A 47 g marble moving at 2.0 m/s strikes a 30 g marble at rest. Assume the collis

Question

A 47 g marble moving at 2.0 m/s strikes a 30 g marble at rest. Assume the collision is perfectly elastic and the marbles collide head-on.
Part A
What is the speed of the first marble immediately after the collision?
Express your answer to two significant figures and include the appropriate units.
V1=???

Part B
What is the speed of the second marble immediately after the collision?
Express your answer to two significant figures and include the appropriate units.
V2=???
*Also I believe I correctly reworked the equations for the specific velocities it is asking for, Im not entirely sure but maybe they could help you to answer the questions faster.
(V1)f=M2(V1i)^2-(V1i)/M2-2M1(V1)i+M1
(V2)f=M(V1)i-Mi(V1)f/M2
Please focus on the question more than the equations, they might be inaccurate.

Explanation / Answer

Masses m = 47 g            M = 30 g Initial velocities u =2 m / s                        U = 0 Let the speed of the first marble after collision be v and second marble be V From Law of conservation of momentum , mu + MU = mv + MV 94 + 0 = 47 v + 30 V 47 v + 30V = 94        ---------( 1) Coeffcient of restitution for elastic collision be 1 ( V - v ) / ( u - U ) = 1    V - v = u - U             = u V - v = 2         V = v + 2     -------( 2) Plug eq( 2) in eq( 1) we get , 47 v + 30(v+2) = 94                                              47 v + 30 v + 60 = 94                                              77v = 34                                                  v = 0.4415 m / s From eq( 2) , V = 2.4415 m / s

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