A player caught a cricket ball of mass $$150\,g$$ moving at a rate of $$20\,m/s$$. If the catching process is completed in $$0.1\,s$$, the force of the blow exerted by the ball on the hand of the player is equal to

Solution & Explanation

1. Identify given parameters and convert to SI units

We are given the following values from the problem statement:

• Mass of the cricket ball ($$m$$): $$150 \,\, \text{g}$$
• Initial velocity ($$u$$): $$20 \,\, \text{m/s}$$
• Final velocity ($$v$$): $$0 \,\, \text{m/s}$$ (since the ball is caught and brought to rest)
• Time interval ($$\Delta t$$): $$0.1 \,\, \text{s}$$

First, we convert the mass from grams to kilograms (SI units):

$$m = \frac{150}{1000} = 0.15 \,\, \text{kg}$$

2. Apply Newton's Second Law of Motion

According to Newton's Second Law of Motion, the average force ($$F$$) exerted on an object is equal to its rate of change of linear momentum ($$\Delta P$$):

$$F = \frac{\Delta P}{\Delta t} = \frac{m \cdot v - m \cdot u}{\Delta t}$$

Since we are calculating the magnitude of the force of the blow:

$$|F| = \frac{m \cdot |v - u|}{\Delta t}$$

3. Compute the Force of the Blow

Substituting the parameters into our equation:

$$|F| = \frac{0.15 \cdot |0 - 20|}{0.1}$$

$$|F| = \frac{0.15 \cdot 20}{0.1}$$

$$|F| = \frac{3.0}{0.1} = 30 \,\, \text{N}$$

Concept Check: Extending the catching process duration ($$\Delta t$$) allows the momentum to change over a longer period, which drastically reduces the impact force felt by the player's hand. If caught instantaneously, the force would be significantly more punishing.

Correct Option Key: 30 N