Two balls A and B of masses 2kg each are moving with speeds 1m/s and 2m/s on a frictionless surface. After colliding, the ball A returns back with speed 0.5m/s, then the impulse of reformation of ball B is
A
−2Ns
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B
0Ns
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C
4Ns
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D
−4Ns
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Solution
The correct option is B0Ns Given, mA=2kg;mB=2kg uA=1m/s;uB=−2m/s vA=−0.5m/s (taking rightward +ve) Applying PCLM, mAuA+mBuB=mAvA+mBvB 2(1)+2(−2)=2(−0.5)+2(vB) ⇒vB=−0.5m/s
Common velocity of balls during collision, v=m1u1+m2u2m1+m2=2(1)+2(−2)4 =−12m/s
Impulse of reformation of ball B Jr=mB(vB−v) =2(−12+12) = 0Ns
This is because the balls A and B after collision move with the same velocity i.e perfectly inelastic collision.