12.0 Oblique collision

Oblique collision is a special case of collision in two dimension. In oblique collision, common normal $(CN)$ of the colliding bodies is inclined with the horizontal.

Let us consider a wedge of mass $M$ inclined at an angle $\theta $ with the horizontal and moving with horizontal velocity $u_1$. A ball of mass $m$ hits the wedge with velocity $v_2$ as shown in the figure.

The coefficient of restitution between the wedge and the ball 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.

Before collision, $$\begin{equation} \begin{aligned} {u_{{2_{CN}}}} = - {u_2}\cos \theta \quad ...(i) \\ {u_{{2_{CT}}}} = {u_2}\sin \theta \quad ...(ii) \\\end{aligned} \end{equation} $$

So, $${v_{{2_{CT}}}} = {u_2}\sin \theta \quad ...(iii)$$

Let us assume the velocity ${v_{{2_{CN}}}}$ along the common normal after collision.

From the coefficient of restitution $(e)$, $$\begin{equation} \begin{aligned} e = \frac{{{v_{{2_{CN}}}} - {v_{{1_{CN}}}}}}{{{u_{{1_{CN}}}} - {u_{{2_{CN}}}}}} \\ e = \frac{{{v_{{2_{CN}}}} - \left( {{v_1}\sin \theta } \right)}}{{0 - \left( { - {u_2}\cos \theta } \right)}} \\ {v_{{2_{CN}}}} = e{u_2}\cos \theta - {v_1}\sin \theta \quad ...(iv) \\\end{aligned} \end{equation} $$

Also, $$\begin{equation} \begin{aligned} {\overrightarrow v _2} = {\overrightarrow v _{{2_{CN}}}} + {\overrightarrow v _{{2_{CT}}}} \\ {\overrightarrow v _2} = \left[ {e{u_2}\cos \theta - {v_1}\sin \theta } \right]\cos \left( {90^\circ - \theta } \right)\widehat i + \left[ {{u_2}\sin \theta } \right]\cos \theta \widehat i \\ {\overrightarrow v _2} = \left( {e{u_2}\cos \theta \sin \theta - {v_1}{{\sin }^2}\theta + {u_2}\sin \theta \cos \theta } \right)\widehat i \\ {\overrightarrow v _2} = \left[ {\left( {e + 1} \right){u_2}\cos \theta \sin \theta - {v_1}{{\sin }^2}\theta } \right]\widehat i\quad ...(v) \\\end{aligned} \end{equation} $$

Initial and final linear momentum of the system are,

$$\begin{equation} \begin{aligned} {\overrightarrow p _i} = M{\overrightarrow u _1} + m{\overrightarrow u _2} = M{u_1}\widehat i - m{u_2}\widehat j \\ {\overrightarrow p _f} = M{\overrightarrow v _1} + m{\overrightarrow v _2} = - M{v_1}\widehat i + m{v_2}\widehat j \\\end{aligned} \end{equation} $$

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

In the vertical $y$ direction normal force due to the ground acts as an external force. Therefore, $$\begin{equation} \begin{aligned} {\overrightarrow p _{{i_x}}} = {\overrightarrow p _{{f_x}}} \\ M{u_1}\widehat i = \left( { - M{v_1} + m{v_2}} \right)\widehat i\quad ...(vi) \\\end{aligned} \end{equation} $$

From equation $(v)$ & $(vi)$ we get,

$$\begin{equation} \begin{aligned} M{u_1} = - M{v_1} + m\left[ {\left( {e + 1} \right){u_2}\cos \theta \sin \theta - {v_1}{{\sin }^2}\theta } \right] \\ {v_1} = \frac{{m\left[ {\left( {e + 1} \right){u_2}\cos \theta \sin \theta - {u_1}} \right]}}{{M + m{{\sin }^2}\theta }} \\\end{aligned} \end{equation} $$

Question 27. A ball of mass $m$ collides perpendicularly with a velocity $v_0$ on a smooth inclined wedge of mass $M$, kept on a smooth horizontal plane as shown in the figure. If the coefficient of restitution is $e$. Then determine the velocity of the wedge after the collision.

Solution: 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.

Before collision, $$\begin{equation} \begin{aligned} {u_{{2_{CN}}}} = - {v_0}\quad ...(i) \\ {u_{{2_{CT}}}} = 0\quad ...(ii) \\\end{aligned} \end{equation} $$

$${v_{{2_{CT}}}} = 0\quad ...(iii)$$

Let us assume the velocity ${v_{{2_{CN}}}}$ along the common normal after collision.

From the coefficient of restitution $(e)$,

$$\begin{equation} \begin{aligned} e = \frac{{{v_{{2_{CN}}}} - {v_{{1_{CN}}}}}}{{{u_{{1_{CN}}}} - {u_{{2_{CN}}}}}} \\ e = \frac{{{v_{{2_{CN}}}} - \left( { - {v_1}\sin \theta } \right)}}{{0 - \left( { - {v_0}} \right)}} \\ {v_{{2_{CN}}}} = e{v_0} - {v_1}\sin \theta \quad ...(iv) \\\end{aligned} \end{equation} $$

Also, $$\begin{equation} \begin{aligned} {\overrightarrow v _2} = {\overrightarrow v _{{2_{CN}}}} + {\overrightarrow v _{{2_{CT}}}} \\ {\overrightarrow v _2} = \left[ {e{v_0} - {v_1}\sin \theta } \right]\cos \left( {90^\circ - \theta } \right)\widehat i + 0 \\ {\overrightarrow v _2} = \left( {e{v_0}\sin \theta - {v_1}{{\sin }^2}\theta } \right)\widehat i\quad ...(v) \\\end{aligned} \end{equation} $$

Initial and final linear momentum of the system are,

$$\begin{equation} \begin{aligned} {\overrightarrow p _i} = M{\overrightarrow u _1} + m{\overrightarrow u _2} = 0\widehat i + m{v_0}\left( { - \sin \theta \widehat i - \cos \theta \widehat j} \right) \\ {\overrightarrow p _f} = M{\overrightarrow v _1} + m{\overrightarrow v _2} = - M{v_1}\widehat i + m{v_2}\widehat i \\\end{aligned} \end{equation} $$

Linear momentum will be conserved only in $x$ direction as no external force acts on the system in $x$ direction. In the vertical $y $ direction normal force due to the ground acts as an external force. Therefore,

$$\begin{equation} \begin{aligned} {\overrightarrow p _{{i_x}}} = {\overrightarrow p _{{f_x}}} \\ - m{v_0}\sin \theta \widehat i = \left( { - M{v_1} + m{v_2}} \right)\widehat i\quad ...(vi) \\\end{aligned} \end{equation} $$

From equation $(iii)$ & $(iv)$ we get, $$\begin{equation} \begin{aligned} - m{v_0}\sin \theta = - M{v_1} + m\left( {e{v_0}\sin \theta - {v_1}{{\sin }^2}\theta } \right) \\ {v_1} = \frac{{m{v_0}\sin \theta \left( {e + 1} \right)}}{{M + m{{\sin }^2}\theta }} \\\end{aligned} \end{equation} $$

Question 28. A wedge of mass $M$ rests on a horizontal surface. The inclination of the wedge is $\alpha $. A ball of mass m moving horizontally with speed $u$ hits the inclined face of the wedge inelastically. After hitting the ball slides up the inclined face of the wedge as seen by the observer sitting on the wedge. Find the velocity of the wedge just after collision.

Solution: Let us assume that the wedge move with velocity ${\overrightarrow v _W}$ after collision in the direction as shown in the figure.

According to the question, the ball slides up along the incline with velocity ${\overrightarrow v _{BW}}$ after the collision.

As we know, $$\begin{equation} \begin{aligned} {\overrightarrow v _{BW}} = {\overrightarrow v _B} - {\overrightarrow v _W} \\ {\overrightarrow v _B} = {\overrightarrow v _{BW}} + {\overrightarrow v _W} \\\end{aligned} \end{equation} $$ or $${\overrightarrow v _B} = {\overrightarrow v _{BW}} - {v_W}\widehat i\quad ...(i)$$

Before collision, $$\begin{equation} \begin{aligned} {u_{{B_{CN}}}} = - u\sin \alpha \quad ...(ii) \\ {u_{{B_{CT}}}} = - u\cos \alpha \quad ...(iii) \\\end{aligned} \end{equation} $$

After collision, $$\begin{equation} \begin{aligned} {v_{{B_{CN}}}} = - {v_W}\sin \alpha \quad ...(iv) \\ {v_{{B_{CT}}}} = - \left( {{v_{BW}} + {v_W}\cos \alpha } \right)\quad ...(v) \\\end{aligned} \end{equation} $$

$$\begin{equation} \begin{aligned} {u_{{B_{CT}}}} = {v_{{B_{CT}}}} \\ - u\cos \alpha = - \left( {{v_{BW}} + {v_W}\cos \alpha } \right) \\ {v_{BW}} = u\cos \alpha - {v_W}\cos \alpha \quad ...(vi) \\\end{aligned} \end{equation} $$

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

In the vertical $y$ direction normal force due to the ground acts as an external force. Therefore,

$$\begin{equation} \begin{aligned} {\overrightarrow p _{{i_x}}} = {\overrightarrow p _{{f_x}}} \\ - mu = - M{v_W} + m\left( { - {v_W} - {v_{BW}}\cos \alpha } \right)\quad ...(vii) \\\end{aligned} \end{equation} $$

From equation $(vi)$ & $(vii)$ we get,

$$\begin{equation} \begin{aligned} - mu = - M{v_W} + m\left[ { - {v_W} - \left( {u\cos \alpha - {v_W}\cos \alpha } \right)\cos \alpha } \right] \\ {v_W} = \frac{{mu{{\sin }^2}\alpha }}{{M + m{{\sin }^2}\theta }} \\\end{aligned} \end{equation} $$