question_answer

A 0.5 kg ball is dropped from rest at a point 1.20 m above the floor. The ball rebounds straight upward to a height of 0.7 m. What is the magnitude and the direction of the impulse of the net force applied to the ball during the collision with the floor?

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.