Question

Two balls, of masses m₁ and m₂ , are suspended side by side from strings of equal lengths. The ball m₁ is pulled to the left to a height ‘h’ and ‘let go’. When it strikes the second ball, it gets stuck to it and the ‘combination’ now rises to the right through a height (h/n). The ratio (m₂/m₁), of the masses of the two balls, equals'

(1) (√n - 1)

(2) √n

(3) (n - 1)

(4) n

Answer 1

(1) (√n - 1)

The velocity, of the ball m₁ , just before it strikes the ball m2 (initially at rest) is √2gh .

The law of conservation of momentum yields.

Where v is the velocity of the ‘combination’ just after the ‘collision’. Hence

The K.E. of the ‘combination’, just after the collision, makes it rise to a height h/n . We, therefore, have