Question

A steel ball of mass $1\text{kg}$ is released from rest as shown and strikes a ${45}^{\circ }$ inclined surface . If the co-efficient of restitution is $0.8$, the distance $S$, where the ball will strike the horizontal plane at $\text{A}$ will be (approximately)

• A. $0.76\text{m}$
• B. $0.86\text{m}$
• C. $0.66\text{m}$
• D. $0.96\text{m}$

Answer 1

The correct option is D $0.96\text{m}$

$\text{FBD}$ for the given problem:Given,
height of the ball above inclined plane, $h=1.5\text{m}$,
mass of the ball, $m=1\text{kg}$,
Co-efficient of restitution, $e=0.8$,
acceleration due to gravity, $g=10{\text{m/s}}^2$
Let us suppose, steel ball strikes the inclined plane with speed $v$.

From the law of conservation of mechanical energy, we can write
$mgh=\frac{1}{2}m{v}^2$

$\Rightarrow 10\times 1.5=\frac{1}{2}\times v^2$

$\Rightarrow v=\sqrt{30}\text{m/s}$

Component of velocity perpendicular to inclined plane before collision,

$v_{\perp }=v\cos {45}^{\circ }$

$=\sqrt{30}\times \frac{1}{\sqrt{2}}$

$=\sqrt{15}\text{m/s}$

Now, component of velocity perpendicular to inclined plane after collision will be

$v_{\perp }^{\prime }=e{v}_{\perp }$
$=0.8\times \sqrt{15}$
$=3.1\text{m/s}$

Component of velocity parallel to inclined plane will not change due to collision
$v_0^{\prime }=v\sin {45}^{\circ }$

$=\sqrt{30}\times \frac{1}{\sqrt{2}}$

$=3.8\text{m/s}$

Now, net velocity in horizontal direction after collision,

$v_x=v_{\perp }^{\prime }\sin {45}^{\circ }+v_0^{\prime }\cos {45}^{\circ }=\frac{3.1+3.8}{\sqrt{2}}=4.9\text{m/s}$

Similarly net velocity along vertical direction,
$v_y=v_{\perp }^{\prime }\cos {45}^{\circ }-v_0^{\prime }\sin {45}^{\circ }=\frac{3.1-3.8}{\sqrt{2}}=-0.5\text{m/s}$ $($downward$)$

Let $t$ be the time, the particle takes from point $\text{C}$ to $\text{A}$,

Then applying equation of motion for vertical motion of the particle from $\text{C}$ to $\text{A}$, we have

$1=0.5t+\frac{1}{2}\times 10\times t^2$

$\Rightarrow 5t^2+0.5t-1=0$

On solving this, we get, $t=0.4\text{s}$ or $t=-0.5\text{s}$.
Since, time cannot be negative.

Now, $DA=V_{x}t$
$=4.9\times 0.4$
$=1.96\text{m}$

Therefore, $S=DA-DE$
$=1.96-1=0.96\text{m}$

$\therefore S=0.96\text{m}$

Hence, option $\left(d\right)$ is correct answer.