A bob of mass $$m$$ suspended by a light string of length $$L$$ is whirled into a vertical circle as shown in figure. What will be the trajectory of the particle, if the string is cut at
(a) point $$B$$ ?
(b) point $$C$$ ?
(c) point $$X$$ ?
When bob is whirled into a vertical circle, the required centripetal force is obtained from the tension in the string. When string is cut, tension in string becomes zero and centripetal force is not provided, hence, bob start to move in a straight line path along the direction of its velocity.
(a) At point $$B$$, the velocity of $$B$$ is vertically downward, therefore, when string is cut at $$B$$, bob moves vertically downward.
(b) At point $$C$$, the velocity is along the horizontal towards right, therefore, when string is cut at $$C$$, bob moves horizontally towards right.
Also, the bob moves under gravity simultaneously with horizontal uniform speed. So, it traversed on a parabolic path with vertex at $$C$$.
(c) At point $$X$$, the velocity of the bob is along the tangent drawn at point $$X$$, therefore when string is cut at point $$C$$, bob moves along the tangent at that point $$X$$.
Also, the bob move under gravity simultaneously with horizontal uniform speed. So, it traversed on a parabolic path with vertex higher than $$C$$.
A graph of potential energy $$V(x)$$ versus $$x$$ is shown in figure. A particle of energy $$E_0$$ is executing motion in it. Draw graph of velocity and kinetic energy versus $$x$$ for one complete cycle $$A F A$$.
$$\begin{aligned} &\text { KE versus } \boldsymbol{x} \text { graph }\\ &\begin{array}{lrl} \text { We know that } & \text { Total } \mathrm{ME} & =\mathrm{KE}+\mathrm{PE} \\ \Rightarrow & E_0 & =\mathrm{KE}+\mathrm{V}(x) \\ \Rightarrow & \mathrm{KE} & =E_0-V(x) \end{array} \end{aligned}$$
$$\text { at } A_1 x=0, V(x)=E_0$$
$$\Rightarrow \quad \mathrm{KE}=E_0-E_0=0$$
$$ \begin{aligned} & \text { at } B_1 V(x)< E_0 \\ & \Rightarrow \quad \text { KE } > 0 \quad \text{(positive)} \end{aligned}$$
at C and $$D_1 V(x)=0$$
$$\Rightarrow \mathrm{KE}$$ is maximum at $$F_1 V(x)=E_0$$
Hence, $$\mathrm{KE}=0$$
The variation is shown in adjacent diagram.
Velocity versus $$x$$ graph
As $$\mathrm{KE}=\frac{1}{2} m v^2$$
$$\therefore$$ At $$A$$ and $$F$$, where $$K E=0, v=0$$.
At $$C$$ and $$D, \mathrm{KE}$$ is maximum. Therefore, $$v$$ is $$\pm \max$$.
At $$B, \mathrm{KE}$$ is positive but not maximum.
Therefore, $$\quad v$$ is $$\pm$$ some value $$\quad$$ (< max.)
The variation is shown in the diagram.
A ball of mass $$m$$, moving with a speed $$2 v_0$$, collides inelastically $$(e>0)$$ with an identical ball at rest. Show that
(a) For head-on collision, both the balls move forward.
(b) For a general collision, the angle between the two velocities of scattered balls is less than $$90^{\circ}$$.
(a) Let $$v_1$$ and $$v_2$$ are velocities of the two balls after collision. Now, by the principle of conservation of linear momentum,
$$\begin{aligned} 2 m v_0 & =m v_1+m v_2 \\ \text{or}\quad 2 v_0 & =v_1+v_2 \\ \text{and}\quad e & =\frac{v_2-v_1}{2 v_0} \end{aligned}$$
$$\begin{array}{lr} \Rightarrow & v_2=v_1+2 v_0 e \\ \therefore & 2 v_1=2 v_0-2 e v_0 \\ \therefore & v_1=v_0(1-e) \end{array}$$
Since, $$e<1 \Rightarrow v_1$$ has the same sign as $$v_0$$, therefore, the ball moves on after collision.
(b) Consider the diagram below for a general collision.
By principle of conservation of linear momentum,
$$P=P_1+P_2$$
For inelastic collision some KE is lost, hence $$\frac{p^2}{2 m}>\frac{p_1^2}{2 m}+\frac{p_2^2}{2 m}$$
$$\therefore \quad p^2>p_1^2+p_2^2$$
Thus, $$p, p_1$$ and $$p_2$$ are related as shown in the figure.
$$\theta$$ is acute (less than 90$$^\circ$$) ($$p^2=p_1^2+p_2^2$$ would given $$\theta=90^{\circ}$$)
Consider a one-dimensional motion of a particle with total energy $$\mathbf{E}$$. There are four regions $$A, B, C$$ and $$D$$ in which the relation between potential energy $$V$$, kinetic energy $$(K)$$ and total energy $$E$$ is as given below
$$\begin{array}{ll} \text { Region A : } V > E & \text { Region B : } V < E \\ \text { Region } \mathbf{C}: K < E & \text { Region D: } V > E \end{array}$$
State with reason in each case whether a particle can be found in the given region or not.
$$\begin{aligned} &\text { We know that }\\ &\begin{aligned} & & \text { Total energy } E & =P E+K E \\ \Rightarrow & & E & =V+K\quad \text{... (i)} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { For region A Given, } V>E \text {, From Eq. (i) }\\ &K=E-V \end{aligned}$$
$$\begin{aligned} &\text { as }\\ &V>E \Rightarrow E-V<0 \end{aligned}$$
Hence, $$K<0$$, this is not possible.
For region B Given, $$V
This is possible because total energy can be greater than PE $$(V)$$.
For region C Given, $$K>E \Rightarrow K-E>0$$
from Eq. (i) $$\mathrm{PE}=V=E-K<0$$
Which is possible, because PE can be negative.
For region D Given, $$V>K$$
This is possible because for a system PE (V) may be greater than KE (K).
The bob $$A$$ of a pendulum released from horizontal to the vertical hits another bob $$B$$ of the same mass at rest on a table as shown in figure.
If the length of the pendulum is 1 m , calculate
(a) the height to which bob A will rise after collision.
(b) the speed with which bob B starts moving.
Neglect the size of the bobs and assume the collision to be elastic.
When ball $$A$$ reaches bottom point its velocity is horizontal, hence, we can apply conservation of linear momentum in the horizontal direction.
(a) Two balls have same mass and the collision between them is elastic, therefore, ball $$A$$ transfers its entire linear momentum to ball $$B$$. Hence, ball $$A$$ will come to at rest after collision and does not rise at all.
$$\begin{aligned} &\text { (b) Speed with which bob } B \text { starts moving }\\ &\begin{aligned} & =\text { Speed with which bob } A \text { hits bob } B \\ & =\sqrt{2 g h} \\ & =\sqrt{2 \times 9.8 \times 1} \\ & =\sqrt{19.6} \\ & =4.42 \mathrm{~m} / \mathrm{s} \end{aligned} \end{aligned}$$