(^570) „„„„„ A Textbook of Engineering Mechanics
Fig. 27.7.
(a) Direction of the body after impact
Let θ = Angle, which the final
velocity makes with the
line of impact, and
v = Final velocity of the body
after impact.
We know from the law of conservation of momentum
that
u sin α = v sin θ
∴ v sin θ = u sin α = 4 sin 60° = 4 × 0.866 = 3.464 ...(i)
We also know from the law of collision of elastic bodies that
v cos θ = e × u cos α = 0.5 × 4 cos 60° = 2 × 0.5 = 1 ...(ii)
Dividing equation (i) by (ii),
sin 3.464
cos 1
v
v
θ

θ
∴ tan θ = 3.464 or θ = 73.9° Ans.
(b) velocity of the body after impact
Substituting the value of θ in equation (ii),
v cos 73.9° = 1
or
11
3.6 m/s
cos 73.9 0.2773
v===
°
Ans.
EXERCISE 27.2

1. A ball is dropped from a height of 25 metres upon a horizontal floor. Find the coefficient
   of restitution between the floor and the ball, if it rebounds to a height of 16 metres.
   (Ans. 0.8)


2. A 1 kg ball traverses a frictionless tube as shown in Fig. 27.8.

Fig. 27.8.

The ball, after falling through a height of 1.2 metres, strikes a 1.5 kg ball hung on a rope.

Find the velocities of the two balls, if the collision is perfectly elastic.

(Ans. v 1 = – 0.97 m/s ; v 2 = 3.88 m/s)

3. A heavy elastic ball drops from the ceiling of a room, and after rebounding twice from the
   floor reaches a height of equal to one-half of the ceiling. Show that the coefficient of
   restitution is (0.5)1/4.