Kinematics Question 6

Question: An object of m kg with speed of v m/s strikes a wall at an angle q and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be

Options:

A) $ 2mv\cos \theta $

B) $ 2mv\sin \theta $

C) 0

D) $ 2mv $

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Answer:

Correct Answer: A

Solution:

$ {{\overrightarrow{P}}_1}=mv\sin \theta \hat{i}-mv\cos \theta \hat{j} $

and $ {{\overrightarrow{P}}_2}=mv\sin \theta \hat{i}+mv\cos \theta \hat{j} $

So change in momentum $ \overrightarrow{\Delta P}={{\overrightarrow{P}}_2}-{{\overrightarrow{P}}_1}=2mv\cos \theta \hat{j}, $

$ |\Delta \overrightarrow{P}|=2mv\cos \theta $



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