A block A of mass 2kg moving with speed of 15m/s, collides with a stationary block B of mass 5kg. If the coefficient of restitution is 0.6, find the magnitude of velocity of blocks A and B after the collision respectively.
A
5.2m/s and 10.5m/s
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B
3.5m/s and 6.85m/s
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C
10.5m/s and 5.2m/s
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D
2.15m/s and 6.85m/s
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Solution
The correct option is D2.15m/s and 6.85m/s Mass of block A, mA=2kg
Mass of block B, mB=5kg
Let uA&uB be the velocity before collision of block A and block B. uA=15m/s,uB=0m/s
Let vA&vB be the velocity after collision of block A and block B.
Applying conservation of linear momentum before and after collision. mAuA+mBuB=mAvA+mBvB 2×15+5×0=2×vA+5×vB 2vA+5vB=30 ... (1)
Coefficient of restitution (e)=velocity of separationvelocity of approach 0.6=vB−vAuA−uB⇒0.6=vB−vA15−0 0.6×15=vB−vA vB−vA=9 ... (2)
Multiplying by 2 in equation (1) 2vB−2vA=18 ... (3)
Adding equation (1) and equation (3), we get 7vB=48 vB=6.85m/s vB=6.85m/s put in equation (2) vA=−2.15m/s ∴ Magnitude of velocities of ball A and ball B after collision is 2.15m/s&6.85m/s
The correct option is (d).