Mechanical Engineering - Engineering Mechanics - Discussion

Momentum before collision = momentum after collision.
m*(v+0) = m*Vnew+m * Vnew
mv = 2(m*Vnew).
Vnew = v/2.

Let's' be mass of ball 1 & 2.
The initial velocity of ball 1 and 2 as 'u1' and 'u2'.
Final velocity of ball 1 and 2 as ' v1' and 'v2'.

According to linear conservation of momentum.

m*u1+m*u2 = m*v1 + m*v2.
Ball 2 is at rest so, u2=0,
Therefore on simplifiaction we get,
u1= v1+v2 ------> eqn 1.

Elasticity is 1, coefficient of restitution 'e' is 1.
e=(v2-v1)/(u1-u2),
1=(v2-v1)/u1,
Therefore, u1=v2-v1 -------> eqn 2.

Solving eqn1 & 2 we get,
v2 = u1 & v1 = 0.
So after impact, the v is transferred to the second ball and the first ball comes to rest.

Note: Actually the way the question is asked is wrong and it is said it has both elasticities as 1 and assumed v1 =v2 which doesn't satisfy the coefficient of restitution formula, it contradicts each other.

For elastic collision, the value of the coefficient of restitution is 1.

If both the balls are moving with the same velocity after the impact, then the relative velocity after the collision will be 0, and the coefficient of restitution also become 0. The coefficient of restitution is 0 only for perfect plastic collision.

The answer given is incorrect.

Reason:
A perfectly elastic ball on striking with another perfectly elastic ball will cause the second ball to move with velocity with which the first ball was approaching and the first ball will come to rest after the collision.

Both Energy, as well as momentum conservation, has to be applied.

Total Kinetic Energy (T.E) = 1/2 mv1^2
Final K.E = T.E
Therefore, 1/2mv1^2 = 1/2mv1'^2 + 1/2mv2'^2 ------> (1)
Also mv1 = mv1' + mv2' ------>(2)

From (2) v1 = v1' + v2' ------> (3)

Substituting (3) in (1)

(v1' + v2')^2 = v1'^2 + v2'^2.
v1'^2 + 2v1'v2' + v2'^2 = v1'^2 + v2'^2.
v1'v2' = 0.

So, either v1'= 0 or v2' = 0.
But since v2' cannot be zero after the collision (since m2 receives an impulse from m1)

Hence, v1' = 0.
On substituting the value of v1' in eqn (3) v2' = v1.

Hence Proved.

Elastic material means that the ability to regain original position.

I think only the second mass will move with velocity we and the first mass will get into rest.

In pool tables, though they aren't perfectly elastic, there can be a few results. So, we can assume for a situation where both the balls have the same velocity v/2 would be their velocity.

m1u1+m2u2=(m1+m2)v2.

u2=0 as it's stationary.

So,
mu/2m=v2,
1u/2=v2.

@ALL.

As per my knowledge, after the collision, both balls are moving with some velocity so that the velocity of balls after the collision is v/2.

Aman is right but in question given that after collision both balls move with the same velocity then use that concept in momentum eq;

mv = mv1+mv1,
so v1=v/2.