Engineering Mechanics

Chapter 27 : Collision of Elastic Bodies „„„„„ 565

We know from the law of collision of elastic bodies that

v 2 cos θ 2 – v 1 cos θ 1 = e (u 1 cos α 1 – u 2 cos α 2 ) = 0.5 (3 cos 30° – 1 cos 30°)

= 0.5 (3 × 0.866 – 1 × 0.866)

v 2 cos θ 2 – v 1 cos θ 1 = 0.866 ...(iv)

Adding equations (iii) and (iv),

3 v 2 cos θ 2 = 5.196

or v 2 cos θ 2 = 1.732 ...(v)

Dividing equation (ii) by (v),

22

0.5

tan 0.2887 or 16.1

1.732

θ= = θ= ° Ans.

(b) Velocity of the 4 kg ball after impact

Substituting the value of θ 2 in equation (ii),

v 2 sin 16.1° = 0.5

∴ 2

0.5 0.5

1.803 m/s

sin 16.1 0.2773

v ===

°

Ans.

(c) Direction, in which the 2 kg ball will move after impact

Substituting the values of θ 2 and v 2 in equation (iv),

1.803 cos 16.1° – v 1 cos θ 1 = 0.866

or v 1 cos θ 1 = 1.803 cos 16.1° – 0.866

= (1.803 × 0.9608) – 0.866 = 0.866 ...(vi)

Dividing equation (i) by (vi)

11

1.5

tan 1.732 or 60

0.866

θ= = θ= °^ Ans.

(d) Velocity of 2 kg ball after impact

Now substituting the value of θ 1 in equation (i),

v 1 sin 60° = 1.5

∴ 1

1.5 1.5

1.732 m/s

sin 60 0.866

v ===

°

Ans.

27.10. DIRECT IMPACT OF A BODY WITH A FIXED PLANE

In the previous articles, we have been discussing the impacts of two bodies. Both these bodies
had some initial velocities, and after impact they had some final velocities (in particular cases, some
of these velocities were zero also). But in the following articles, we shall discuss the impact of a body
with a fixed plane.

Now consider a body having a direct impact on a fixed plane.
Let u = Initial velocity of the body,
v = Final velocity of the body, and
e = Coefficient of restitution.
We know that the fixed plane will not move even after impact. Thus the velocity of approach
is equal to (u) and velocity of separation is equal to (v). The Newton’s Law of Collision of Elastic
Bodies also holds good for this type of impact. i.e.
v = eu