Smooth spheres P and Q are travelling towards each other on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2021

Calculating the initial kinetic energy:

$KE_{initial} = \frac{1}{2} \cdot 7 \cdot (1)^2 + \frac{1}{2} \cdot 3 \cdot (5)^2 = 41 J$

Calculating the final kinetic energy:

$KE_{final} = \frac{1}{2} \cdot 7 \cdot (-2)^2 + \frac{1}{2} \cdot 3 \cdot (2)^2 = 20 J$

Thus, the loss is:

$\Delta KE = KE_{initial} - KE_{final} = 41 - 20 = 21 J$

Using the impulse formula:

$I = m(v_1 - u_1)$

For sphere P:

$I = 7(-2 - 1) = 7(-3) = -21$

Thus,

Magnitude of impulse = 21 N·s.

Using the equation of motion:

$v^2 = u^2 + 2as$

Here, $u = 0$, $a = 10 \, m/s^2$, $s = 11.25$. So we have:

$v^2 = 0 + 2 \cdot 10 \cdot 11.25$

Simplifying gives:

$v^2 = 225$

Thus: $v = \sqrt{225} = 15 \, m/s$

Using the coefficient of restitution:

$e = \frac{v_{rebound}}{v_{impact}}$

Where: $\frac{2}{3} = \frac{v_{rebound}}{15}$

Thus, solving gives: $v_{rebound} = 10 \, m/s$

Now, applying the second equation of motion again:

$v^2 = u^2 + 2as$

Where $u = 10 \, m/s$, $v = 0$, and $a = -10 \, m/s^2$. Then:

$0 = (10)^2 + 2(-10)(h)$

So: