Civil Engineering - Applied Mechanics - Discussion

23. 

The masses of two balls are in the ratio of 2 : 1 and their respective velocities are in the ratio of 1 : 2 but in opposite direction before impact. If the coefficient of restitution is , the velocities of separation of the balls will be equal to

[A]. original velocity in the same direction
[B]. half the original velocity in the same direction
[C]. half the original velocity in the opposite direction
[D]. original velocity in the opposite direction

Answer: Option D

Explanation:

No answer description available for this question.

Hari Har said: (Jul 21, 2016)  
Someone please explain this to me.

Snehal Wankhede said: (Aug 30, 2016)  
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.

Kush Kakkar said: (Jan 17, 2017)  
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).

Asha said: (May 9, 2017)  
I think Option C is correct.

Lasith said: (Jul 27, 2018)  
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.

Keshav said: (Jun 17, 2019)  
C should be the answer, since v2-v1= -u/2 ie half the original velocity of the first ball in opposite direction.

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