A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g=10 m/s2) nearly

The correct option is 4.2 kg m/s.

Let us break down the physical process step-by-step to find the magnitude of the impulse imparted to the ball.

Step 1: Identify the given values
- Mass of the ball, m = 0.15 kg
- Initial drop height, h = 10 m
- Acceleration due to gravity, g = 10 m/s²
- Rebound height, h' = 10 m (the ball rebounds to the same height)

Step 2: Calculate the velocity of the ball just before striking the ground
As the ball is dropped from a height h, its initial velocity is zero. Using the equations of motion (or conservation of energy), the velocity v₁ just before it hits the ground is given by:
$v_1^2=u^2+2gh$
Since the initial velocity u = 0:
$v_1=\sqrt{2gh}$
Substituting the given values:
$v_1=\sqrt{2\times 10\times 10}=\sqrt{200}\approx 14.14\text{m/s}$
Taking the downward direction as negative, we can write the velocity vector just before collision as:
$\overrightarrow{v_1}=-14.14\hat{j}\text{m/s}$

Step 3: Calculate the velocity of the ball just after rebounding
Since the ball rebounds back to the same height h = 10 m, it must leave the ground with the same speed but in the opposite (upward) direction. Let this velocity be v₂:
$v_2=\sqrt{2gh}\approx 14.14\text{m/s}$
Taking the upward direction as positive:
$\overrightarrow{v_2}=+14.14\hat{j}\text{m/s}$

Step 4: Calculate the magnitude of impulse
Impulse is defined as the change in momentum of the ball:
$\overrightarrow{J}=\Delta \overrightarrow{p}=m(\overrightarrow{v_2}-\overrightarrow{v_1})$
Substituting the velocity vectors:
$\overrightarrow{J}=0.15\times (14.14\hat{j}-(-14.14\hat{j}\left)\right)$
$\overrightarrow{J}=0.15\times 2\times 14.14\hat{j}$
$\overrightarrow{J}=0.3\times 14.14\hat{j}\approx 4.24\hat{j}\text{kg m/s}$
Thus, the magnitude of the impulse imparted to the ball is:
$J\approx 4.2\text{kg m/s}$