Problem Analysis

We are given a ball projected horizontally from a large inclined plane with angle $\theta$. The ball undergoes successive collisions with the plane. We need to find the conditions under which the ball follows identical trajectories between every two successive bounces.

Let the velocity of projection be $\vec{u}$. For the trajectories to be identical, the velocity vector relative to the incline immediately after the first impact must be identical (in magnitude and direction) to the initial velocity vector $\vec{u}$ relative to the incline.

Step 1: Analyzing Initial Velocity Components

Let us define a coordinate system with the $x$-axis pointing down the incline and the $y$-axis perpendicular to the incline (upwards). The initial velocity $\vec{u}$ is horizontal.

The components of the initial velocity $\vec{u}$ along these axes are:

• Parallel to incline ($x$-direction): $u_x = u \cos \theta$
• Perpendicular to incline ($y$-direction): $u_y = u \sin \theta$

The acceleration due to gravity has components: $g_x = g \sin \theta$ and $g_y = -g \cos \theta$.

Step 2: Kinematics of Flight

First, we calculate the time of flight $T$ between the launch and the first impact. The displacement in the $y$-direction becomes zero at the point of impact.

$$ y = u_y t + \frac{1}{2} a_y t^2 $$ $$ 0 = (u \sin \theta) T – \frac{1}{2} (g \cos \theta) T^2 $$

Solving for $T$ ($T \neq 0$): $$ T = \frac{2 u \sin \theta}{g \cos \theta} $$

Now, let’s find the velocity components just before impact ($v^-$):

• Perpendicular component ($v_y^-$): $$ v_y^- = u_y + a_y T = u \sin \theta – (g \cos \theta) \left( \frac{2 u \sin \theta}{g \cos \theta} \right) = -u \sin \theta $$
• Parallel component ($v_x^-$): $$ v_x^- = u_x + a_x T = u \cos \theta + (g \sin \theta) \left( \frac{2 u \sin \theta}{g \cos \theta} \right) $$ $$ v_x^- = u \cos \theta + \frac{2 u \sin^2 \theta}{\cos \theta} $$

Step 3: Calculating Coefficient of Restitution ($e$)

For the trajectory to repeat, the velocity component perpendicular to the plane just after impact ($v_y^+$) must be equal to the initial perpendicular velocity ($u \sin \theta$).

Using the definition of the coefficient of restitution ($e$): $$ e = \frac{\text{velocity of separation}}{\text{velocity of approach}} = \frac{v_y^+}{|v_y^-|} $$

We require $v_y^+ = u \sin \theta$. Since we found $|v_y^-| = u \sin \theta$, we get: $$ e = \frac{u \sin \theta}{u \sin \theta} = 1 $$

Step 4: Calculating Inclination Angle ($\theta$)

Similarly, for the trajectory to repeat, the parallel velocity component just after impact ($v_x^+$) must be equal to the initial parallel velocity ($u \cos \theta$). However, during flight, gravity increased the parallel velocity. Friction must act during the impact to reduce this velocity back to its initial value.

We use the Impulse-Momentum theorem along the $x$-direction. The impulsive force is friction $f$. Since there is slipping, the friction impulse is related to the normal impulse by the coefficient of friction $\mu$.

Normal Impulse ($J_N$): $$ J_N = \Delta p_y = m(v_y^+ – v_y^-) = m(u \sin \theta – (-u \sin \theta)) = 2mu \sin \theta $$

Friction Impulse ($J_f$): $$ J_f = \mu J_N = \mu (2mu \sin \theta) $$ This impulse acts up the incline (opposing motion).

Momentum change along $x$: $$ \Delta p_x = m(v_x^+ – v_x^-) = -J_f $$ We require $v_x^+ = u \cos \theta$. Substituting the value of $v_x^-$ from Step 2: $$ m \left[ u \cos \theta – \left( u \cos \theta + \frac{2 u \sin^2 \theta}{\cos \theta} \right) \right] = -2 \mu m u \sin \theta $$ Simplifying: $$ – m \left( \frac{2 u \sin^2 \theta}{\cos \theta} \right) = -2 \mu m u \sin \theta $$ $$ \frac{\sin \theta}{\cos \theta} = \mu $$ $$ \tan \theta = \mu $$

Final Answer

For the ball to follow identical trajectories between every two successive bounces:

• The coefficient of restitution is $e = 1$.
• The inclination of the plane is given by $\tan \theta = \mu$ (or $\theta = \tan^{-1}\mu$).