A bomb of mass 21kgs kept at rest explodes in two parts in ratio of masses 2:5, the lighter part travels with a speed of 50m/s then heavier mass travels with a speed (in m/s) of
Answers
Given : a bomb of mass 21 kgs kept at rest explodes in two parts in ratio of masses 2:5 ,
the lighter part travels with a speed of 50 m/s
To Find : heavier mass travels with a speed (in m/s)
Solution:
bomb of mass 21 kgs
ratio of masses 2:5 ,
lighter part = (2/7)21 = 6 kg
Heavier part = (5/7)21 = 15 kg
Initial Momentum = 0 as it is at rest
Final momentum = 6 * 50 + 15v
Using law of conservation of momentum
6 * 50 + 15v = 0
=> 15v = - 300
=> v = - 20
Hence heavier mass travels with a speed 20 m/s in opposite direction to lighter weight
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