# Answer to Question #13432 in Mechanics | Relativity for harisha

Question #13432

A block of mass 2*1000g moving at 2*100m/s collides with another block of same mass at rest. if actual loss of K.E. is half of the maximum then the coefficient of restitution is

Expert's answer

KE1max = m * V^2 / 2 = 2 * 200^2 / 2 = 40000 J

Loss is KE1 = KE2 = KE1max / 2

= 20000 J.

So, after the impact V1 = V2.

The coefficient of restitution is

given by

C = (Vb - Va) / (Ua - Ub)

for two colliding objects, where

Va

is the final velocity of the first object after impact

Vb is the final

velocity of the second object after impact

Ua is the initial velocity of the

first object before impact

Ub is the initial velocity of the second object

before impact

C = (V2 - V1) / (200 - 0) = 0

Loss is KE1 = KE2 = KE1max / 2

= 20000 J.

So, after the impact V1 = V2.

The coefficient of restitution is

given by

C = (Vb - Va) / (Ua - Ub)

for two colliding objects, where

Va

is the final velocity of the first object after impact

Vb is the final

velocity of the second object after impact

Ua is the initial velocity of the

first object before impact

Ub is the initial velocity of the second object

before impact

C = (V2 - V1) / (200 - 0) = 0

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