The upper ball is released from rest, collides with the stationary lower ball, and sticks to it. The strings are both$\mathbf{50}\mathbf{}\mathbf{cm}$long. The upper ball has mass 2.00 kg, and it is initially 10.0 cm higher than the lower ball, which has mass 3.00 kg. Find the frequency and maximum angular displacement of the motion after the collision.

The term force may be defined as the product of mass and its acceleration.

Consider the given data as below.

The length of both strings L=50 cm

The mass of upper is $m_1=2kg$

The mass of lower ball is$m_2=3kg$

The initial height of the ball is $h=10cm$

The angular momentum of the simple pendulum with small amplitude is calculated as

$\omega =2\pi f$

$$\begin{gathered} f=\frac{1}{2\pi }\sqrt{\frac{g}{L}} \\ =\frac{1}{2\times 3.14}\sqrt{\frac{9.8}{50\times {10}^{-2}}} \\ =0.7\mathrm{Hz} \end{gathered}$$

The potential energy converted in kinetic energy at the point of maximum speed

$$\begin{gathered} \mathrm{mgh}=\frac{1}{2}{\mathrm{mv}}_{max}^2 \\ {\mathrm{v}}_{max}=\sqrt{2\mathrm{gh}} \\ {\mathrm{v}}_{max}=\sqrt{2\times 9.8\times 10\times {10}^{-2}} \\ {\mathrm{v}}_{max}=1.4\mathrm{m}/\mathrm{s} \end{gathered}$$

Consider before collision the speed of mass$m_1$is${\mu }_1$and the speed of mass $m_2$is ${\mu }_2$and after collision the speed of mass $m_1$is $v_1$and the speed of mass $m_2$is $v_2$. So according to law of conservation of mass,

${\mathrm{m}}_1{\mathrm{u}}_1+{\mathrm{m}}_2{\mathrm{u}}_2={\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2$

Before collision:

$$\begin{gathered} {\mathrm{u}}_1={\mathrm{v}}_{\max } \\ {\mathrm{v}}_{\max }=1.4\mathrm{m}/\mathrm{s} \\ {\mathrm{u}}_2=0\mathrm{m}/\mathrm{s} \end{gathered}$$

After collision:

$$\begin{gathered} v_1=v_2 \\ V=v_2 \end{gathered}$$

Than

$m_1{v}_{max}+m_2\times 0=(m_1+m_2)V$

$\begin{array}{r}V=\frac{m_1{v}_{max}}{m_1+m_2}\\ =\frac{2\times 1.4}{2+3}\\ =0.56\mathrm{m}/\mathrm{s}\end{array}$

Now the maximum displacement after collision the kinetic energy converted into potential energy so

$Mg{h}_1=\frac{1}{2}M{V}^2$

From $h_1$calculated as

$$\begin{gathered} {\mathrm{h}}_1=\frac{{\mathrm{V}}^2}{2\mathrm{g}} \\ =\frac{(0.56{)}^2}{2\times 9.8} \\ =0.016\mathrm{m} \\ =1.6\mathrm{cm} \end{gathered}$$

Hence, the maximum displacement after collision is

Now calculated the angular displacement,

$$\begin{gathered} cos(\theta )=\frac{base}{hypotenuse} \\ =\frac{48.4}{50} \\ =co{s}^{-1}\left(\frac{48.4}{50}\right) \\ =0.254rad \\ cos\left(\theta \right)=0.254rad\times \frac{{180}^{\circ }}{\pi rad} \\ =14{.53}^{\circ } \end{gathered}$$

Hence, the frequency and maximum angular displacement of the motion after the collision is f =0.7 Hz and $\theta =0.254rad$respectively