5. (a) Three identical small smooth spheres A, B and C, each of mass m, lie in a straight line on a smooth horizontal surface with B between A and C - Leaving Cert Applied Maths - Question 5 - 2018

Using the speeds from the previous step, we can analyze the condition for e:

From equation (2) we derived:

$e = \frac{v_B - v_A}{3u}$

Plug in our expressions for $v_A$ and $v_B$, we get:

$e = \frac{\frac{(5 + e)u}{2} - \frac{(5 - e)u}{2}}{3u}$

This simplifies to:

$e = \frac{eu}{3u}$

Thus, simplifying gives:

$\frac{(5 + e) - (5 - e)}{3} = e$

You can rearrange this inequality to show that:

$e > \frac{5}{7}$

Substituting $e = \frac{3}{7}$ into our expressions for the speeds after the collision:

1. Calculate $v_A$ and $v_B$:

   • $v_A = \frac{(5 - \frac{3}{7})u}{2} = \frac{32}{14}u$
   • $v_B = \frac{(5 + \frac{3}{7})u}{2} = \frac{38}{14}u$

2. Now we find the distance each sphere will travel after the collision. For B to collide with A again, $v_B$ must be greater than $v_A$, that is not the case here since:

$\frac{38}{14}u > \frac{32}{14}u$

Since $v_B$ keeps increasing while A’s is slower, B will not collide with A again.

Using conservation of momentum and the coefficient of restitution:

1. Conservation of Momentum:

$2m(4u) + 3m(u) = 2mv_P + 3mv_Q$

Which gives:

$8u + 3u = 2v_P + 3v_Q$ (1)

2. Coefficient of Restitution:

Using the angle α for both spheres:

$e = \frac{v_Q - v_P}{4u - u}$

Which gives:

$e = \frac{v_Q - v_P}{3u}$ (2)

3. Substitute the known expressions and then solve for speeds $v_P$ and $v_Q$ after the collision, leading to:

• $v_P = \frac{2}{3}u cos(α)$
• $v_Q = \frac{4}{3}u sin(α)$

Given that after the collision the speed of P is twice that of Q:

$v_P = 2v_Q$

We can set the expressions from the previous steps:

$\frac{2}{3}u cos(α) = 2\left(\frac{4}{3}u sin(α)\right)$

This simplifies and reduces to yield

Using trigonometric identities we can find α such that:

$tan(α) = \frac{16}{25}$

Thus calculate:

$α = 38.66°$