Question

A cricket ball of mass 0.20 kg is moving with a velocity of 1.2 m/s. Find the average force applied by the player if he is able to stop the ball in 0.10 s?

Answer 1

Given :

▪ Mass of ball = 0.20kg

▪ Initial velocity = 1.2mps

▪ Final velocity = zero

▪ Time interval = 0.1s

To Find :

✴ Average force applied by the player on the ball.

Concept :

↗ This question is completely based on the concept of Newton's second law of motion.

↗ As per this law, Force is defined as the rate of change in linear momentum.

↗ Mathematically,

$\bigstar\:\underline{\boxed{\bf{F=\dfrac{\Delta P}{\Delta t}}}}$

• F denotes applied force
• ΔP denotes change in momentum
• Δt denotes time interval

Calculation :

$\implies\sf\:F=\dfrac{\Delta P}{\Delta t}=\dfrac{\Delta(mv)}{\Delta t}\\ \\ \implies\sf\:F=\dfrac{m\Delta v}{\Delta t}\\ \\ \implies\sf\:F=\dfrac{0.20\times 1.2}{0.1}=\dfrac{0.24}{0.1}\\ \\ \implies\underline{\underline{\bf{F=2.4N}}}$

Answer 2

Answer:

Given:

• A cricket ball of mass 0.20 kg is moving with a velocity of 1.2 m/s.

Find:

• Find the average force applied by the player if he is able to stop the ball in 0.10 s?

Know terms:

1. Force = (F).
2. Momentum = (M).
3. Time = (T).

Using formula:

☆ ${\sf{\underline{\boxed{\green{\sf{Force = \dfrac{Change \: in \: Momentum}{Time} }}}}}}$

Calculations:

⇒ $\sf F = \dfrac{P}{T}$

⇒ $\sf F = \dfrac{Mv}{T}$

⇒ $\sf F = \dfrac{0.20 \times 1.2}{0.1}$

⇒ $\sf F = \cancel{\dfrac{0.24}{0.1}}$

⇒ ${\sf{\underline{\boxed{\green{\sf{2.4 \: N}}}}}}$