Mechanical Engineering - Engineering Mechanics - Discussion

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).

@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.

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'.

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.

Cconservation of momentam.
mv=m1v1+m2v2.
mv=mv+mv.
v=v/2.

If the second mass it at rest initially. After collision 1st mass gets rest second mass move with a velocity ' V'. NOT By both ' V/2'.

We can not take velocity only based on momentum. we need to take in condition loss of kinetic energy which is zero in this case.

So we have two equation
V = V1 + V2 --------------> 1

Square of V =Square of V1 +Square of V2 --------------> 2
Solve this, you will get V1=0 and V2=V.

@ALL.

It is the fact is it is a perfectly elastic collision so no kinetic energy losses. So the velocity of the 1st body will transfer to next body. And the answer will be V.

Ans. : V/2 is correct.