63
Question
A bomb at rests explodes into three parts of equal masses. Two parts fly off at night angles to each other with velocities of 9 m/s and 12 m/s. Third part is moving opposite to above system. find the velocity of third part.
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Solution
Since there is no external force on the system its momemtum will be conserved⇒ Initial Momemtum = Final Momentum⇒0=m3(9^i)+m3(12^j)+m3→v
⇒m3(9^i)+m3(12^j)=−m3→v
⇒→v=−9^i−12^j
⇒ Initial Momemtum = Final Momentum
⇒0=m3(9^i)+m3(12^j)+m3→v
⇒m3(9^i)+m3(12^j)=−m3→v
⇒→v=−9^i−12^j
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