Question

A cricket ball of mass 100g moving with a velocity of 20m/s. A batsman stops the ball in 0.5 seconds. Find the force applied.



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Answer 1

Given,

• Mass of the ball (m) = 100 g = 0.1 kg

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

• Final velocity of the ball (v) = 0 m/s

• Time (t) = 0.5 seconds

To calculate,

• Force applied to stop the ball (F)

Clarification,

We'll have to know the concept of the Newton's second law of motion in order to tackle this question.

According to the Newton's second law of motion, the rate of change in the momentum of a body is directly proportional to the applied force and it takes place in the direction in which force acts. So,

$\longrightarrow \sf {F \propto \dfrac{m(v-u)}{t}}$

The formula of acceleration (a) is v-u/t. So, we can write it as :-

$\longrightarrow \sf {F \propto ma}$

$\longrightarrow \sf {F = k \times ma}$

Where k is constant. In SI units, the value of is 1. So,

$\sf { \longrightarrow \underline { \boxed{\sf{F = ma }}} \: \red{ \bigstar} }$

We'll use this formula to solve this question!

Step-by-step explanation,

As we know that,

$\sf { \longrightarrow \underline { \boxed{\sf{F = ma }}} \: \red{ \bigstar} }$

Calculating acceleration :

→ a = $\sf { \dfrac{v - u}{t} }$ m/s²

→ a = $\sf { \dfrac{0 - 20}{0.5} }$ m/s²

→ a = $\sf { \dfrac{- 20}{0.5} }$ m/s²

→ a = $\sf { \dfrac{- 20 \times 10}{5} }$ m/s²

→ a = $\sf { \dfrac{- 200}{5} }$ m/s²

→ a = -40 m/s²

Calculating force applied :

→ F = ma

→ F = 0.1 kg × -40 m/s²

→ F = $\sf { \dfrac{ 1}{10} }$ kg × -40 m/s²

→ F = 1 kg × -4 m/s²

→ F = -4 kg m/s²

1 kg × 1 m/s² = 1 Newton, So,

→ F = - 4 N

Negative sign means it was applied in the opposite direction in which the body was moving.

Therefore , 4 N of force was applied in the opposite direction by the batsman.

Answer 2

Hi diya, here is your answer

Hope it helps you out