Question

A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.6 kg and 2.3 kg, and the length of the wire is 1.21 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

Answer 1

(a) Here, use conservation of energy to find out the velocity of the ball just before the collision -

v = ?(2gh) = ?(2 * 9.8m/s² * 1.21m) = 4.9 m/s
See drawing for direction. We will call it positive.

(b) Now, as mentioned in the problem, the collision is an elastic head-on collision, we know (from CoE) that
the relative velocity of approach = relative velocity of separation,

so -  
4.9 m/s = v - u
v = 4.9 + u ----------------------------------(i)
where v, u are the post-collision velocities of the block, ball respectively

Conserve momentum: initial p = final p
1.6kg * 4.9 m/s = 1.6kg * u + 2.3 kg * v

substitute the value of v from (i) above -

7.84 = 1.6u + 2.3*(4.9 + u) = 11.3 + 3.9u

=> -3.46 = 3.9u
=> u = -0.89 m/s (I'll call this "negative") ? ball
v = 4.9 + u = 4.01 m/s ("positive") ? block

Therefore, velocity of the ball just after the collision is -0.89 m/s. The negative sign indicates that its direction is opposite to the initial direction.