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JEE MainPhysicsModerateCentre of Mass

A particle of mass 1 kg is thrown vertically upwards with speed
100 m/s. After 5s it explodes into twoparts. One part of mass 400g comes
back with speed 25 m/s, what is the speed of other part just
afterexplosion?

Show answer and solution
Velocity of particle after 5 s
v = u - gt
v = 100 - 10 × 5
= 100 - 50 = 50 m/s (upwards)
Conservation of linear momentum gives
Mv = m1v1 + m2v2......(i)
Taking upward direction positive, the velocity v_1 will be negative.
∴v1=−25m/s,v=50m/s\therefore v_{1} = - 25m/s,v = 50m/s
 \ Also M=1kg,m_1=400g=0.4kg
and m_2=(M - m_1)=1 - 0.4=0.6kg
Thus,Eq.(i) becomes,
1∗50=0.4×(−25)+0.6v21*50 = 0.4 \times ( - 25) + 0.6v_{2}
or 5\ 5`<!- ->`0= - 10+0.6v_2
or 0.6v_2=60
orv2=600.6=100m/sv_{2} = \frac{60}{0.6} = 100m/s
As v_2 is positive, therefore the other part will move upwards with a
velocity 100 m/s.

Source: Grass-G / P-07 Physics paper 17 October.docx

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