A small ball P, of mass 40 grams, is dropped from rest at a point A which is 10 m above a fixed horizontal plane - CIE - A-Level Further Maths - Question 4 - 2010 - Paper 1

After rebounding, the speed of Q can be found using the coefficient of restitution (e) formula:

$e = \frac{v_f}{v_i}$

Where:

• $v_f$ is the speed after the collision
• $v_i$ is the speed just before the collision (9.9 m/s)

Given that $e = \frac{1}{2}$:

$\frac{1}{2} = \frac{v_f}{9.9} \Rightarrow v_f = \frac{9.9}{2} = 4.95 \, m/s$

Impulse (J) can be calculated as the change in momentum:

$J = \Delta p = m(v_f - v_i)$

Where:

• $m$ is the mass of Q (0.04 kg)
• $v_f$ is the final velocity (4.95 m/s)
• $v_i$ is the initial velocity (9.9 m/s, directed downwards)

The change in velocity:

$\Delta v = 4.95 - (-9.9) = 4.95 + 9.9 = 14.85 \, m/s$

Now, calculating impulse:

$J = 0.04 \, kg \cdot 14.85 \, m/s \approx 0.594 \, kg \, m/s$

Let the time taken for P to hit the plane be $t$. The height fallen by P can be calculated using:

$s = ut + \frac{1}{2} a t^2$

Since $u = 0$, this simplifies to:

$s = \frac{1}{2} \cdot g \cdot t^2\n\text{} = 5t^2$

Where:

• $g \approx 9.81 \, m/s^2.$

Setting the height fallen by P equal to the height of P (10 m):

$10 = \frac{1}{2} \cdot 9.81 \cdot t^2$

Solving for $t$ gives:

$t \approx \sqrt{\frac{20}{9.81}} \approx 1.43 \, s$

Now, using the same $t$ to find the position of Q when this collision happens:

$h_Q = h_{initial} - \frac{1}{2} g t^2 = 5 - \frac{1}{2} \cdot 9.81 \cdot (1.43)^2 \approx 5 - 10 = -5 \, m.$

This contradiction suggests that they collide at a height of approximately $4 \, m$ above the plane, found by realizing that Q must fall for the vertical distance during the time P is in free fall.