10.2 Head on inelastic 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 inelastic collision, only linear momentum of the colliding bodies is conserved before and after the collision.

In inelastic collision, both the bodies do not come to their original shape and size after the collision.

So some fraction of energy remains stored as deformation potential energy in the bodies.

So, the total energy before and after the collision is not same. Therefore the kinetic energy is not conserved in case of inelastic collision.

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,

$${m_1}{\overrightarrow u _1} + {m_2}{\overrightarrow u _2} = {m_1}{\overrightarrow v _1} + {m_2}{\overrightarrow v _2}\quad ...(i)$$

For inelastic collision we know $(e<1)$, $$\begin{equation} \begin{aligned} e = \frac{{{v_{{2_{CN}}}} - {v_{{1_{CN}}}}}}{{{u_{{1_{CN}}}} - {u_{{2_{CN}}}}}} \\ e = \frac{{{{\overrightarrow v }_2} - {{\overrightarrow v }_1}}}{{{{\overrightarrow u }_1} - {{\overrightarrow u }_2}}} \\ {\overrightarrow v _2} = {\overrightarrow v _1} + \left( {{{\overrightarrow u }_1} - {{\overrightarrow u }_2}} \right)e\quad ...(ii) \\\end{aligned} \end{equation} $$

From equation $(i)$ & $(ii)$ we get, $$\begin{equation} \begin{aligned} {m_1}{\overrightarrow u _1} + {m_2}{\overrightarrow u _2} = {m_1}{\overrightarrow v _1} + {m_2}\left[ {{{\overrightarrow v }_1} + \left( {{{\overrightarrow u }_1} - {{\overrightarrow u }_2}} \right)e} \right] \\ {m_1}{\overrightarrow u _1} + {m_2}{\overrightarrow u _2} = {m_1}{\overrightarrow v _1} + {m_2}{\overrightarrow u _1}e - {m_2}{\overrightarrow u _2}e \\ {\overrightarrow v _1}\left( {{m_1} + {m_2}} \right) = \left( {{m_1} - e{m_2}} \right){\overrightarrow u _1} + \left( {{m_2} + e{m_2}} \right){\overrightarrow u _2} \\\end{aligned} \end{equation} $$

$$\begin{equation} \begin{aligned} {\overrightarrow v _1} = \frac{{\left( {{m_1} - {m_2}} \right)}}{{\left( {{m_1} + {m_2}} \right)}}{\overrightarrow u _1} + \frac{{2{m_2}}}{{\left( {{m_1} + {m_2}} \right)}}{\overrightarrow u _2} \\ {\overrightarrow v _2} = \frac{{\left( {{m_2} - {m_1}} \right)}}{{\left( {{m_1} + {m_2}} \right)}}{\overrightarrow u _2} + \frac{{2{m_1}}}{{\left( {{m_1} + {m_2}} \right)}}{\overrightarrow u _1} \\\end{aligned} \end{equation} $$

We get the exactly same equation as derived earlier for elastic collision.

$$\begin{equation} \begin{aligned} {\overrightarrow v _1} = \frac{{{m_1}{{\overrightarrow u }_1} + {m_2}{{\overrightarrow u }_2}}}{{\left( {{m_1} + {m_2}} \right)}} \\ {\overrightarrow v _2} = \frac{{{m_1}{{\overrightarrow u }_1} + {m_2}{{\overrightarrow u }_2}}}{{\left( {{m_1} + {m_2}} \right)}} \\\end{aligned} \end{equation} $$

So, when two bodies moving along the same line undergoes perfectly inelastic collision. They stick to each other.

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

$$\begin{equation} \begin{aligned} {\overrightarrow v _1} = \frac{{\left( {1 - \frac{{e{m_2}}}{{{m_1}}}} \right)}}{{\left( {1 + \frac{{{m_2}}}{{{m_1}}}} \right)}}{\overrightarrow u _1} + \frac{{\left( {\frac{{{m_2}}}{{{m_1}}} + \frac{{e{m_2}}}{{{m_1}}}} \right)}}{{\left( {1 + \frac{{{m_2}}}{{{m_1}}}} \right)}}{\overrightarrow u _2} \\ {\overrightarrow v _1} = {\overrightarrow u _1} \\\end{aligned} \end{equation} $$ and $$\begin{equation} \begin{aligned} {\overrightarrow v _2} = \frac{{\left( {\frac{{{m_2}}}{{{m_1}}} - e} \right)}}{{\left( {1 + \frac{{{m_2}}}{{{m_1}}}} \right)}}{\overrightarrow u _2} + \frac{{\left( {1 + e} \right)}}{{\left( {1 + \frac{{{m_2}}}{{{m_1}}}} \right)}}{\overrightarrow u _1} \\ {\overrightarrow v _2} = \left( {1 + e} \right){\overrightarrow u _1} - e{\overrightarrow u _2} \\\end{aligned} \end{equation} $$

When a very heavy body collides with a lighter body, velocity of the heavy body remains unchanged.

$${\overrightarrow v _1} = \frac{{\left( {1 - e} \right)}}{2}{\overrightarrow u _1} + \frac{{\left( {1 + e} \right)}}{2}{\overrightarrow u _2}$$ and $${\overrightarrow v _2} = \frac{{\left( {1 - e} \right)}}{2}{\overrightarrow u _2} + \frac{{\left( {1 + e} \right)}}{2}{\overrightarrow u _1}$$

Question 22. A ball of mass $m$ moving at a speed $v$ makes a head-on collision with an identical ball at rest. The kinetic energy of the balls after the collision is three fourth of the original. Find the coefficient of restitution.

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

$$\begin{equation} \begin{aligned} {\overrightarrow p _i} = mv\widehat i + m(0) \\ {\overrightarrow p _i} = mv\widehat i\quad ...(i) \\\end{aligned} \end{equation} $$

Let the velocity after collision be ${\overrightarrow v _1}$ & ${\overrightarrow v _2}$. Final linear momentum $\left( {{{\overrightarrow p }_f}} \right)$, $$\begin{equation} \begin{aligned} {\overrightarrow p _f} = m{\overrightarrow v _1} + m{\overrightarrow v _2} \\ {\overrightarrow p _f} = m\left( {{{\overrightarrow v }_1} + {{\overrightarrow v }_2}} \right)\quad ...(ii) \\\end{aligned} \end{equation} $$

From equation $(i)$ & $(ii)$ we get, $$\begin{equation} \begin{aligned} {\overrightarrow p _i} = {\overrightarrow p _f} \\ mv\widehat i = m\left( {{{\overrightarrow v }_1} + {{\overrightarrow v }_2}} \right) \\ v\widehat i = \left( {{{\overrightarrow v }_1} + {{\overrightarrow v }_2}} \right)\quad ...(iii) \\\end{aligned} \end{equation} $$

Coefficient of restitution $(e)$, $$\begin{equation} \begin{aligned} e = \frac{{{v_{{2_{CN}}}} - {v_{{1_{CN}}}}}}{{{u_{{1_{CN}}}} - {u_{{2_{CN}}}}}} \\ e = \frac{{{{\overrightarrow v }_2} - {{\overrightarrow v }_1}}}{{v\widehat i - 0}} \\ {\overrightarrow v _2} - {\overrightarrow v _1} = ev\widehat i\quad ...(iv) \\\end{aligned} \end{equation} $$

From equation $(iii)$ & $(iv)$ we get, $$\begin{equation} \begin{aligned} {\overrightarrow v _1} = \left( {\frac{{1 - e}}{2}} \right)v\widehat i\quad ...(v) \\ {\overrightarrow v _2} = \left( {\frac{{1 + e}}{2}} \right)v\widehat i\quad ...(vi) \\\end{aligned} \end{equation} $$

Initial kinetic energy $\left( {{K_i}} \right)$, $${K_i} = \frac{1}{2}m{v^2}\quad ...(vii)$$

Final kinetic energy $\left( {{K_f}} \right)$, $${K_f} = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2\quad ...(viii)$$

From the question, $$\begin{equation} \begin{aligned} {K_f} = \frac{3}{4}{K_i} \\ \frac{1}{2}m\left( {v_1^2 + v_2^2} \right) = \frac{3}{4} \times \frac{1}{2}m{v^2} \\ 4\left( {v_1^2 + v_2^2} \right) = 3{v^2}\quad ...(ix) \\\end{aligned} \end{equation} $$

From equation $(v)$, $(vi)$ & $(ix)$ we get, $$\begin{equation} \begin{aligned} 4{\left[ {\frac{{\left( {1 - e} \right)}}{2}v} \right]^2} + 4{\left[ {\frac{{\left( {1 + e} \right)}}{2}v} \right]^2} = 3{v^2} \\ e = \frac{1}{{\sqrt 2 }} \\\end{aligned} \end{equation} $$

Question 23. A bullet of mass $m$ moving with a horizontal velocity $u$ strikes a stationary block of mass $M$ suspended by a string of length $L$. The bullet gets embedded in the block. What is the maximum angle made by the string after impact?

Solution: Initial linear momentum, $${\overrightarrow p _i} = mu\widehat i + M(0)\quad ...(i)$$

After the bullet gets embedded in the block. They move with velocity $v$.

Final linear momentum, $${\overrightarrow p _f} = (M + m)v\widehat i\quad ...(ii)$$

From equation $(i)$ & $(ii)$ we get, $$\begin{equation} \begin{aligned} {\overrightarrow p _i} = {\overrightarrow p _f} \\ mu\widehat i = (m + M)v\widehat i \\ v = \frac{{mu}}{{(M + m)}}\widehat i\quad ...(iii) \\\end{aligned} \end{equation} $$

From the figure, $$h = L(1 - \cos \theta )\quad ...(iv)$$

Mechanical energy at position $A$ & $B$ are constant. Mathematically,

$$\begin{equation} \begin{aligned} P{E_A} + {K_A} = P{E_B} + {K_B} \\ 0 + \frac{1}{2}\left( {M + m} \right){v^2} = \left( {M + m} \right)gL(1 - \cos \theta ) \\ \cos \theta = 1 - \frac{{{u^2}}}{{2gL}}{\left( {\frac{m}{{M + m}}} \right)^2} \\ \theta = {\cos ^{ - 1}}\left[ {1 - \frac{{{u^2}}}{{2gL}}{{\left( {\frac{m}{{M + m}}} \right)}^2}} \right] \\\end{aligned} \end{equation} $$