10.1 Head on elastic collision

1.1 Position of centre of mass for a system of two particles

2.1 Rod
2.2 Semicircular ring
2.3 Semicircular disc
2.4 Solid sphere
2.5 Hemispherical Shell
2.6 Rectangular Plate
2.7 Square Plate
2.8 Circular Plate
2.9 Solid Cone
2.10 Hollow Cone
2.11 Questions
2.12 Centre of mass of a rigid complex bodies

3.1 Motion of mass of the system

5.1 Rocket Propulsion

6.1 Instantaneous impulse

9.1 For direct collision
9.2 For collision in two dimension
9.3 For oblique collision

10.1 Head on elastic collision
10.2 Head on inelastic collision

In elastic collision, linear momentum and kinetic energy of the colliding bodies are conserved before and after the collision.

In elastic collision, both the bodies comes to their original shape and size after the collision.

So no energy remains stored as deformation potential energy in the bodies.

So, the total energy before and after the collision remains same.

Therefore the kinetic energy is also conserved in addition to the linear momentum.

Let us consider two balls $A$ & $B$ of masses $m_1$ & $m_2$ respectively. Their velocities are $u_1$ & $u_2$ $\left( {{u_1} > {u_2}} \right)$ before the collision and $v_1$ & $v_2$ after the collision along the same line.

Conservation of linear momentum along the horizontal direction,

$${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$$ or $${m_1}\left( {{u_1} - {v_1}} \right) = {m_2}\left( {{v_2} - {u_2}} \right)\quad ...(i)$$

Conservation of total kinetic energy before and after the collision,

$$\frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2 = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2$$ or $$\begin{equation} \begin{aligned} \frac{1}{2}{m_1}\left( {u_1^2 - v_1^2} \right) = \frac{1}{2}{m_2}\left( {v_2^2 - u_2^2} \right) \\ {m_1}\left( {{u_1} - {v_1}} \right)\left( {{u_1} + {v_1}} \right) = {m_2}\left( {{v_2} - {u_2}} \right)\left( {{v_2} + {u_2}} \right)\quad ...(ii) \\\end{aligned} \end{equation} $$

Dividing equation $(ii)$ by $(i)$ we get, $$\left( {{u_1} + {v_1}} \right) = \left( {{v_2} + {u_2}} \right)$$ or $${v_2} - {v_1} = {u_1} - {u_2}$$ or $$\frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}} = 1 = e$$

Also, $${v_2} = {v_1} + {u_1} - {u_2}\quad ...(iii)$$

From equation $(i)$ & $(iii)$ we get, $$\begin{equation} \begin{aligned} {m_1}\left( {{u_1} - {v_1}} \right) = {m_2}\left( {{v_1} + {u_1} - {u_2} - {u_2}} \right) \\ {m_1}\left( {{u_1} - {v_1}} \right) = {m_2}\left( {{v_1} + {u_1} - 2{u_2}} \right) \\ {v_1}\left( {{m_1} + {m_2}} \right) = \left( {{m_1} - {m_2}} \right){u_1} + 2{m_2}{u_2} \\\end{aligned} \end{equation} $$ $${v_1} = \frac{{\left( {{m_1} - {m_2}} \right)}}{{\left( {{m_1} + {m_2}} \right)}}{u_1} + \frac{{2{m_2}}}{{\left( {{m_1} + {m_2}} \right)}}{u_2}\quad ...(v)$$

Similarly, $${v_2} = \frac{{\left( {{m_2} - {m_1}} \right)}}{{\left( {{m_1} + {m_2}} \right)}}{u_2} + \frac{{2{m_1}}}{{\left( {{m_1} + {m_2}} \right)}}{u_1}\quad ...(vi)$$

Substituting the value in equation $(v)$ & $(vi)$ we get, $${v_1} = {u_2}\quad \& \quad {v_2} = {u_1}$$

When ${m_1} \gg {m_2}$, then $\left( {\frac{{{m_2}}}{{{m_1}}}} \right) \approx 0$.

Therefore substituting the condition in equation $(v)$ & $(vi)$ we get, $$\begin{equation} \begin{aligned} {v_1} = \frac{{\left( {1 - \frac{{{m_2}}}{{{m_1}}}} \right)}}{{\left( {1 + \frac{{{m_2}}}{{{m_1}}}} \right)}}{u_1} + \frac{{2\frac{{{m_2}}}{{{m_1}}}}}{{\left( {1 + \frac{{{m_2}}}{{{m_1}}}} \right)}}{u_2} \\ {v_1} \approx {u_1} \\\end{aligned} \end{equation} $$

Similarly, $$\begin{equation} \begin{aligned} {v_2} = \frac{{\left( {\frac{{{m_2}}}{{{m_1}}} - 1} \right)}}{{\left( {1 + \frac{{{m_2}}}{{{m_1}}}} \right)}}{u_2} + \frac{2}{{\left( {1 + \frac{{{m_2}}}{{{m_1}}}} \right)}}{u_1} \\ {v_2} \approx 2{u_1} - {u_2} \\\end{aligned} \end{equation} $$

When ${m_2} \gg {m_1}$, then $\left( {\frac{{{m_1}}}{{{m_2}}}} \right) \approx 0$.

Therefore substituting the condition in equation $(v)$ & $(vi)$ we get, $$\begin{equation} \begin{aligned} {v_1} = \frac{{\left( {\frac{{{m_1}}}{{{m_2}}} - 1} \right)}}{{\left( {\frac{{{m_1}}}{{{m_2}}} + 1} \right)}}{u_1} + \frac{2}{{\left( {\frac{{{m_1}}}{{{m_2}}} + 1} \right)}}{u_2} \\ {v_1} \approx 2{u_2} - {u_1} \\\end{aligned} \end{equation} $$

Similarly, $$\begin{equation} \begin{aligned} {v_2} = \frac{{\left( {1 - \frac{{{m_1}}}{{{m_2}}}} \right)}}{{\left( {\frac{{{m_1}}}{{{m_2}}} + 1} \right)}}{u_2} + \frac{{2\frac{{{m_1}}}{{{m_2}}}}}{{\left( {\frac{{{m_1}}}{{{m_2}}} + 1} \right)}}{u_1} \\ {v_2} \approx {u_2} \\\end{aligned} \end{equation} $$

Question 21. Two bodies of mass $m$ & $3m$ moving in opposite directions collide elastically with velocities $v$ & $2v$. Find their velocities after collision.

Solution: Initial linear momentum $\left( {{{\overrightarrow p }_i}} \right)$,

$$\begin{equation} \begin{aligned} {\overrightarrow p _i} = mv\widehat i + 3m( - 2v)\widehat i \\ {\overrightarrow p _1} = - 5mv\widehat i\quad ...(i) \\\end{aligned} \end{equation} $$

Let us assume the velocities of the two bodies after collision. Then the final linear momentum $\left( {{{\overrightarrow p }_f}} \right)$,

$${\overrightarrow p _f} = m{\overrightarrow v _1} + 3m{\overrightarrow v _2}\quad ...(ii)$$

As the collision is elastic. Therefore, $$\begin{equation} \begin{aligned} e = \frac{{{v_{{2_{CN}}}} - {v_{{1_{CN}}}}}}{{{u_{{1_{CN}}}} - {u_{{2_{CN}}}}}} \\ 1 = \frac{{{{\overrightarrow v }_2} - {{\overrightarrow v }_1}}}{{v\widehat i - ( - 2v)\widehat i}} \\ {\overrightarrow v _2} - {\overrightarrow v _1} = 3v\widehat i \\ {\overrightarrow v _2} = {\overrightarrow v _1} + 3v\widehat i\quad ...(iii) \\\end{aligned} \end{equation} $$

During collision, the linear momentum is conserved. So, from equation $(i)$, $(ii)$ & $(iii)$ we get,

$$\begin{equation} \begin{aligned} {\overrightarrow p _i} = {\overrightarrow p _f} \\ - 5mv\widehat i = m{\overrightarrow v _1} + 3m\left( {{{\overrightarrow v }_1} + 3v\widehat i} \right) \\ {\overrightarrow v _1} = - \frac{{7v}}{2}\widehat i\quad ...(iv) \\\end{aligned} \end{equation} $$

From equation $(iii)$ & $(iv)$ we get, $${\overrightarrow v _2} = - \frac{v}{2}\widehat i$$