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 velocity
v
v/2
v/4
v/8
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
70 comments Page 5 of 7.

Sampad kumar Rout said:   1 decade ago
For perfectly elastic body coefficient of restitution =1 so first body will stop and 2nd will move in a velocity v.

Ramana said:   1 decade ago
Two balls are same mass and same nature with the contact motion so each one have half of the original velocity.

B.Lavanya said:   4 years ago
Momentum before collision = momentum after collision.
m*(v+0) = m*Vnew+m * Vnew
mv = 2(m*Vnew).
Vnew = v/2.
(73)

Manohar said:   1 decade ago
Total momentum before and after collision is constant. Hence,

m(v+0) = m(v'+v').

mv = 2mv'.

v' = v/2.

Anoima said:   7 years ago
I think only the second mass will move with velocity we and the first mass will get into rest.
(2)

Sandeep said:   1 decade ago
There is no energy loss in perfectly elastic collision and hence momentum will be conserved.

Pavan said:   9 years ago
mv1 + mv2 = mv'1+mv'2.
m is equal.

v1 + 0 = 2v' because v1 = 0 and v'1 = v'2.
v' = v/2.

Uzzal said:   9 years ago
mv1_+mv2 = mv'1+mv'2.

m is equal.

v1+0 = 2v' because v1 = 0 and v'1 = v'2.

v' = v/2.

Nazik said:   1 decade ago
mv1_+mv2 = mv'1+mv'2.
m is equal.

v1+0 = 2v' because v1 = 0 and v'1 = v'2.
v' = v/2.

Deepak said:   10 years ago
The total energy of momentum of force is directly impact v = 1/v, it is impact force.


Post your comments here:

Your comments will be displayed after verification.