Physics Principles & Diagram

This problem involves projectile motion under gravity and the physics of collisions defined by the coefficient of restitution.

• Regular Manner: The trajectory repeats for every step. The displacement between two successive impacts is $\Delta x = h$ and $\Delta y = -h$.
• Coefficient of Restitution ($e$): Relates the vertical component of velocity just after impact ($u_y$) to the velocity just before impact ($v_y$) via $u_y = e v_y$.

Question 27: Height $H$ attained

Let $u_y$ be the vertical velocity just after bouncing, and $v_y$ be the vertical velocity just before hitting the next step.

From kinematics, the maximum height $H$ reached above the bounce point is related to the initial vertical velocity $u_y$ by:

$$ u_y = \sqrt{2gH} \quad \text{…(1)} $$

The ball falls from the peak height $H$ down to the next step, which is a vertical distance of $H + h$. The velocity just before impact ($v_y$) is attained by falling through this height:

$$ v_y = \sqrt{2g(H+h)} \quad \text{…(2)} $$

Using the coefficient of restitution equation $u_y = e v_y$:

$$ \sqrt{2gH} = e \sqrt{2g(H+h)} $$

Squaring both sides and solving for $H$:

$$ 2gH = e^2 [2g(H+h)] $$ $$ H = e^2 H + e^2 h $$ $$ H(1 – e^2) = e^2 h $$ $$ H = \frac{e^2 h}{1 – e^2} $$

Question 25: Air time between two successive bounces

The total time of flight $T$ is the time taken to rise to the peak height $H$ plus the time taken to fall from the peak height to the next step (height $H+h$).

$$ T = t_{\text{rise}} + t_{\text{fall}} $$ $$ T = \sqrt{\frac{2H}{g}} + \sqrt{\frac{2(H+h)}{g}} $$

We substitute $H = \frac{e^2 h}{1 – e^2}$. First, let’s simplify the term $(H+h)$:

$$ H+h = \frac{e^2 h}{1 – e^2} + h = \frac{e^2 h + h(1 – e^2)}{1 – e^2} = \frac{h}{1 – e^2} $$

Now substitute $H$ and $(H+h)$ into the time equation:

$$ T = \sqrt{\frac{2}{g} \left( \frac{e^2 h}{1 – e^2} \right)} + \sqrt{\frac{2}{g} \left( \frac{h}{1 – e^2} \right)} $$ $$ T = e \sqrt{\frac{2h}{g(1 – e^2)}} + \sqrt{\frac{2h}{g(1 – e^2)}} $$ $$ T = (1+e) \sqrt{\frac{2h}{g(1 – e^2)}} $$

Using the identity $1 – e^2 = (1-e)(1+e)$:

$$ T = (1+e) \sqrt{\frac{2h}{g(1-e)(1+e)}} = \sqrt{\frac{2h (1+e)^2}{g(1-e)(1+e)}} $$ $$ T = \sqrt{\frac{2h(1+e)}{g(1-e)}} $$

Question 26: Horizontal component of velocity

The horizontal velocity $u_x$ is constant throughout the motion. In the time interval $T$ (time between bounces), the marble covers a horizontal distance equal to the width of the step, $h$.

$$ u_x \cdot T = h $$ $$ u_x = \frac{h}{T} $$

Substitute the expression for $T$ derived in Q25:

$$ u_x = \frac{h}{\sqrt{\frac{2h(1+e)}{g(1-e)}}} $$ $$ u_x = \sqrt{\frac{h^2 \cdot g(1-e)}{2h(1+e)}} $$ $$ u_x = \sqrt{\frac{gh(1-e)}{2(1+e)}} $$