Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?
According to work-energy theorem,
Change in $$\mathrm{KE}=$$ Work done by the retarding force
KE of the body $$=$$ Retarding force $\times$ Displacement
As KE of the bodies and retarding forces applied on them are same, therefore, both bodies will travel equal distances before coming to rest.
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).