91Ó°ÊÓ

Mass of the ball moving with speed v₀= 20 g

Mass of the motionless ball = 100 g

Length of the string = 1 m

The collision of the ball is an elastic collision which means mechanical energy is conserved.

As per the formula for final velocity after elastic collision, final velocity of the large ball v_f.2 :

$v_{f.2}=\frac{2m_1}{m_1+m_2}{v}_{i.1}$

(v_i.1 is the initial velocity of the smaller ball)

It is given that v_i.1 = v₀

Therefore,

$$\begin{gathered} v_{f.2}=\frac{2\left(20g\right)}{20g+100g}{v}_0 \\ {v}_{f.2}=\frac{40g}{120g}{v}_0=\frac{v_0}{3} \end{gathered}$$

When collision occurs, The large ball reaches to a height y as shown in figure.

When large ball reaches the height, its Kinetic energy gets converted into potential energy.

Using Law of conservation of Energy,

Kinetic Energy = Potential Energy

$$\begin{gathered} \frac{1}{2}{m}_2{v_{f.2}}^2=m_{2}gy \\ \frac{1}{2}{\left(\frac{v_0}{3}\right)}^2=gy1. \end{gathered}$$

Height, y can be calculated as

$$\begin{gathered} \cos \mathrm{θ}=\frac{L-y}{L} \\ L-y=L\cos \mathrm{θ} \\ y=L-L\cos \mathrm{θ} \\ y=L(1-\cos \mathrm{θ}) \\ Puttingthevalueofyinequation1. \\ \frac{1}{2}{\left(\frac{v_0}{3}\right)}^2=gL(1-\cos \mathrm{θ}) \\ {\left(\frac{v_0}{3}\right)}^2=2gL(1-\cos \mathrm{θ}) \\ \left(\frac{v_0}{3}\right)=\sqrt{2gL(1-\cos \mathrm{θ})} \\ {v}_0=3\sqrt{2gL(1-\cos \mathrm{θ})} \\ Puttingthevaluesofg,Land\mathrm{θ} \\ {v}_0=3\sqrt{2\left(9.8m/s^2\right)(1m)(1-\cos 50°)} \\ {v}_0=7.9m/s \end{gathered}$$