a) A smooth sphere A of mass 4m, moving with speed u on a smooth horizontal table collides directly with a smooth sphere B of mass m, moving in the opposite direction with speed u - Leaving Cert Applied Maths - Question 5 - 2021

From the impulse-momentum theorem:

$T = mv_2 - m(-u)$

Substituting for $v_2$:

$T = m\left(\frac{(1 + 3e)u}{5}\right)$

This leads us to:

$\frac{8mu}{5} \leq T \leq \frac{16mu}{5}$

confirming the calculation within the limits established.

Using conservation of momentum for the x-direction:

$2m\frac{w}{\sqrt{3}} - ku = 2mv_1$.

Also for the y-direction:

$0 + 0 = 0 + mv_2$.

Solving these gives the expression for $k$ as:

$k = \frac{\sqrt{3(1-e)}}{2(1+e)}$.

Using the derived expressions from momentum conservation:

$3v_2 = \frac{u\sqrt{3}}{6} + ev_1$.

Thus, the speed of Q after the collision becomes:

$v_2 = \frac{3u(1+e)}{6(1+e)}$.

This details the final velocity of Q post-collision.