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Question: A particle of mass m is projected with a velocity \(6\widehat{i} + 8\widehat{j}\). Find the change i...

A particle of mass m is projected with a velocity 6i^+8j^6\widehat{i} + 8\widehat{j}. Find the change in momentum when it just touches ground.

A

0

B

12 m

C

16 m

D

20 m

Answer

16 m

Explanation

Solution

∆p = m(vf – vi) = m[(6i^−8j^)−(6i^+8j^)]=−16mj^\left\lbrack \left( 6\widehat{i} - 8\widehat{j} \right) - \left( 6\widehat{i} + 8\widehat{j} \right) \right\rbrack = - 16m\widehat{j}.

|∆p|= 16 m