A player takes 0.1s I catching a ball of mass 150 g moving with velocity of 20 m/s. The force imparted by the bal on the hands of the player is

Options

A. 0.3 N
B. 3 N
C. 30 N
D. 300 N

Answer

To solve the problem, we need to find the force imparted by the ball on the hands of the player when catching it. Here's a step-by-step breakdown of the solution:

### Step 1: Identify the given data
- Mass of the ball (m) = 150 g = 150 × 10^(-3) kg = 0.15 kg
- Initial velocity of the ball (u) = 20 m/s
- Final velocity of the ball (v) = 0 m/s (since the player catches the ball)
- Time taken to catch the ball (t) = 0.1 s

### Step 2: Calculate the acceleration (a)
Using the formula for acceleration:
\[ a = \frac{v - u}{t} \]

Substituting the values:
\[ a = \frac{0 - 20}{0.1} \]
\[ a = \frac{-20}{0.1} \]
\[ a = -200 \, \text{m/s}^2 \]

(Note: The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which is expected since the ball is being brought to rest.)

### Step 3: Calculate the force (F)
Using Newton's second law of motion:
\[ F = m \cdot a \]

Substituting the values:
\[ F = 0.15 \, \text{kg} \cdot (-200 \, \text{m/s}^2) \]
\[ F = -30 \, \text{N} \]

The negative sign indicates that the force is acting in the opposite direction of the ball's initial motion. However, we are interested in the magnitude of the force:
\[ |F| = 30 \, \text{N} \]

### Final Answer
The force imparted by the ball on the hands of the player is **30 N**.

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