Two particles P and Q have masses 0.3 kg and m kg respectively - Edexcel - A-Level Maths Mechanics - Question 6 - 2011 - Paper 1

Using the equations of motion, we can analyze the forces acting on mass Q.

The weight of mass Q (mg) is given by:

$mg - T = ma$

where:

• $T$ is the tension in the string
• $a$ is the acceleration (1.4 m/s²)

For mass P on the inclined plane, the forces can be expressed as:

$T - 0.3g \sin(\alpha) = 0.3a$

Eliminating T using these two equations, we have:

$mg = 0.3 \cdot 9.8 \cdot \sin(\alpha) + 0.3 imes 1.4$

Solving for m yields:

$m = 0.4 ext{ kg}$.

Using the kinematic formula for particle P when it moves down the incline,

$v = u + at$

Initially, $u = 0$ and the acceleration $a$ can be derived from:

$-0.3g \sin(\alpha) = F$

Substituting the given values, we have:

$v = 1.4 \times 0.5 = 0.7 ext{ m/s}$

When the string breaks, we calculate the deceleration:

0 = v + a(t)\; 0 = 0.7 - 9.8t\; t = \frac{0.7}{9.8} \approx 0.0714 ext{ s}$$ Thus, the further time until P comes to instantaneous rest is approximately 0.0714 s.