Question

A particle A of mass $m$ initially at rest slides down a height of 1.25 m on a friction-less ramp, collides with and sticks to an identical particle B of mass $m$ at rest as shown in figure.
Then, particle A and B together collide elastically with particle C of mass 2 m at rest. The speed of particle C after the collision with combined body (A = B) would be $(Take,g=10m{s}^{-2})$

• A. $2m{s}^{-1}$
• B. $1.25m{s}^{-1}$
• C. $2.5m{s}^{-1}$
• D. $5m{s}^{-1}$

Answer 1

The correct option is C $2.5m{s}^{-1}$

Velocity of A just before collision$=\sqrt{2gh}=\sqrt{2\times 10\times 1.25}=5m{s}^{-1}$

Velocity of (A+B) just after collision $=5/2=2.5m{s}^{-1}$

(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$m{s}^{-1}$

Hence option 'C' is correct.