this situation from the reference frame of the truck, who's mass is >> than the balls. Then, knowing that kinetic energy is conserved, you can solve for the resultant velocity of the ball treating the mass of the truck as infinitely large. This will yield a very good approximation.
 

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