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 $
BETA
