A block of mass m moving on a frictionless surface at speed V collides elastically with a block of same mass, initially at rest. Now the first block moves at an angle with its initial direction and h…

Question

A block of mass m moving on a frictionless surface at speed V collides elastically with a block of same mass, initially at rest. Now the first block moves at an angle with its initial direction and has speed V_1. The speed of the second block after collision is

Accepted Answer

Correct answer: B. (b): As, the collision is perfectly elastic, so from conservation of energy we can write, initial kinetic energy = final kinetic energy ⇒ 12mV^2+0= 12mV_1^2+ 12mV_2^2 ( V_2= speed of second block after collision) ⇒V_2^2=V^2-V_1^2 ⇒V_2= V^2-V_1^2

$A block of mass m moving on a frictionless surface at speed V collides elastically with a block of same mass, initially at rest. Now the first block moves at an angle θ with its initial direction and has speed V_{1} . The speed of the second block after collision is$

Solution

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