Two particles P and Q, of mass 4 kg and 7 kg respectively, are lying 0.5 m apart on a smooth horizontal table - Leaving Cert Applied Maths - Question 4 - 2015

Using the equation of motion: $v^2 = u^2 + 2as$

Where:

• $u = 0$ (initial speed)
• $a = \frac{3g}{10}$ (initial acceleration from part (i))
• $s = 3$ m

Substituting these values: $v^2 = 0 + 2 \left(\frac{3g}{10}\right)(3)$

Calculating: $v^2 = \frac{18g}{10} = 1.8g$

Taking the square root: $v = \sqrt{1.8g}$ $v ≈ 4.24 ext{ m/s}$

Using the relationship of accelerations in a connected system:

• Let $a_1$ be the acceleration of P: $a_1 = \frac{(4 + 7 + 3)a_2}{14}$

Since the total mass being accelerated is 14 kg, we set up: $14a = (4+7+3)a_2$

From part (i), $a_2 = \frac{4g}{10}$. Thus: $v_1 = \sqrt{2s a_1}$ Replacing $s$ with 3 m and simplifying gives the speed: $v_1 ≈ 3 ext{ m/s}$

For the wedge system, we set:

1. For mass on the incline (5 kg at 30°): $T - 5g \sin 30° = 5a \rightarrow T = 5g \sin 30° + 5a$

2. For the hanging mass (7 kg at 45°): $7g \sin 45° - T = 7a$

Substituting the value for $T$ into the second equation: $7g \sin 45° - 5g \sin 30° - 5a = 7a$

Solving yields: $a = 2 ext{ m/s}²$

To find the vertical force on the ground, we calculate: $F = \sum{all vertical forces}$ For the wedge system:

• Weight of the wedge: 11g
• Forces for both masses multiplied by the cosines of their angles: $F = 5g \cos 30° + 7g \cos 45° + 11g$ This evaluates to: $F \approx 220.5 ext{ N}$.