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Question

# A bullet of mass 50 gm is fired from below onto a bob of mass 450 gm of a long simple pendulum as shown in the figure. The bullet remains inside the bob and the bob rises through a height of 1.8 m vertically upwards. Find the speed of the bullet.

A
60 m/s
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B
80 m/s
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C
40 m/s
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D
20 m/s
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Solution

## The correct option is A 60 m/sLet the speed of the bullet be v just before collision with the bob and velocity of system after collision=v′ Given m=50 gm=mass of bullet and M=450 gm= mass of bob T=0 in the string after collision, Considering (bullet+bob) as a system and ignoring the external force of gravity during the very short time interval of collision. Hence, we can apply conservation of momentum in vertical direction for colliding systems: ⇒Pi=Pf mv+(M×0)=(M+m)v′ ∴v′=mvM+m=50×v450+50=0.1v...(i) Now, taking vertically upwards as +ve y axis, S=+1.8 m, vy=0, uy=v′ ay=−g m/s2 Applying kinematic equation from initial position to final position of (bullet+bob) system: v2y=u2y+2ayS ⇒02=v′2−2g×1.8 ⇒(0.1v)2=36 ∴v=√3600=60 m/s

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