A) \[2m\,v\,\cos \theta \]
B) \[2\,m\,v\,\sin \theta \]
C) 0
D) \[2\,m\,v\]
Correct Answer: A
Solution :
\[{{\overrightarrow{P}}_{1}}=m\,v\,\sin \theta \,\hat{i}-m\,v\,\cos \theta \,\hat{j}\] and \[{{\overrightarrow{P}}_{2}}=m\,v\,\sin \theta \,\hat{i}+\,m\,v\,\cos \theta \,\hat{j}\] So change in momentum \[\overrightarrow{\Delta P}={{\overrightarrow{P}}_{2}}-{{\overrightarrow{P}}_{1}}=2\,m\,v\,\cos \theta \,\hat{j},\]\[|\Delta \overrightarrow{P}|\,=2\,m\,v\,\cos \theta \]
You need to login to perform this action.
You will be redirected in
3 sec
