11.0 Collision in two dimension

Two bodies $A$ & $B$ of masses $m_1$ & $m_2$ moving with velocity ${\overrightarrow u _1}$ & ${\overrightarrow u _2}$ $\left( {{{\overrightarrow u }_1}\cos \theta > {{\overrightarrow u }_2}} \right)$ collides with each other.

The coefficient of restitution of the colliding bodies is $e$.

Let us assume that these bodies move with velocities ${\overrightarrow v _1}$ & ${\overrightarrow v _2}$ after collision in the direction as shown in the figure.

Let $NN'$ & $TT'$ be the direction along common normal $(CN)$ & common tangent $(CT)$ respectively.

From the FBD ${\overrightarrow F _C}$ is the force which acts during collision along the common normal $(CN)$ direction $(NN')$ & no force acts along the common tangent $(CT)$ direction $TT'$ during collision.

So, the linear momentum of the individual particles (if mass is constant) are conserved along the common tangent $(CT)$.

Conservation of linear momentum of individual particles along common tangent $(CT)$.

For body $A$, $$\begin{equation} \begin{aligned} {\overrightarrow p _{{i_{C{T_A}}}}} = {\overrightarrow p _{{f_{C{T_A}}}}} \\ {m_1}{u_1}\sin \theta = {m_1}{v_1}\sin \alpha \\ {u_1}\sin \theta = {v_1}\sin \alpha \quad ...(i) \\\end{aligned} \end{equation} $$

For body $B$, $$\begin{equation} \begin{aligned} {\overrightarrow p _{{i_{C{T_B}}}}} = {\overrightarrow p _{{f_{C{T_B}}}}} \\ 0 = - {m_2}{v_2}\sin \beta \\ {v_2} = 0\quad ...(ii) \\\end{aligned} \end{equation} $$

Initial and final linear momentum of the system are,

$$\begin{equation} \begin{aligned} {\overrightarrow p _i} = {m_1}{\overrightarrow u _1} + {m_2}{\overrightarrow u _2} = {m_1}{u_1}\left( {\cos \theta \widehat i + \sin \theta \widehat j} \right) + {m_2}{u_2}\widehat i \\ {\overrightarrow p _f} = {m_1}{\overrightarrow v _1} + {m_2}{\overrightarrow v _2} = {m_1}{v_1}\left( {\cos \alpha \widehat i + \sin \alpha \widehat j} \right) + {m_2}{v_2}\left( {\cos \beta \widehat i - \sin \beta \widehat j} \right) \\\end{aligned} \end{equation} $$

Linear momentum will be conserved in both $x$ & $y$ direction as no external force acts on the system in these directions. Therefore,

$$\begin{equation} \begin{aligned} {\overrightarrow p _i} = {\overrightarrow p _f} \\ {m_1}{u_1}\left( {\cos \theta \widehat i + \sin \theta \widehat j} \right) + {m_2}{u_2}\widehat i = {m_1}{v_1}\left( {\cos \alpha \widehat i + \sin \alpha \widehat j} \right) + {m_2}{v_2}\left( {\cos \beta \widehat i - \sin \beta \widehat j} \right) \\ \left( {{m_1}{u_1}\cos \theta + {m_2}{u_2}} \right)\widehat i + {m_1}{u_1}\sin \theta \widehat j = \left( {{m_1}{v_1}\cos \alpha + {m_2}{v_2}\cos \beta } \right)\widehat i + \left( {{m_1}{v_1}\sin \alpha - {m_2}{v_2}\sin \beta } \right)\widehat j \\\end{aligned} \end{equation} $$

Equating LHS & RHS we get, $$\begin{equation} \begin{aligned} \left( {{m_1}{u_1}\cos \theta + {m_2}{u_2}} \right) = \left( {{m_1}{v_1}\cos \alpha + {m_2}{v_2}\cos \beta } \right) = \left( {{m_1}{v_1}\cos \alpha + {m_2}{v_2}\cos \beta } \right)\quad ...(iii) \\ {m_1}{u_1}\sin \theta = \left( {{m_1}{v_1}\sin \alpha - {m_2}{v_2}\sin \beta } \right)\quad ...(iv) \\\end{aligned} \end{equation} $$

Coefficient of restitution $(e)$, $$\begin{equation} \begin{aligned} e = \frac{{{v_{{2_{CN}}}} - {v_{{1_{CN}}}}}}{{{u_{{1_{CN}}}} - {u_{{2_{CN}}}}}} \\ e = \frac{{{v_2}\cos \beta - {v_1}\cos \alpha }}{{{u_1}\cos \theta - {u_2}}}\quad ...(v) \\\end{aligned} \end{equation} $$

Using equations $(i)$ to $(v)$, any problem related to collision in two dimensions can be solved.

Let us now understand the above concept better with the help of an example.

Question 24. A ball of mass $m$ hits a floor with a speed $u$ making an angle of incidence $\alpha $ with the normal. The coefficient of restitution is $'e'$. Find the speed of the reflected ball and the angle of reflection of the ball.

Solution: Let the velocity of the ball of mass $m$ be $v$ after collision and makes an angle $\beta $ with the normal.

Initial and final linear momentum are, $$\begin{equation} \begin{aligned} {\overrightarrow p _i} = mu\sin \alpha \widehat i - mu\cos \alpha \widehat j \\ {\overrightarrow p _f} = mv\sin \beta \widehat i + mv\cos \beta \widehat j \\\end{aligned} \end{equation} $$

Linear momentum will be conserved only in $x$ direction as no external force acts on the system in $x$ directions.

$$\begin{equation} \begin{aligned} {\overrightarrow p _{{i_x}}} = {\overrightarrow p _{{f_x}}} \\ mu\sin \alpha \widehat i = mv\sin \beta \widehat i \\ u\sin \alpha = v\sin \beta \quad ...(iii) \\\end{aligned} \end{equation} $$

Coefficient of restitution $(e)$, $$\begin{equation} \begin{aligned} e = \frac{{{v_{{2_{CN}}}} - {v_{{1_{CN}}}}}}{{{u_{{1_{CN}}}} - {u_{{2_{CN}}}}}} \\ e = \frac{{0 - \left( {v\cos \beta \widehat j} \right)}}{{ - u\cos \alpha \widehat j - 0}} \\ eu\cos \alpha = v\cos \beta \quad ...(iv) \\\end{aligned} \end{equation} $$

Dividing equation $(iii)$ by $(iv)$ we get, $$\begin{equation} \begin{aligned} \frac{{\tan \alpha }}{e} = \tan \beta \\ \beta = {\tan ^{ - 1}}\left( {\frac{{\tan \alpha }}{e}} \right) \\\end{aligned} \end{equation} $$

Squaring and adding equation $(iii)$ & $(iv)$ we get,

$$\begin{equation} \begin{aligned} {v^2}\left( {{{\sin }^2}\beta + {{\cos }^2}\beta } \right) = {\left( {u\sin \alpha } \right)^2} + {\left( {eu\cos \alpha } \right)^2} \\ v = \sqrt {{{\left( {u\sin \alpha } \right)}^2} + {{\left( {eu\cos \alpha } \right)}^2}} \\\end{aligned} \end{equation} $$

Question 25. Why we conserve linear momentum of individual particles along common tangent in two dimensional collision.

Solution: We conserve linear momentum of individual particles along common tangent in two dimensional collision because during the collision, force $\left( {{{\overrightarrow F }_C}} \right)$ acts only along common normal.

So, the velocity along the common tangent of the individual particles remain unchanged and hence the linear momentum along the common tangent.

Question 26. A ball is projected from the ground at some angle with the horizontal. Coefficient of restitution between the ball and the ground is $e$. The ball undergoes a motion as shown in the figure.

Find the ratio of the time of flight, horizontal range and maximum height in two successive paths.

Solution: Let the ball is projected with a velocity $v$ at an angle $\theta $ with the horizontal.

Due to the collision, only the velocity in the vertical direction (along common normal) will be affected. There will be no change in the velocity in horizontal direction.

At point $P$, collision takes place. Let assume ball as object 1 and ground as object 2.

Since the ground is always at rest therefore, $u_2=0$ & $v_2=0$.

From the newton’s law of restitution we get, $$\begin{equation} \begin{aligned} e = \frac{{{v_{{2_{CN}}}} - {v_{{1_{CN}}}}}}{{{u_{{1_{CN}}}} - {u_{{2_{CN}}}}}} \\ e = \frac{{0 - {v_y}\widehat j}}{{ - v\sin \theta \widehat j - 0}} \\ {v_y} = ev\sin \theta \\\end{aligned} \end{equation} $$