SATHEE: Laws Of Motion Question 265

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 $[g = 10 m / s^{2}]$ [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

$v_{1} = \sqrt{2 g h_{1}} = \sqrt{2 \times 10 \times 5}$

or $\vec{v_{1}} = + 10 m / s = 10 m / s$

velocity of rebound $v_{2} = \sqrt{2 g h_{2}} = \sqrt{2 \times 10 \times 20} = 20 m / s$ or $\vec{v_{2}} = - 20 m / s$

$F = m \frac{d v}{d t} = \frac{m (\vec{v_{2}} - \vec{v_{1}})}{d t} = \frac{0.4 (- 20 - 10)}{d t} = 100 N$ by solving $d t = 0.12 \sec$

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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/s2] [MP PMT 1999]

Options:

Answer:

Solution:

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