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Q.

A particle of mass 1 kg is thrown vertically upward with speed 100 ms-1 . After 5 s, it explodes into twoparts. One part of mass 400 g emerges with speed 25 ms-1in downward direction, what is the velocity of other part just after explosion? (Take, g = 10 ms-2)

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a

100 ms-1 upward

b

600 ms-1 upward

c

100 ms-1 downward

d

300 ms-1 upward

answer is A.

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Detailed Solution

Velocity of particle after 5 s,               v=u=gt=100-10×5                  =100-50=50 ms-1                           (upwards)Conservation of linear momentum gives               Mv = m1v1+m2v2                                             …(i)Taking upward direction positive,                   v1=-25ms-1, v=50 ms-1                   M = 1kg, m1=400 g = 0.4 kg                   m2 = M-m1=1-0.4=0.6 kgFrom Eq. (i), we get           1×50=0.4×(-25)+0.6v2  or v2 = 100 ms-1  (upwards)
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