A body of mass 2kg has kinetic energy of 400J. A constant force of 10N is applied in the opposite direction of velocity. Find the magnitude of the momentum of the body 2s after application of the force.
A
40kg m/s
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B
60kg m/s
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C
30kg m/s
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D
20kg m/s
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Solution
The correct option is D20kg m/s Let the initial velocity of the object is u0
So, K.E.=12mu20 400=12×2×u20 u0=√400 u0=20m/s
For finding the acceleration of body due to opposite force F=10N, −F=ma [Force is in opposite direction] −10=2×a a=−5m/s2
Applying 1st equation of motion v=u0+at v=20+(−5)×2 v=10m/s ∴The momentum P is P=mv=10×2=20kg m/s