Elastic Collision with a Softball

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In an elastic collision problem, a 0.220 kg softball moving at 8.4 m/s collides head-on with a stationary ball, bouncing back at 3.4 m/s. The initial attempt to solve the problem incorrectly assumed both balls had the same mass, leading to an incorrect final velocity calculation. A suggestion was made to utilize the conservation of momentum and energy principles, specifically the rule that the relative speed of the colliding bodies after the collision is the negative of their relative speeds before the collision. After applying this advice, the correct answer was achieved. The discussion highlights the importance of correctly identifying mass differences and utilizing conservation laws in collision problems.
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Homework Statement



A softball of mass 0.220 kg that is moving with a speed of 8.4 m/s collides head-on and elastically with another ball initially at rest. Afterward the incoming softball bounces backward with a speed of 3.4 m/s.

Homework Equations



KE final = KE initial

The Attempt at a Solution



0.5(-3.4 m/s)^2 + 0.5(V)^2 = 0.5(8.4 m/s)^2 + 0.5(0 m/s)^2
5.78 m^2/s^2 + 0.5(V)^2 = 35.28 m^2/s^2
0.5 V^2 = 29.5 m^2/s^2
v^2 = 59 m^2/s^2
v = 7.68 m/s

Which is incorrect and I'm not sure why.
 
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You seem to have assumed that the second ball has the same mass as the first baseball. The problem doesn't explicitly say that, so I think you have to assume that they are not the same.

If I may make a small suggestion; In problems where you need to apply both conservation laws (momentum, energy) to figure out the unknowns, consider this equivalent rule to the conservation of energy: For elastic collisions the relative speed of the colliding bodies after collision is equal to the negative of their relative speeds before collision. So if v1 and v2 represent the initial speeds of bodies 1 and 2, and if u1 and u2 represent their speeds after collision, then

(v1 - v2) = -(u1 - u2)

This may make your mathematical life easier :smile:
 
Thank you for the advice. I tried something else and got the right answer.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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