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Question

A cracker is thrown into air with a velocity of 10 m/s at an angle of 45o with the vertical. When it is at a height of 0.5 m from the ground, it explodes into a number of pieces which follow different parabolic paths. What is the velocity of centre of mass, when it is at a height of 1 m from the ground? (g=10m/s2)

A
45m/s
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B
25m/s
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C
44m/s
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D
10 m/s
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Solution

The correct option is A 45m/s
Since cracker explodes in mid air it does not encounter any external force so velocity of COM will remain unchanged due to explosion.
Horizontal component =10sin45o=102=52m/s
From third equation of motion, at a height of 1 m, vertical component of velocity of center of mass =(52)22×10×1=5020=30m/s
Hence, net velocity =(52)2+(30)2=80=45m/s

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