Question

6. A base ball of 200 g is thrown to a batsman with a velocity of 20 m/s. The batsman hits harder and the
ball moves with a velocity of 50 m/s in the opposite direction. Find the force exerted on the ball, if the
force acts for 0.01 s?​

Answer 1

Mass of the base ball (m) = 200 g = 0.2 kg

Initial velocity of the ball (u) = - 20 m/s

[ ∵ Ball moving against the action of force ]

Final velocity of the ball (v) = 50 m/s

Time taken (t) = 0.01 s

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★ Force : It is defined as product of mass and acceleration

➙ F = ma

➳ Acceleration : It is defined as rate of change in velocity

➤ a = (v - u) / t

★ ═══════════════════ ★

➠ $\sf F=m \bigg( \dfrac{v-u}{t} \bigg)$

➠ $\sf F=0.2 \bigg( \dfrac{50-(-20)}{0.01} \bigg)$

➠ $\sf F=0.2 \bigg( \dfrac{70}{0.01} \bigg)$

➠ $\sf F=0.2(7000)$

➠ $\sf F= 2(700)$

➠ $\sf F=1400\ N\ \; \pink{\bigstar}$

Force exerted on the ball is 1400 Newtons .

Answer 2

Answer:

⠀⠀⠀⌬ Mass (m) = 200 g

⠀⠀⠀⌬ Initial Velocity (u) = - 20 m/s

⠀⠀⠀⌬ Final Velocity (v) = 50 m/s

⠀⠀⠀⌬ Time (t) = 0.01 s

⠀⠀⠀⌬ Force exerted (F) = ?

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\Longrightarrow\sf Force = Mass \times Acceleration\\\\\\:\Longrightarrow\sf Force = 200\:g \times \bigg\lgroup\dfrac{(Final\ -Initial)\ Velocity}{Time}\bigg\rgroup\\\\\\:\Longrightarrow\sf Force = \dfrac{200}{1000} \times \bigg\lgroup\dfrac{50 - ( - 20)}{0.01}\bigg\rgroup\\\\\\:\Longrightarrow\sf Force = \dfrac{2}{10} \times \dfrac{(50 + 20)}{ \frac{1}{100} }\\\\\\:\Longrightarrow\sf Force = \dfrac{2}{10} \times70 \times 100\\\\\\:\Longrightarrow\sf Force = 2 \times 7 \times 100\\\\\\:\Longrightarrow\sf Force = 1400 \:N$

⠀

$\therefore\:\underline{\textsf{Force exerted on the base ball is \textbf{1400 N}}}.$