Mechanical Engineering - Engineering Mechanics - Discussion

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.

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.

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.

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'.

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 w = u - v ----->(1).

From conservation of kinetic energy,

1/2 mu2 = 1/2 mv2 + 1/2 mw2,

u2 = v2 + w2 ----->(2).

(1) in (2),

u2 = v2 + (u - v)2.

Simplifying gives 2 uv = 2 v2.

v = u.

In (1), w = 2u.

After collision ball 1 return with the same velocity and other attain a twice velocity.

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 w = u - v ----->(1).

From conservation of kinetic energy,

1/2 mu2 = 1/2 mv2 + 1/2 mw2,

u2 = v2 + w2 ----->(2).

(1) in (2),

u2 = v2 + (u - v)2.

Simplifying gives 2 uv = 2 v2.

v = u.

In (1), w = 2u.

After collision ball 1 return with the same velocity and other attain a twice velocity.

Velocity after striking will be v/2.
Because it is a perfectly elastic body so after striking both are will move in an opp direction so by applying momentum equation.

Total momentum before striking = Total momentum After strike.

m(v1i+v2i) = m(v1f+v2f).
@ first V2i = 0.

mV = m(v1+V2) Perfectly elastic so V1f=V2F.
V=2Vf.
V(final)=V/2.

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.

You are right guyz. It will follow conservation of momentum. But if it is perfectly elastic then the 1st ball should come at rest after colliding with the 2nd one. That means final velocity of 2nd ball will change from 0 to V. How it is v/2? Maybe I'm missing something. Please correct me guyz.

For two perfectly elastic body, the velocity after collided become same i.e. velocity 1st ball after collide equal to velocity of 2nd ball after collide.

So final velocity of the ball half of the initial velocity. It will get from the equation of conservation of momentum after and before.