Question

9
A ball of mass 50 g is dropped from a height of
20 m. A boy on the ground hits the ball vertically
upwards with a bat with an average force of 200
N, so that it attains a vertical height of 45 m.
The time for which the ball remains in contact with
the bat is

Answer 1

GIVEN:-

• Force = 200N

• Mass of Ball = 50g = 0.05kg

For Dropping:-

• Initial Velocity while dropping = 0m/s

• Height = 20m

• Acceleration due to gravity = +10m/s

After being hit:-

• Final Velocity = 0m/s

• Height attained = 45m

• Acceleration due to gravity = -10m/s

TO FIND:-

• The Time for the ball remain in Contact with the bat.

FORMULAE USED :-

• ${\huge{\boxed{\rm{ v^2 - u^2 = 2as}}}}$

• ${\huge{\boxed{\rm{ Impulse = F\times{t}}}}}$

CONCEPT USED:

-

• When body is at height height it's final velocity is zero.

• Change in momentum is equal to the rate if impulse.

Now,

For Dropping

$\implies\rm{ v^2 - u^2 = 2as}$

$\implies\rm{ v^2 - (0)^2 = 2gh}$

$\implies\rm{ v^2 = 2\times{10}{20}}$

$\implies\rm{ v^2 = 400}$

$\implies\rm{ v = -20m/s^{-1}}$

• When the ball was in contact with bat it's velocity was 20m/s^{-1}.

So,Initial Momentum = mu = 0.05 × -20 = -1kgm/s.

After being contact from the bat.

$\implies\rm{ v^2 - u^2 = 2as}$

$\implies\rm{ -u^2 = 2gh}$

$\implies\rm{ -u^2 = 2\times{-10}\times{45}}$

$\implies\rm{ -u^2 = -900}$

$\implies\rm{ u = 30m/s^{-1}}$

So, Final Momentum = mv = 0.05 × 30 = 1.5kg

$\implies\rm{ Change\:in\: Momentum = mv - mu}$

$\implies\rm{ Impulse = 1.5 + 1}$

$\implies\rm{ Impulse = 2.5}$.

Therefore

$\implies\rm{Impulse = F\times{t}}$

$\implies\rm{ 200\times{t} = 2.5}$

$\implies\rm{ t =\dfrac{2.5}{200}}$

$\implies\rm{ t = 0.0125s}$.

Hence, Time for the ball remain in Contact with the bat is 0.0125s