Question

A box of mass $2\text{kg}$ slides from a height of $8\text{m}$ on the inclined plane and struck the ball. After collision, the ball sticks on the box. If mass of the ball is $200\text{g}$, then what will be the velocity of the box after collision? Assuming all surfaces are frictionless and initially the ball is at rest. (take $g=10m/s^2)$

• A. $2.3\text{m/s}$
• B. $4.5\text{m/s}$
• C. $8.5\text{m/s}$
• D. $11.5\text{m/s}$

Answer 1

The correct option is D $11.5\text{m/s}$

Let mass of the box = $m_1$
mass of the ball = $m_2$
Given height h = $8\text{m}$
Let velocity of box at point C, just before the collision = $u_1$
Velocity of ball before collision $u_2=0$ (At rest)

Initially box has potential energy at point $A$ and after reaching at the point $C$, just before the collision, potential energy will convert into kinetic energy.
Applying energy conservation at point $A$ and $B$.
$(P.E{)}_{\text{at point A}}=(K.E{)}_{\text{at point C}}$.
$m_{1}gh=\frac{1}{2}{m}_1{u}_1^2$
$u_1=\sqrt{2gh}$
$u_1=\sqrt{2\times 10\times 8}$
$u_1=\sqrt{160}\text{m/s}$
$u_1=12.65\text{m/s}$.
(Velocity of box in horizontal direction just before the collision)
Let the velocity of box and ball after collision be $v$ and applying the conservation of momentum before collision and after collision,
$m_1{u}_2+m_2{u}_2=(m_1+m_2)v$
$\Rightarrow 2\times 12.65+\left(0.200\right)\times 0=(2+0.200)v$
$\Rightarrow v=\frac{2\times 12.65}{2.2}$
$\Rightarrow v=11.5\text{m/s}$
So, velocity of box after collision is $11.5\text{m/s}$.