First slide

Third law

Question

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