Mechanical Engineering - Engineering Mechanics - Discussion

Two balls have same mass and elastic property.

So e = 1,

Momentum before impact = momentum after impact.

mv+m*0 = mv1+mv1 {after impact v1 = v2}.

Hence, v1 = v/2.

As here one velocity is responsible for motion of two equal mass body so, the velocity will distribute.

As we know the co-efficient of restitution(e=1) for perfectly body.

mv = mv1+mv2.
mv = mv'+mv'.

v' = v/2.

Conservation of momentom
mv=m1v1+m2v2
mv=mv`+mv`
v`=v/2

Initial momentum = final momentum.

Let the initial velocity of first mass be u.
Final velocities attained by first and second masses be v and w respectively.

From conservation of momentum,
mu = mv + mw
We can have v = u - w ........(1).

From conservation of kinetic energy,

1/2mu2 = 1/2mv2 + 1/2mw2 NB: u2 is read u square.
u2 = v2 + w2................(2).
(1) in (2),
u2 = (u - w)2 + w2.
simplifying gives 2uw = 2w2.

w = u.
in (1), v = 0.

This simply means that when the first mass collides with the second, it comes to rest and the second mass takes off with the initial velocity of the first.

The analysis is simple. This was proven by Newton and the experiment he used is called Newton's Craddle.

From conservation of momentum.

Momentum before impact = momentum after impact.

mv+m*0 = mv1+mv1 { after impact v1=v2}.

Hence,
v1 = v/2.

For perfectly elastic body coefficient of restitution =1 so first body will stop and 2nd will move in a velocity v.

We know that total momentum remains constant i.e., m1*v1=m2*v2 given that mass is constant m1=m2 then v1=v2 so the 1st ball gives half of its velocity to 2nd.

After collision the second body will move by velocity 'v' because first body will remain rest and second will move.

One of the body which was moving initially with velocity V would come to rest and second body will move with velocity v.

mv +m *0 = m*0 + mv.

For perfectly elastic collision velocity gets exchanged and e = 1.