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Question

# A bullet of mass M is fired with a velocity of 50 ms−1 at an angle θ with the horizontal. At the highest point of trajectory it collides with a bob of mass 3M suspended vertically by a massless string of length 103m and gets embedded into it. After the collision the string moves through an angle 120∘, what is the angle of throw θ.

A
cos125
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B
cos135
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C
cos145
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D
cos115
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Solution

## The correct option is B cos−145Before collision, bullet momentum at A=M(50)cosθAfter collision, bob momentum at A=(3M+M)vaccording to conservation of linear momentum M(50)cosθ=(3M+M)vorv=50cosθ4velocity v is responsible for rotating the bob in vertical circle and raises it to height =h=103+103sin30=5mAlso kinetic energy at A is getting converted into potential energy at B, so according to law of conservation of energy, 12(4M)v2=(4M)ghor12(50cosθ4)2=10×5 orcos2θ=1625orθ=cos−145

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