A moving block having mass m , collides with another stationary block having mass 4m . The lighter block comes to rest after collision. When the initi - Physics MCQ NEET (UG) 2018 2026 | ExamTest.live

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A moving block having mass m, collides with another stationary block having mass 4 m . 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:

Solution & Step-by-step Explanation

1. Conservation of Linear Momentum: m v + 4 m (0) = m (0) + 4 m v^{'} m v = 4 m v^{'} ⟹ v^{'} = \frac{v}{4} .2. Coefficient of restitution (e) is: e = \frac{Velocity of separation}{Velocity of approach} e = \frac{v ^{'} - 0}{v - 0} = \frac{v /4}{v} = \frac{1}{4} = 0.25

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A moving block having mass m, collides with another stationary block having mass 4 m . 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:

0.8

0.25

0.5

0.4

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