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Question

A block A of mass 2 kg moving with speed of 15 m/s, collides with a stationary block B of mass 5 kg. If the coefficient of restitution is 0.6, find the magnitude of velocity of blocks A and B after the collision respectively.

A
5.2 m/s and 10.5 m/s
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B
3.5 m/s and 6.85 m/s
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C
10.5 m/s and 5.2 m/s
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D
2.15 m/s and 6.85 m/s
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Solution

The correct option is D 2.15 m/s and 6.85 m/s
Mass of block A, mA=2 kg
Mass of block B, mB=5 kg
Let uA & uB be the velocity before collision of block A and block B.
uA=15 m/s,uB=0 m/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=vBvAuAuB0.6=vBvA150
0.6×15=vBvA
vBvA=9 ... (2)
Multiplying by 2 in equation (1)
2vB2vA=18 ... (3)
Adding equation (1) and equation (3), we get
7vB=48
vB=6.85 m/s
vB=6.85 m/s put in equation (2)
vA=2.15 m/s
Magnitude of velocities of ball A and ball B after collision is 2.15 m/s & 6.85 m/s
The correct option is (d).

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