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areverseay

- 4

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- Homework Statement
- A particle A of mass 2m travelling with speed u undergoes an inelastic collision with a

stationary particle B of mass 10m. After the collision, A travels with a speed vA at 90 degrees to its original direction of motion and B travels with speed vB at an angle θ = arcsin(3/5) to the original direction of motion of A. Express vA and vB in terms of u; what fraction of kinetic energy is lost as a result of the collision?

- Relevant Equations
- p = mv

KE = (p^2)/2m

vA = 3u/4 and vB = u/4, and 1/8 KE is lost. I can't get to these answers however: for the first part, I got to u = vA + 3vB using conservation of momentum, and the fact that particle B is at an angle, hence I would think its momentum should be 10mvBsin(arcsin(3/5)). Doing the same for A with angle sin90 gives 2mu = 2mvA + 6mvB, hence the equation I first listed, but I don't know how to isolate vA and vB in terms of u. For the second part, even with vA and vB in terms of u, I can't get to 1/8 lost: using KE = (p^2)/2m and summing the KE for particles A and B, I get (54mu^2)/80.

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