# A boy is rolling a 0.5kg ball on the frictionless floor

Question:

A boy is rolling a $0.5 \mathrm{~kg}$ ball on the frictionless floor with the speed of

$20 \mathrm{~ms}^{-1}$. The ball gets deflected by an obstacle on the way. After

deflection it moves with $5 \%$ of its initial kinetic

energy. What is the speed of the ball now?

1. $19.0 \mathrm{~ms}^{-1}$

2. $4.47 \mathrm{~ms}^{-1}$

3. $14.41 \mathrm{~ms}^{-1}$

4. $1.00 \mathrm{~ms}^{-1}$

Correct Option: , 2

Solution:

(2)

Given, $\mathrm{m}=0.5 \mathrm{~kg}$ and $\mathrm{u}=20 \mathrm{~m} / \mathrm{s}$

Initial kinetic energy $\left(\mathrm{k}_{\mathrm{i}}\right)=\frac{1}{2} \mathrm{mu}^{2}$

$=\frac{1}{2} \times 0.5 \times 20 \times 20=100 \mathrm{~J}$

After deflection it moves with $5 \%$ of $\mathrm{k}_{\mathrm{i}}$

$\therefore \mathrm{k}_{\mathrm{f}}=\frac{5}{100} \times \mathrm{k}_{\mathrm{i}} \Rightarrow \frac{5}{100} \times 100$

$\Rightarrow \mathrm{k}_{\mathrm{f}}=5 \mathrm{~J}$

Now, let the final speed be ' $v^{\prime} \mathrm{m} / \mathrm{s}$, then :

$k_{\mathrm{f}}=5=\frac{1}{2} \mathrm{mv}^{2}$

$\Rightarrow \mathrm{v}^{2}=20$

$\Rightarrow v=\sqrt{20}=4.47 \mathrm{~m} / \mathrm{s}$