Question

A ball of mass $6\text{kg}$ hits the floor with a speed $2{\text{ms}}^{-1}$ making an angle of incidence ${30}^{\circ }$ with the normal. The coefficient of restitution of $\frac{1}{3}.$ Find the kinetic energy of ball just after reflection.

• A. $4\text{J}$
• B. $48\text{J}$
• C. $17\text{J}$
• D. $13\text{J}$

Answer 1

The correct option is A $4\text{J}$



During the collision, floor exert a normal force on the ball.

There is no external force parallel to the floor. Thus, the parallel component of the velocity of the ball remains unchanged.

$\Rightarrow 2\sin {30}^{\circ }=v\sin \theta$

$\Rightarrow v\sin \theta =1......\left(1\right)$

Using Newton's law of impact normal to the floor, we get,

$e=\frac{\text{velocity of separation}}{\text{velocity of approach}}=\frac{v_2-v_1}{u_1-u_2}$

Here, $(u_1{)}_y=2\cos {30}^{\circ };(u_2{)}_y=0{\text{ms}}^{-1}(v_1{)}_y=v\cos \theta ;(v_2{)}_y=0{\text{ms}}^{-1}$

$\Rightarrow \frac{1}{3}=\frac{v\cos \theta }{2\times \cos {30}^{\circ }}$

$\Rightarrow v\cos \theta =\frac{2}{3}\cos {30}^{\circ }=\frac{1}{\sqrt{3}}......\left(2\right)$

From equation (1) and (2) we get,

$v^2{\sin }^2\theta +v^2{\cos }^2\theta =(1{)}^2+{\left(\frac{1}{\sqrt{3}}\right)}^2$

$\Rightarrow v^2=\frac{4}{3}$

Now, the kinetic energy of the ball is,

$K.E=\frac{1}{2}m{v}^2$

$\Rightarrow K.E=\frac{1}{2}\times 6\times \frac{4}{3}=4\text{J}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.

Why this question ? Caution: In bouncing of a ball from the floor like questions, $v^{{}^{\prime }}=ev$ directly can be applied if the ball is colliding perpendicular to the floor. When this is not the case, this formula can be used for vertical component only.