Civil Engineering - Applied Mechanics - Discussion

The correct answer is option C i.e. half the original velocity in the opposite direction.

In Question Given that opposite direction i.e head on Collision.

Let m1, u1 for 1st body moving towards second body and
m2,u2 for a second body moving towards 1st body

Therefore after colliding final velocities are v1 & v2 respectively
As per the law of conservation:

Initial change in momentum = Final change in momentum
m1u1-m2u2 = m2v2-m1v1.
2m2u1-m2u2 = m2v2-2m2v1 (given m1=2m2).
m2(2u1-u2) = m2(v2-2v1).
2u1-2u1 = v2-2v1 (given u2=2u1).
0 = v2-2v1.
v1/v2 = 1/2 ie equal to ratio of original velocity.

So, Option D is correct.

C should be the answer, since v2-v1= -u/2 ie half the original velocity of the first ball in opposite direction.

The mass of 1st ball (m1)=2M, mass of 2nd ball (m2)= M
initial velocity of 1st ball (u1)= U, initial velocity of 2nd ball (u2)= -2U { -ve sign because of opp direction.

The coefficient of restitution=1/2,
let, v1 = final velocity of 1st ball, v2= final velocity of the 2nd ball
from the law of conservation of momentum,
m1u1+m2u2=m1v1+m2v2.
2MU+M(-2U)=2Mv1+Mv2.
0=2Mv1+Mv2.
v2=-2v1.

From the law of collision of elastic bodies that;
(v2-v1)=e(u1-u2).
On solving v1=-1/2 U.

So negative sign indicates that the direction of v1 is opposite to that of U thus 1st ball will move back with 1/2 of its original velocity.

I think Option C is correct.

Coefficient of restitution = Relative velocity before collision / relative velocity after collision
1:2 = 1:2/ relative velocity after collision.

The relative velocity after collision = (1:1).

I think it's option C.
Using,
1) m1v1 + m2v2 = m1v1' + m2v2'.
2) e = -(v2' - v1')/(v2 - v1).

Answer v2' = -v2/2, v1' = -v1/w2.

Someone please explain this to me.