Two particles P and Q have mass 0.5 kg and m kg respectively, where m < 0.5 - Edexcel - A-Level Maths Mechanics - Question 6 - 2007 - Paper 1

Using Newton's second law for particle P:

$0.5g - T = 0.5a$

Here, ( g \approx 9.8 \text{ m/s}^2 ) and ( a = 2.8 \text{ m/s}^2 ).

Substituting these values in:

$0.5 \cdot 9.8 - T = 0.5 \cdot 2.8$

From which we derive:

$T = 0.5 \cdot 9.8 - 0.5 \cdot 2.8 = 3.5 ext{ N}$

Applying Newton's second law for particle Q:

$T - mg = m(-a)$

Substituting ( T = 3.5 \text{ N} ) and ( g \approx 9.8 ext{ m/s}^2 ):

$3.5 - m \cdot 9.8 = -m \cdot 2.8$

Rearranging gives:

$3.5 = m (9.8 - 2.8) \Rightarrow m = \frac{3.5}{7} = \frac{5}{18}$

The inextensibility of the string indicates that both particles P and Q have the same magnitude of acceleration (descent of P and ascent of Q).

When P strikes the ground, it does not rebound. Particle Q, now free, moves under gravity.

Using kinematic equations, we have:

Let ( h ) be the distance that Q falls before the string becomes taut again. After P strikes the ground, the distance to the pulley is ( 3.15 ) m. We calculate:

$v^2 = u^2 + 2as$

where ( u = 0 ), ( a = g ), and ( s = 3.15 ):

$v^2 = 0 + 2 \cdot 9.8 \cdot 3.15 = 61.68$

Thus, ( v = \sqrt{61.68} \approx 7.85 ext{ m/s}$$

Next, we find time ( t ) using:

$v = u + at \Rightarrow t = \frac{v}{g} = \frac{7.85}{9.8} \approx 0.8 ext{ s}$

This is the time from when P strikes the ground to when Q's string becomes taut.