# Mechanical Engineering - Engineering Mechanics - Discussion

Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 3)

3.

Two balls of equal mass and of perfectly elastic material are lying on the floor. One of the ball with velocity

*v*is made to struck the second ball. Both the balls after impact will move with a velocityDiscussion:

70 comments Page 1 of 7.
VVG said:
1 year ago

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.

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.

(6)

B.Lavanya said:
2 years ago

Momentum before collision = momentum after collision.

m*(v+0) = m*Vnew+m * Vnew

mv = 2(m*Vnew).

Vnew = v/2.

m*(v+0) = m*Vnew+m * Vnew

mv = 2(m*Vnew).

Vnew = v/2.

(43)

Akif said:
4 years ago

In pool tables, though they aren't perfectly elastic, there can be a few results. So, we can assume for a situation where both the balls have the same velocity v/2 would be their velocity.

(1)

Arpit said:
4 years ago

Elastic material means that the ability to regain original position.

(1)

Kottalamuthu said:
4 years ago

What is an elastic material?

Ravichandrakumar.K.B said:
5 years ago

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.

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.

(13)

Dhruvil Prajapati said:
5 years ago

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.

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.

(3)

Anoima said:
5 years ago

I think only the second mass will move with velocity we and the first mass will get into rest.

(2)

AJEESH said:
5 years ago

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.

Wravin111 said:
6 years ago

m1u1+m2u2=(m1+m2)v2.

u2=0 as it's stationary.

So,

mu/2m=v2,

1u/2=v2.

u2=0 as it's stationary.

So,

mu/2m=v2,

1u/2=v2.

(1)

Post your comments here:

Your comments will be displayed after verification.

Quick links

Quantitative Aptitude

Verbal (English)

Reasoning

Programming

Interview

Placement Papers

© IndiaBIX™ Technologies