Question

A mass m is placed at the rim of a frictionless

Answer 1

Answer:

Explanation:

at collision, m has a velocity of √2gR

The center of mass of the system has a velocity of

m(√2gR) + 3m(0) = (m + 3m)v

v = ¼√2gR

The stationary mass (3m) will depart at twice the system center of mass velocity

v(3m) = ½√2gR   ANSWER B

The relative velocity before the collision will be the negative of the relative velocity after the collision.

v(m) = ½√2gR - √2gR = -½√2gh ANSWER A

m will rise to a height of v²/2g = (-½√2gR)²/2g = 0.25R