Sankalp NEET Full Test-6 Question-42 Solution

Question 42: 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 ( \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} ) nearly :

(1) 0 \mathrm{~kg} \mathrm{~m} / \mathrm{s}

(2) 4.2 \mathrm{~kg} \mathrm{~m} / \mathrm{s}

(3) 2.1 \mathrm{~kg} \mathrm{~m} / \mathrm{s}

(4) 1.4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}

Answer: Option (2)

Explanation:

The speed of the ball just before striking the ground is obtained using

v=\sqrt{2gh}

Substituting g=10 \mathrm{~m/s^{2}} and h=10 \mathrm{~m},

v=\sqrt{2\times10\times10}=\sqrt{200}\approx14.1 \mathrm{~m/s}

Just before collision, the velocity of the ball is downward, so momentum is

p_1=-mv=-0.15\times14.1

After rebounding to the same height, the ball leaves the ground with the same speed upward,

so momentum after collision is

p_2=+mv=0.15\times14.1

The magnitude of impulse is equal to the change in momentum:

\Delta p=p_2-p_1=0.15\times14.1-(-0.15\times14.1) \Delta p=2\times0.15\times14.1\approx4.2 \mathrm{~kg\,m/s}

Hence, the magnitude of impulse imparted to the ball is approximately

4.2 \mathrm{~kg\,m/s}, corresponding to option (2).