5. (a) Two small smooth spheres A, of mass 2 kg, and B, of mass 3 kg, are suspended by light strings from a ceiling as shown in the diagram - Leaving Cert Applied Maths - Question 5 - 2016

From these equations, solve for $v_1$ and $v_2$. On solving

1. $v_1 = -0.22 \, m/s$
2. $v_2 = 3.10 \, m/s$

Using the vertical motion components of sphere B post-collision to find θ, we can utilize:

$\cos(θ) = \frac{\text{height}}{\text{length}} = \frac{2(2 - \text{cos} θ)}{3}$

After simplifying the final equational form to find θ: $\cos θ = 0.7548 \Rightarrow θ \approx 40.99°$

Using conservation of momentum: $m \cdot 3\text{i} + m \cdot 2\text{j} = m(\text{velocity of P}) + m(\text{velocity of Q}).$ Given:

• For P: before impact = $3\text{i} + u_j$
• For Q: before impact = $2\text{i} + v_j$

Using some initial calculations, we derive,

• Velocity of P before impact: $v_{0} = 2\text{i} + 3\text{j}$
• Velocity of Q before impact: $u_{1} = 3\text{i} + 2 \text{j}$.

Using the direction vectors: $\tan(\alpha) = \frac{\text{rise}}{\text{run}} = \frac{u_j}{3}$

Upon solving we find: $\alpha \approx 33.69°$

Adding the angle through which the motion of P is deflected: $\beta \approx 18.43° \Rightarrow \beta + \alpha = 52°$.