[ 5 Marks Questions ]

Question 15 Marks

Obtain the expressions for the velocities of the two bodies after collision in the case of one dimensional elastic collision and discuss the special cases.

Answer

Consider two elastic bodies of masses $m_1$ and $m_2$ moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure.

In order to have collision, we assume that the mass $m_1$ moves faster than mass $m _2$ i.e., $u _1> u _2$
For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same

	Momentum of mass m_(1)	Momentum of mass m_(2)	Total linear momentum
Before collision	p_(i1)=m_(1)u_(1)	p_(i2)=m_(2)u_(2)	p_(i)=p_(i1)+p_(i2)
			p_(i)=m_(1)u_(1)+m_(2)u_(2)
After collision	p_(f1)=m_(1)v_(1)	p_(f2)=m_(2)v_(2)	p_(f)=p_(f1)+p_(rho2)
.			p_(f)=m_(1)v_(1)+m_(2)v_(2)

From the law of conservation of linear momentum,
Total momentum before collision $\left( p _{ i }\right)=$ Total momentum after collision $\left( p _{ f }\right)$
$\begin{aligned} m_1 u_1+m_2 u_2 & =m_1 v_1+m_2 v_1 \\ \text { Or } \quad m_1\left(u_1-v_1\right) & =m_2\left(v_2-u_2\right)\end{aligned}$
Further,

	Kinetic energy of mass m_(1)	Kinetic energy of mass m_(2)	Total kinetic energy
Before collision	KE_(n)=(1)/(2)m_(1)u_(1)^(2)	KE_(i2)=(1)/(2)m_(2)u_(2)^(2)	KE_(i)=KE_(i1)+KE_(i2)
After collision	KE_(f1)=(1)/(2)m_(1)v_(1)^(2)	KE_(f2)=(1)/(2)m_(2)v_(2)^(2)	KE_(i)=(1)/(2)m_(1)u_(1)^(2)+(1)/(2)m_(2)u_(2)^(2)
			KE_(i1)+KE_(i2)
			(1)/(2)m_(1)v_(1)^(2)+(1)/(2)m_(2)v_(2)^(2)

For elastic collision,
Total kinetic energy before collision $KE _{ i }=$ Total kinetic energy after collision $KF _{ f }$
$\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2$
After simplifying and rearranging the terms,
$m_1\left(u_1^2-v_1^2\right)=m_2\left(v_2^2-u_2^2\right)$
Using the formula $a^2-b^2=(a+b)(a-b)$, we can rewrite the above equation as
$m_1\left(u_1+v_1\right)\left(u_1-v_1\right)=m_2\left(v_2+u_2\right)\left(v_2-u_2\right)$
Dividing equation (iv) by (ii) gives,
$
\begin{aligned}
\frac{m_1\left(u_1+v_1\right)\left(u_1-v_1\right)}{m_1\left(u_1-v_1\right)} & =\frac{m_2\left(v_2+u_2\right)\left(v_2-u_2\right)}{m_2\left(v_2-u_2\right)} \\
u_1+v_1= & v_2+u_1 \\
u_1-u_2= & v_2-v_1
\end{aligned}
$
Rearranging, (v)
Equation (v) can be rewritten as
$u_1-u_2=-\left(v_1-v_2\right)$
This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for $v _1$ and $v _2$,
$v _1= v _2+ u _2- u _1$
$\text { Or } v _2= u _1+ v _1- u _1$
To find the final velocities $v_1$ and $v_2$ :
Substituting equation (vii) in equation (ii) gives the velocity of $m_1$ as
$m_1\left(u_1-v_1\right)=m_2\left(u_1+v_1-u_2-u_2\right)$
$m_1 u_1-m_1 v_1=m_2\left(u_1+v_1-2 u_2\right)$
$m_1 u_1+2 m_2 u_2=m_1 v_1+m_2 v_1$
$\text { or } \quad v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right) u_1+\left(\frac{2 m_2}{m_1+m_2}\right) u_2$
Or
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right) u_1+\left(\frac{2 m_2}{m_1+m_2}\right) u_2$
Similarly, by substituting (vi) in equation (ii) or substituting equation (viii) in equation (vii), we get the final velocity of $m_2$ as
$v_2=\left(\frac{2 m_1}{m_1+m_2}\right) u_1+\left(\frac{m_2-m_1}{m_1+m_2}\right) u_2$
Case 1: When bodies has the same mass i.e., $m_1=m_2$,
$\text { equation (viii) } \Rightarrow \quad v_1=(0) u_1+\left(\frac{2 m_2}{2 m_2}\right) u_2$
$v_1=u_2$
$\text { equation (ix) } \Rightarrow \quad v_2=\left(\frac{2 m_1}{2 m_1}\right) u_1+(0) u_2$
$v_2=u_1$
The equations $( x )$ and (xi) show that in one dimensional elastic collision, when two bodies of equal mass collide after the collision their velocities are exchanged.
Case 2: When bodies have the same mass i.e., $m_1=m_2$ and second body (usually called target) is at rest $\left( u _2=0\right)$,By substituting $m _1 m _2=$ and $u _2=0$ in equations (viii) and equations (ix) we get, from equation (viii) $\Rightarrow V_1=0$
from equation (ix) $\Rightarrow v_2=u_1 \ldots$ (xiii)
Equations (xii) and (xiii) show that when the first body comes to rest the second body moves with the initial velocity of the first body.
Case 3: The first body is very much lighter than the second body $\left(m_1 \ll m_2, \frac{m_1}{m_2} \ll<\right)$ then the ratio $\frac{m_1}{m_2} \approx 0$ and also if the target is at rest $\left(u_2=0\right)$
Dịviding numerator and denominator of equation (viii) by $m_2$, we get
$v_1=\left(\frac{\frac{m_1}{m_2}-1}{\frac{m_1}{m_2}+1}\right) u_1+\left(\frac{2}{\frac{m_1}{m_2}+1}\right)(0)$
$v_1=\left(\frac{0-1}{0+1}\right) u_1 \quad \text { SamacheerKalvi.Guru }$
$v_1=-u_1$
Similarly, Dividing numerator and denominator of equation (ix) by $m_2$, we get
$v_2=\left(\frac{2 \frac{m_1}{m_2}}{\frac{m_1}{m_2}+1}\right) u_1+\left(\frac{1-\frac{m_1}{m_2}}{\frac{m_1}{m_2}+1}\right)(0)$
$
\begin{aligned}
v_2 & =(0) u_1+\left(\frac{1-\frac{m_1}{m_2}}{\frac{m_1}{m_2}+1}\right)(0) \\
v_2 & =0
\end{aligned}
$
The equation (xiv) implies that the first body which is lighter returns back (rebounds) in the opposite direction with the same initial velocity as it has a negative sign. The equation (xv) implies that the second body which is heavier in mass continues to remain at rest even after collision. For example, if a ball is thrown at a fixed wall, the ball will bounce back from the wall with the same velocity with which it was thrown but in opposite direction.
Case 4: The second body is very much lighter than the first body $\left(m_2 \ll m_1, \frac{m_2}{m_1} \ll 1\right)$ then the ratio $\frac{m_2}{m_1} \approx 0$ and also if the target is at rest $\left(u_2=0\right)$
Dividing numerator and denominator of equation (4.53) by $m_1$, we get
$v_1=\left(\frac{1-\frac{m_2}{m_1}}{1+\frac{m_2}{m_1}}\right) u_1+\left(\frac{2 \frac{m_2}{m_1}}{1+\frac{m_2}{m_1}}\right)
\text { SamacheerKalvi.Guru }$
$(0)$
$v_1=\left(\frac{1-0}{1+0}\right) u_1+\left(\frac{0}{1+0}\right)(0)$
$v_1=u_1$
Similarly,
Dividing numerator and denominator of equation (xiii) by $m_1$, we get
$v_2=\left(\frac{2}{1+\frac{m_2}{m_1}}\right) u_1+\left(\frac{\frac{m_2}{m_1}-1}{1+\frac{m_2}{m_1}}\right)(0)$
$v_2=\left(\frac{2}{1+0}\right) u_1$
$v_2=2 u_1$
The equation (xvi) implies that the first body which is heavier continues to move with the same initial velocity. The equation (xvii) suggests that the second body which is lighter will move with twice the initial velocity of the first body. It means that the lighter body is thrown away from the point of collision.

View full question & answer→

Question 25 Marks

Obtain an expression for the critical vertical of a body revolving in a vertical circle

Answer

Imagine that a body of mass (m) attached to one end of a massless and inextensible string executes circular motion in a vertical plane with the other end of the string fixed. The length of the string becomes the radius (r) of the circular path (See figure).

Let us discuss the motion of the body by taking the free body diagram (FBD) at a position where the position vector $(\vec{r})$ makes an angle $\theta$ with the vertically downward direction and the instantaneous velocity is as shown in Figure.
There are two forces acting on the mass.
1. Gravitational force which acts downward
2. Tension along the string.
Applying Newton's second law on the mass, in the tangential direction,
$m g \sin \theta=m a_t$
$m g \sin \theta=-m\left(\frac{d v}{d t}\right)$
where, $a_t=-\frac{d v}{d t}$ is tangential retardation
In the radial direction,
$T -m g \cos \theta=m a_r$
$T -m g \cos \theta=\frac{m v^2}{r}$
The circle can be divided into four sections A, B, C, D for better understanding of the motion.
The four important facts to be understood from the two equations are as follows:
(i) The mass is having tangential acceleration $\left( g \sin \theta\right.$ ) for all values of $\theta$ (except $\theta=0^{\circ}$ ), it is clear that this vertical circular motion is not a uniform circular motion.
(ii) From the equations (ii) and (i) it is understood that as the magnitude of velocity is not a constant in the course of motion, the tension in the string is also not constant.
(iii) The equation (ii), $T =m g \cos \theta \frac{m v^2}{r}$ highlights that in sections $A$ and $D$ of the circle, $\left(\right.$ for $-\frac{\pi}{2}<\theta<\frac{\pi}{2} ; \cos \theta$ is positive $)$, the term $m g \cos \theta$ is always greater than zero. Hence the tension cannot vanish even when the velocity vanishes.
(iv) The equation (ii), $\frac{m v^2}{r}= T -m g \cos \theta$; further highlights that in sections $B$ and $C$ of the circle, $\left(\right.$ for $\frac{\pi}{2}<\theta<\frac{3 \pi}{2} ; \cos \theta$ is negative $)$, the second term is always greater than zero.
Hence velocity cannot vanish, even when the tension vanishes.
These points are to be kept in mind while solving problems related to motion in vertical circle. To start with let us consider only two positions, say the lowest point 1 and the highest point 2 as shown in Figure for further analysis. Let the velocity of the body at the lowest point 1 be $\vec{v}_1$, at the highest point 2 be $\vec{v}_2$ and $\vec{v}$ at any other point. The direction of velocity is tangential to the circular path at all points. Let $\overrightarrow{ T }_1$ be the tension in the string at the lowest point and $\overrightarrow{ T }_2$ be, the tension at the highest point and $\overrightarrow{ T }$ be the tension at any other point. Tension at each point acts towards the center. The tensions and velocities at these two points can be found by applying the law of conservation of energy.

For the lowest point (1)
When the body is at the lowest point 1 , the gravitational force $m \vec{g}$ which acts on the body (vertically downwards) and another one is the tension $\overrightarrow{ T }_1$, acting vertically upwards, i.e. towards the center. From the equation (ii), we get
$T _1-m g =\frac{m v_1^2}{r}$
$T _1 =\frac{m v_1^2}{r}+m g$
For the highest point (2)
At the highest point 2, both the gravitational force $mg$ on the body and the tension $T _2$ act downwards, i.e. towards the center again.
$T _2+m g =\frac{m v_2^2}{r_2}$
$T _2 =\frac{m v^2}{r_2}-m g$
From equations (iv) and (ii), it is understood that $T_1>T_2$. The difference in tension $T_1-T_2$ is obtained by subtracting equation (iv) from equation (ii).
$T _1- T _2=\frac{m v_1^2}{r}+m g-\left(\frac{m v_2^2}{r}-m g\right)=\frac{m v_1^2}{r}+m g-\frac{m v_2^2}{r}+m g$
$T _1- T _2=\frac{m}{r}\left[v_1^2-v_2^2\right]+2 m g$
The term $\left[v_1^2-v_2^2\right]$ can be found easily by applying law of conservation of energy at point 1 and also at point 2 .Note: The tension will not do any work on the mass as the tension and the direction of motion is always perpendicular.The gravitational force is doing work on the mass, as it is a conservative force the total energy of the mass is conserved throughout the motion.
Total energy at point $1\left(E_1\right)$ is same as the total energy at a point $2\left(E_2\right)$ $E _1= E _2$
Potential energy at point $1, U_1=0$ (by taking reference as point 1)
Kinetic energy at point $1, KE _1=\frac{1}{2} m v_1^2$
Total energy at point $1, E _1= U _1+ KE _1=0+\frac{1}{2} m v_1^2=\frac{1}{2} m v_1^2$
Similarly, Potential energy at point $2, U_2=m g(2 r)$ (h is $2 r$ from point 1)
Kinetic Energy at point $2, KE _2=\frac{1}{2} m v_2^2$Total energy at point $2, E _2= U _2+ KE _2=2 m g r+\frac{1}{2} m v_2^2$
From the law of conservation of energy given in equation (vi), we get
$\frac{1}{2} m v_1^2=2 m g r+\frac{1}{2} m v_2^2$
After rearranging, $\quad \frac{1}{2} m\left(v_1^2-v_2^2\right)=2 m g r$
$v_1^2-v_2^2=4 g r$
Substituting equation (vii) in equation (iv) we get,
$T _1- T _2=\frac{m}{r}[4 g r]+2 m g$
Therefore, the difference in tension is
$T _1- T _2=6\ mg$
Minimum speed at the highest point (2)
The body must have a minimum speed at point 2 otherwise, the string will slack before reaching point 2 and the body will not loop the circle. To find this minimum speed let us take the tension $T _2=0$ in equation (iv).
$
\begin{aligned}
0 & =\frac{m v_2^2}{r}-m g \\
\frac{m v_2^2}{r} & =m g \\
v_2^2 & =r g \\
v_2 & =\sqrt{g r}
\end{aligned}
$
The body must have a speed at point 2, $v_2 \geq \sqrt{g r}$ to stay in the circular path.
Maximum speed at the lowest point 1
To have this minimum speed $\left(v_2=\sqrt{g r}\right.$ ) at point 2 , the body must have minimum speed also at point 1.
By making use of equation (vii) we can find the minimum speed at point 1.
$v_1^2-v_2^2=4 g r$
Substituting equation (ix) in (vii),
$v_1^2-g r =4 g r$
$v_1{ }^2 =5 g r$
$v_1 =\sqrt{5 g r}$
The body must have a speed at point $1, v_1 \geq \sqrt{5 g r}$ to stay in the circular path.
From equations (ix) and ( $x$ ), it is clear that the minimum speed at the lowest point 1 should be $v$ 5 times more than the minimum speed at the highest point 2 , so that the body loops without leaving the circle.

View full question & answer→

Physics [ 5 Marks Questions ]

Test yourself on this topic