Question

Two boys A and B are standing at two positions as shown in the diagram. Each step is 20 cm high and 20 cm wide. The potential energy is assumed to be zero at the ground. Consider $g=10m/s^2$ and assume that there is no loss of energy.


If B releases a ball of mass 500 grams from the height as shown in the diagram, what will be the velocity of the ball when it reaches the level of the third step from the ground?

[1 mark]

• A. $\sqrt{20}m/s$
• B. $\sqrt{30}m/s$
• C. $\sqrt{24}m/s$
• D. $\sqrt{34}m/s$

Answer 1

The correct option is B $\sqrt{30}m/s$

Initial position of the ball with respect to the ground,
$h_i=50+(20\times 8)=210cm=2.1m$

Final position of the ball with respect to the ground, $h_f=Levelofthethirdstepabove$
the ground
$h_f=(20\times 3)=60cm=0.6m$

At the point of release, the initial velocity will be zero. So,

$v_i=0$
Let the velocity of the ball be v when it reaches the level of the third step from the ground. So, $v_f=v$

Let us apply the principle of conservation of mechanical energy at the initial and final positions.

$\Rightarrow \frac{1}{2}m{v_i}^2+mg{h}_i=\frac{1}{2}m{v_f}^2+mg{h}_f\Rightarrow 0+(m\times g\times 2.1)\Rightarrow \frac{1}{2}m{v}^2+(m\times g\times 0.6)\Rightarrow v=\sqrt{2g(2.1-0.6)}m/s\Rightarrow v=\sqrt{30}m/s$