Question

A ball of mass 500 grams is suspended with the help of a light, inextensible string of length 100 cm. The maximum value of to which the string rotates from the mean position A to reach the extreme position C is ${60}^{\circ }$. Consider $g=10m/s^2$


ii. What will be the speed of the ball when it reaches half the value of the maximum height it can achieve?

[1 mark]

• A. v = $\sqrt{2.5}$ m/s
• B. v = $\sqrt{3.5}$ m/s
• C. v = $\sqrt{1.5}$ m/s
• D. v = $\sqrt{5}$ m/s

Answer 1

The correct option is D v = $\sqrt{5}$ m/s

From the previous part, we know the potential energy at F is equal to 1.25 J.
Let the velocity at point F be
v.

The potential energy at F, $P.E{.}_F=m\times g\times AE=0.5\times 10\times 0.25=1.25J$
Kinetic energy at F, $K.E{.}_F=\frac{1}{2}m{v}^2$

Using the principle of conservation of energy,
$\Rightarrow TotalenergyatC=TotalenergyatF\Rightarrow m\times g\times AD=\frac{1}{2}m{v}^2+(m\times g\times AE)\Rightarrow \frac{1}{2}m{v}^2=m\times g\times (AD-AE)\Rightarrow v=\sqrt{2g(AD-AE)}\Rightarrow v=\sqrt{2g(0.50-0.25)}m/s\Rightarrow v=\sqrt{5}m/s$