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 'θ' with its initial direction and has speed 'v₁'. The speed of the second block after collision is ______.

विकल्प

• `sqrt("v"_1^2 - "v"^2)`

• `sqrt("v"^2 - "v"_1^2)`

• `sqrt("v"^2 + "v"_1^2)`

• `sqrt("v" - "v"_1)`

Answer 1

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 'v₁'. The speed of the second block after collision is `underline(sqrt("v"^2 - "v"_1^2))`.

Explanation:

The situation can be shown as

Applying law of conservation of kinetic energy,

KE (before collision)= KE (after collision)

`1/2 "mv"^2 + 1/2 "m"(0)^2 = 1/2"mv"_1^2 + 1/2 "mv"_2^2`

`=> "v"^2 = "v"_1^2 + "v"_2^2`

`=> "v"_2 = sqrt("v"^2 - "v"_1^2)`

Thus, the velocity of second block after collision is `sqrt("v"^2 - "v"_1^2)`.