First slide

Law of conservation of momentum and it's applications

Question

A bomb of mass 9kg explodes into 2 pieces of mass 3kg and 6kg. The velocity of mass 3kg is 1.6 m/s, the K.E. of mass 6kg is

Moderate

A

3.84 J
B

9.6 J
C

1.92 J
D

2.92 J

Solution

As the bomb initially was at rest therefore
Initial momentum of bomb = 0
Final momentum of system = $m_{1} v_{1} + m_{2} v_{2}$
As there is no external force
$∴ m_{1} v_{1} + m_{2} v_{2} = 0 \Rightarrow 3 \times 1 .6 + 6 \times v_{2} = 0$
velocity of 6 kg mass $v_{2} = 0 .8 m / s$ (numerically)
Its kinetic energy $= \frac{1}{2} m_{2} v_{2}^{2} = \frac{1}{2} \times 6 \times {(0 .8)}^{2} = 1 .92 J$

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