A) 24.5 N-s
B) 49.0 N-s
C) 98.0 N-s
D) 50.0 N-s
Correct Answer: B
Solution :
From figure, it is obvious that only the vertical component of the velocity of projectile is hanging Change in velocity in vertical direction \[=-u\sin \theta -u\sin \theta =-2u\sin \theta .\] Therefore the change in momentum \[=-2\,mu\,\sin \theta \] \[=-2\times 0.5\times 98\times \sin {{30}^{o}}\] \[=-2\times 0.5\times 98\times \frac{1}{2}=-49\,N-s\] Therefore the magnitude of change in its momentum\[=49\,N-s\] Hence, the correction option is [b].You need to login to perform this action.
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