Impulse and Momentum, Energy Formulations

Let '\theta' be the angle about line of impact through which ball will move often the impact.
Let 'u' be the vertical downward velocity of the ball before striking and 'v' be the velocity of ball after the impact which make an angle '\theta' with the line of impact. As ball fall freely under the gravity from height \mathrm{h}=4.9 \mathrm{m}, hence downward velocity '\mathrm{u}' at the instance of striking the rigid body
u=\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 4.9}
or, u=9.8 \mathrm{~m} / \mathrm{sec}
for perfectly elastic collision, e=1
As e=\frac{\text { relative velocity of seperation }}{\text { relative velocity of approach }}=1\;\;...(i)
As block is rigid, so block velocity =0.
So, along the line of impact,
relative velocity of approach =u \cos 30^{\circ}-0 =u \cos 30^{\circ}
relative velocity of separation along the line of impact =v \cos \theta-0=v \cos \theta
so, by equation (i)
u \cos 30^{\circ}=v \cos \theta \;\;\;...(ii)
In the direction normal to the impact, the component velocity is not affected so,
u \sin 30^{\circ}=v \sin \theta \;\;\;...(iii)
So, by equation (ii) and (iii)
V=\sqrt{u^{2} \cos ^{2} 30^{\circ}+u^{2} \sin ^{2} 30^{\circ}}
or, V=9.8 \mathrm{~m} / \mathrm{sec}
by equation (ii) and (iii)
\tan \theta=\tan 30^{\circ}, \theta=30^{\circ}
So, inclination to the plane for of the section
=90^{\circ}-30^{\circ}=60^{\circ}
So, momentum equation is given by during seperation:

\vec{P}_{s}=m v \cos 30^{\circ} \hat{i}+m v \cos 30^{\circ} \hat{j} =17 \hat{i}+9.8 \hat{j}

Let V_{1} is the speed of 3 kg mass after collision
\mathrm{V}_{2} is the speed of m kg mass after collision
\begin{array}{l} e=I=\frac{V_{2}-V_{1}}{4} \\ \Rightarrow V_{2}-V_{1}=4 \end{array}
By linear momentum conservation
\begin{array}{l} 3 \times 4=3 V_{1}+m V_{2}\\ \frac{1}{2} \times 3 \times V_{1}^{2}=6 \\ \Rightarrow V_{1}=\pm 2 \end{array}
By using (1), (2) \& (3) we get
m = 1 kg (or) 9 kg

\begin{aligned} (1) \rightarrow \qquad t&=? \\ v&=u+a t \\ 0&=4-10 t \\ t&=\frac{4}{10}=0.4 s \\ (2) \rightarrow \qquad t^{\prime}&=? \\ u^{\prime}&=0.8 \times u \\ &=0.8 \times 4=3.2 \mathrm{m} / \mathrm{s} \\ \mathrm{v}^{\prime}&=U^{\prime}+\mathrm{al}^{\prime} \\ 0&=3.2-10 t^{\prime} \\ t^{\prime}&=\frac{3.2}{10}=0.32 \mathrm{s} \\ (3) \rightarrow \qquad t^{\prime \prime}&=? \\ U^{\prime \prime} &=0.8 u^{\prime} \\ &=0.8 \times 3.2=2.56 \mathrm{m} / \mathrm{s} \\ V^{\prime \prime} &=U^{\prime \prime}+a t^{\prime} \\ 0 &=2.56-10 t^{\prime \prime} \\ t^{\prime \prime} &=0.256 s \end{aligned}
So, t, t^{\prime}, t^{\prime \prime} are forming a GP series
\begin{array}{l} \text{So, total time }=2\left(t+t^{\prime}+t^{\prime \prime}+\ldots . .0\right)\\ =2[0.4+0.32+0.256+\ldots .0] \\ =2 \times \frac{0.4}{1-0.8}=2 \times 2=4 s \end{array}

m_{A}=m_{B}=m
Outer circumferential radius of both disc A
and B is also same i.e. R
\begin{aligned} I_{A}&=m R^{2} \\ I_{B}&=\frac{m R^{2}}{2} \end{aligned}

\frac{V_{A}}{V_{B}} at bottom of plane =?
m g h=(K \cdot E) of disc A=K \cdot E of disc B
\begin{aligned} \frac{1}{2} m V_{A}^{2}+\frac{1}{2} I_{A} \omega_{A}^{2} &=\frac{1}{2} m V_{B}^{2}+\frac{1}{2} I_{B} \omega_{B}^{2} \\ \frac{1}{2} m V_{A}^{2}+\frac{1}{2} m R^{2} \frac{V_{A}^{2}}{R^{2}} &=\frac{1}{2} m V_{B}^{2}+\frac{1}{2} \frac{m R^{2}}{2} \times \frac{V_{B}^{2}}{R^{2}} \\ \frac{V_{A}^{2}}{2}+\frac{V_{A}^{2}}{2} &=\frac{V_{B}^{2}}{2}+\frac{V_{B}^{2}}{4} \\ V_{A}^{2} &=\frac{3}{4} V_{B}^{2} \\ \frac{V_{A}}{V_{B}} &=\sqrt{\frac{3}{4}} \end{aligned}

\begin{aligned} r &=t^{2}: \theta=t \\ \mathrm{K.E} &=\frac{1}{2} m v^{2}=? \text { at } t=2 \mathrm{sec} \\ \Rightarrow m &=1 \mathrm{kg} \\ \vec{V} &=r \omega(\hat{t})+\frac{d r}{d t} \dot{r}=t^{2} \times 1(\hat{t})+2 t \hat{r} \\ \therefore \frac{d \theta}{d t}&=\omega=1 \\ \vec{V} &=t^{2}(\hat{t})+2 t(\hat{r}) \\ \text{at }\qquad t &=2 s \\ \vec{V} &=4(\hat{t})+4(\hat{r}) \\ |\vec{V}| &=\sqrt{16+16}=\sqrt{32} \\ K.E. &=\frac{1}{2}mv^{2}=\frac{1}{2}\times1\times32=16 \end{aligned}