A body of mass 300 g kept at rest breaks into two parts due to internal forces. One part of mass 200 g is found to move at a speed of 12 m/s towards the east. What will be the velocity of the other part?

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Solution

Dear Student ,
We know that the momentum is always conserved if there is no external force
Here all the forces are internal so we can apply momentum conservation
initially the mass was M=300 g and initial speed , u=0
Now the mass M is broken into two parts M₁and M₂
And the velocities respectively is v₁ =12m/s towards east and v₂
where M₁=200g and M₂=100 g
Now apply the conservation principle ,
$P_{i} = P_{f} M * 0 = M_{1} v_{1} + M_{2} * v_{2} M_{1} v_{1} = - M_{2} v_{2} 100 * v_{2} = - 200 * 12 m / s (t o w a r d s e a s t) S o, v_{2} = - 24 m / s t o w a r d s e a s t$
Here negative sign indicates that the velocity of second particle is opposite to the direction of first particle .
So the second particle is moving at a speed of 24m/s towards west .

Regards

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