Alley S.

asked • 07/25/23

Elastic Collision

A 42.0 g marble moving at 2.20 m/s strikes a 22.0 g marble at rest. Note that the collision is elastic and that it is a "head-on" collision so all motion is along a line.




What is the speed of 42.0 g marble immediately after the collision?

AND

What is the speed of 22.0 g marble immediately after the collision?



1 Expert Answer

By:

Alley S.

I have momentum before=momentum after is .0924=.42(v1f)+.022(v2f) For KE I have .10164=.021(v1f^2)+.011(v2f^2) How do I solve for the speed of each marble?
Report

07/26/23

Cole M.

tutor
Your numbers for momentum and KE seem correct! From here, it is just solving a system of equations. Next, try solving the momentum equation for v1f, you should get something in terms of v2f. You can then plug this expression into the energy equation and solve for v2f. Once you have determined v2f, you can solve for v1f. Let me know if you have any more questions!
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07/26/23

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