A particle A of mass m initially at rest slides down a height of 1.25 m on a frictionless ramp, collides with and sticks to an identical particles B of mass m at rest as shown in the figure.

Then, particles A and B together collide elastically with particle G of mass 2m at rest. The speed of particle G after the collision with combined body (A+B) would be (Take, g`=10 ms^(-2))`

Options

A. `2ms^(-1)`
B. `1.25 ms^(-1)`
C. `2.5 ms^(-1)`
D. `5 ms^(-1)`

Answer

C

Explanation

Velocity of A just before collision
`=sqrt(2gh) = sqrt(2 xx 10 xx 1.25` = `5ms^(-1)`
Velocity of `(A+B)` just after collision = 5/2 = 2/5ms^(-1)
(From conservation of linear momentum)
In elastic collision between two bodies of equal masses velocities are interchanged.
Hence, velocity of C will becomes `2.5 ms^(-1)`