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 va…

Question

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 (NEET2018)

Accepted Answer

Correct answer: B. (b): Let final velocity of the block of mass 4m=v^ ' Initial velocity of block of mass 4 m=0 Final velocity of block of mass m=0 According to law of conservation of linear momentum, mv+4m 0=4mv^ '+0 ⇒v^ '=v/4 Coefficient of restitution, e= Relativevelocityofseparation Relativevelocityofapproach= v/4v=0.25

$A moving block having mass m, collides with another stationary block having mass 4 m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be (NEET2018)$

Solution

More Collision & Variable Mass PYQs