Question

# A bomb of mass 12 kg is dropped by a fighter plane moving horizontally with a speed of 100 msâˆ’1 from a height of 1 km from the ground. The bomb exploded after 10 s into two pieces of masses in the ratio 1:5. If the small part started moving horizontally with a speed of 600 msâˆ’1, the speed of bigger part will be (g=10 msâˆ’2)

A
100 ms1
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B
1065 ms1
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C
120 ms1
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D
1002 ms1
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Solution

## The correct option is C 120 ms−1Given that, Initial horizontal velocity of bomb ux=100 m/s Initial vertical velocity of the bomb uy=0 m/s Vertical velocity of bomb after 10 s will be vy=uy+gt vy=gt=100 m/s m1:m2=1:5 m1+m2=12 kg ∴ m1=2 kg and m2=10 kg Applying conservation of linear momentum in X−direction mux=m1vx1+m2vx2 12 kg×100 m/s=2 kg×600 m/s+10 kg vx2 ⇒vx2=0 Applying conservation of linear momentum in Y− direction, mvy=m1vy2 12kg×100 m/s=2 kg×0+10 vy2 vy2=120 m/s,v2=√v2x2+v2y2=120 m/s ∴(c)

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