Mechanical Engineering - Engineering Mechanics - Discussion

Ans. : V/2 is correct.

@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.

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.

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'.

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

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'.

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.

@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.

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).