Question

8. A cricket ball of mass 100g moves with a velocity 20m/s. A batsman Stops the ball in 0.05
second. Find the force applied?
​

Answer 1

Answer:

• 40 N of force is applied.

Explanation:

Given that:

• A cricket ball of mass 100g moves with a velocity 20 m/s.
• A batsman stops the ball in 0.05 second.

To Find:

• The force applied.

Formula used:

1. F = ma
2. a = (v - u)/t

Where,

• F = Force
• m = Mass = 100g = 0.1 kg
• a = Acceleration
• v = Final velocity = 0
• u = Initial velocity = 20 m/s
• t = Time = 0.05 second

Finding the acceleration:

⟶ a = (0 - 20)/0.05

⟶ a = - 20/0.05

⟶ a = - 400

∴ Acceleration = 400 m/s²

Finding the force applied:

⟶ F = 0.1 × 400

⟶ F = 40

∴ Force applied = 40 N

Answer 2

Answer:

Given :-

• A cricket ball of mass 100 g moves with a velocity 20 m/s.
• A batsman stops the ball in 0.05 second.

To Find :-

• What is the force applied.

Formula Used :-

$\clubsuit$ Acceleration Formula :

$\longmapsto \sf\boxed{\bold{\pink{a =\: \dfrac{v - u}{t}}}}$

where,

• a = Acceleration
• v = Final Velocity
• u = Initial Velocity

$\clubsuit$ Force Formula :

$\longmapsto \sf\boxed{\bold{\pink{F =\: ma}}}$

where,

• F = Force
• m = Mass
• a = Acceleration

Solution :

First, we have to convert g to kg :

$\implies \sf Mass =\: \dfrac{1}{1000}$

$\implies \sf\bold{\green{Mass =\: 0.1\: kg}}$

Now, we have to find the acceleration:

Given :

• Final Velocity (v) = 0 m/s
• Initial Velocity (u) = 20 m/s
• Time (t) = 0.05 second

According to the question by using the formula we get,

$\implies \sf a =\: \dfrac{0 - 20}{0.05}$

$\implies \sf a =\: \dfrac{- 20 \times 100}{5}$

$\implies \sf a =\: \dfrac{- \cancel{2000}}{\cancel{5}}$

$\implies \sf\bold{\purple{a =\: 400\: m/{s}^{2}}}$

Now, we have to find the force applied :

Given :

• Mass = 0.1 kg
• Acceleration = 400 m/s²

According to the question by using the formula we get,

$\implies \sf Force\: applied =\: 0.1 \times 400$

$\implies \sf Force\: applied =\: \dfrac{1}{10} \times 400$

$\implies \sf Force\: applied =\: \dfrac{ 40\cancel{0}}{1\cancel{0}}$

$\implies \sf\bold{\red{Force\: applied =\: 40\: N}}$

$\therefore$ The force applied is 40 N .