Laws Of Motion Question 265
Question: A ball of mass 400 gm is dropped from a height of 5m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 100 newton so that it attains a vertical height of 20 m. The time for which the ball remains in contact with the bat is $ [g=10m/s^{2}] $ [MP PMT 1999]
Options:
A) 0.12s
B) 0.08 s
C) 0.04 s
D) 12 s
Show Answer
Answer:
Correct Answer: A
Solution:
Velocity by which the ball hits the bat
$ v _{1}=\sqrt{2gh _{1}}=\sqrt{2\times 10\times 5} $
or $ \overrightarrow{v _{1}}=+10m/s=10m/s $
velocity of rebound $ v _{2}=\sqrt{2gh _{2}}=\sqrt{2\times 10\times 20}=20m/s $ or $ \overrightarrow{v _{2}}=-20m/s $
$ F=m\frac{dv}{dt}=\frac{m(\overrightarrow{v _{2}}-\overrightarrow{v _{1}})}{dt}=\frac{0.4(-20-10)}{dt}=100N $ by solving $ dt=0.12\sec $
BETA
