(8.3) Two equal masses are each traveling with the same speed v, as shown in the

Question

(8.3) Two equal masses are each traveling with the same speed v, as shown in the figure to the right. After colliding with each other, they stick together and move with a speed of v/2. (a) Determine the required angle that will produce this final state, (b) Calculate the kinetic energy of the system (both masses together) before and after the collision. (c) What percentage of the original kinetic energy remains as kinetic energy after the collision? Answers: (a) 60°, (b) mu2.mu2, (c) 25% U) Ly 2) Tn

Explanation / Answer

As the object starts moving along horizontal after collision means the final momentum of system along y axis becomes zero.

Now applying law of conservation of momentum before and after collision

m1 (v cos theta) + m2 (v cos theta) = (m1 + m2) (v/2)

(m1 = m2 = m)

cos theta + cos theta = 2/2

2 cos theta = 1

Theta = cos-1(1/2) = 60°

B) kinetic energy of system before collision

0.5 mv2 + 0.5 mv2 = mv2

Kinetic energy of system after collision

0.5 (2m) (v/2)2 = mv2/4

C) % of kinetic energy = (final KE/ initial KE) x 100 = ((1/4) mv2)/(mv2) x 100 = 25%

Please rate my answer if you find it helpful. Good luck...

Please rate my answer if you find it helpful. Good luck...

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