Mechanical Engineering - Engineering Mechanics - Discussion

Conservation of momentum

Consider
U-initial velocity
V-final velocity
m1u1+m2u2=m1v1+m2v2
m(u1+0)=m1(v1+v2)
u1=v1+v2

Therefore finial velocity equally distributed

So u=2v.
V=u/2.

See, there is a concept of coefficient of restitution(e).

i.e e= Velocity of separation/ velocity of approach.
While here velocity of approach is V.
and for the perfectly elastic case, e=1;
Now, let's assume after impact velocity of ball 1 and ball 2 id V1 and V2.
Then, by the coefficient of restitution concept: velocity of separation = velocity of approach.
i.e V = V2 - V1 -----> (1)
and by conservation of momentum; mV = mV1 + mV2 => V1 + V2 = V -----> (2)
after solving eq(1) and eq (2);
you will get V2 = V and V1 =0 ;

In Nutshell, after collision ball 1 will come to rest and ball 2 will gain the velocity 'V'.

I think it would be v/√2.

If the second body before struck were in rest. Must answer should be v/√2. (kinetic energy should be equal before and after collision).

This is considered as inelastic collision so both will move in same velocity, V/2, If the collision was in the vacuum, ie perfectly elastic collision, V1F will be 0 and V2F will be V.

What is an elastic material?

Two balls are same mass and same nature with the contact motion so each one have half of the original velocity.

Velocity of first moving ball is v after impact its velocity becomes half therefor every action the is equal and opposite reaction so velocity of 2nd ball is also v/2.

There is no energy loss in perfectly elastic collision and hence momentum will be conserved.

The first ball will be at rest and the second will have a velocity v unless the collision angle is not zero, then in a certain angle of collision the two balls will have the same velocity.

Total mass=mv'+mv' therefore v=2v' and v'=v/2.