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Question

Two balls A and B of masses 2 kg each are moving with speeds 1 m/s and 2 m/s on a frictionless surface. After colliding, the ball A returns back with speed 0.5 m/s, then the impulse of reformation of ball B is



Your Answer
A
2 Ns
Correct Answer
B
0 Ns
Your Answer
C
4 Ns
Your Answer
D
4 Ns

Solution

The correct option is B 0 Ns
Given, 
mA=2 kg;mB=2 kg
uA=1 m/s;uB=2 m/s
vA=0.5 m/s
(taking rightward +ve)
Applying PCLM,
mAuA+mBuB=mAvA+mBvB
2(1)+2(2)=2(0.5)+2(vB)
vB=0.5 m/s

Common velocity of balls during collision, v=m1u1+m2u2m1+m2=2(1)+2(2)4
=12 m/s

Impulse of reformation of ball B
Jr=mB(vBv)
=2(12+12)
= 0 Ns

This is because the balls A and B after collision move with the same velocity i.e perfectly inelastic collision.

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