A) 4.28 N-s, upward
B) 4.28 N-s, downward
C) 8.56 N-s, upward
D) 8.56 N-s, downward
Correct Answer: A
Solution :
The velocity of ball just before collision \[=\sqrt{2g\times 1.2}=\sqrt{2.4g}\,\text{m/s}\]in downward direction. The velocity of ball just after collision \[=\sqrt{2g\times 0.7}=\sqrt{1.4g}\,\text{m/s}\]in up ward direction. Impulse\[={{\overrightarrow{p}}_{f}}-{{\overrightarrow{p}}_{i}}\] \[=m[\sqrt{1.4g}-(-\sqrt{2.4g})]\]\[=4.28N-s\]in the direction of final momentum, ie, upward.You need to login to perform this action.
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