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A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

Asked in NEET 2018 · Coefficient of restitution and rebound

Answer: (2) 0.25

Step-by-step solution

Given: m at speed v strikes 4m at rest; the lighter block stops dead.

Idea: momentum fixes the heavy block's speed, and the coefficient of restitution then compares separation with approach.

mv+0=m(0)+4mv', so v'=v/4.

Velocity of approach =v-0=v.

Velocity of separation =v'-0=v/4.

e=(v/4)/v=0.25.

Why the other options are wrong

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