Ball B is hanging from an inextensible cord. An identical ball A is released from rest when it is just touching the cord and drops through the vertical distance hA = 8 in. before striking ball B. Assuming e = 0.9 and no friction, determine the resulting maximum vertical displacement hB of ball B.

The image of the ball hanging from the chord is missing, so i have attached it.

hb = 0.1016 m

Explanation:

We are given;

Height from which block A is dropped;hA = 8 in = 0.2032 m

Coefficient of restitution;e = 0.9

Now, let us make v_a and v_b the velocities of balls A and B respectively after collision.

If we assume that both balls have the same masses, then from conservation of momentum, v_a = -v_b

Thus;

½m(v_a)² + ½m(v_b)² = m•g•hA

m will cancel out, also, putting -v_b for v_a, we have;

½(-v_b)² + ½(v_b)² = g•hA

½(v_b² + v_b²) = g•hA

½(2v_b²) = g•hA

v_b² = g•hA

v_b = √g•hA

v_b = √(9.81 × 0.2032)

v_b = 1.412 m/s

Now, using conservation of total mechanical energy, we have;

m•g•hb = ½mv_b²

Making hb the subject, we have;

hb = ½v_b²/g

hb = 1.412²/(2 × 9.81)

hb = 0.1016 m

measuring things 1 litre of water