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Question

Two blocks of masses m1 and m2 are connected by a light inextensible string which passes over a fixed pulley. Initially, mass m1 moves with a velocity v0 when the string is not taut. Neglect friction at all contact surfaces. Find the velocity of the blocks just after the string is taut, if m1=2 kg, m2=3 kg and v0=5 m/s


A
2 m/s
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B
3 m/s
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C
5 m/s
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D
0 m/s
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Solution

The correct option is A 2 m/s
Tension developed in both parts of the string will be same just after the string become taut.
Hence, impulsive forces acting on both blocks will be the same
J1=J2=J ..(i)

Because of string constraint, speed of both blocks will be equal. Let it be v just after the string becomes taut.


Applying impulse-momentum theorem (taking positive direction rightward):
J=m Δv
For block m1:
J=m(vfvi)=2[(+v)(+v0)]=2(vv0)
J=2(v5) ..(ii)

For block m2:
J=m2(vfvi)=3(v0)=3v ...(iii)

Equating Eq. (ii) and (iii):
2v10=3v
v=105=2 m/s

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