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

For particle P, apply Newton's second law:

$0.5g - T = 0.5a$

Substituting known values:

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

Calculating gives:

$4.9 - T = 1.4$

Thus,

$T = 4.9 - 1.4 = 3.5 \text{ N}$

For particle Q:

Applying Newton's second law:

$T - mg = ma$

Given that ( a = 2.8 \text{ m s}^{-2} ) and substituting for T:

$3.5 - mg = m \times 2.8$

This implies:

$mg + 2.8m = 3.5$

Factoring out m gives:

$m(g + 2.8) = 3.5$

Substituting ( g = 9.8 ):

$m(9.8 + 2.8) = 3.5 \implies m \times 12.6 = 3.5 \implies m = \frac{3.5}{12.6} = \frac{5}{18}$

The information that the string is inextensible allows us to assert that both particles P and Q have the same magnitude of acceleration at all times. This means that when P descends a certain distance, Q must correspondingly ascend or descend the same distance without any slack in the string.

After P strikes the ground, it does not rebound. For Q, it moves freely under gravity. Using the second equation of motion:

$s = ut + \frac{1}{2} g t^2$

Where:

• ( s = 3.15 ) m
• ( u = 0 ) (initial velocity of Q)
• ( g = 9.8 ) m/s²

Calculating:

$3.15 = 0 + \frac{1}{2} \times 9.8 \times t^2$

This leads to:

$t^2 = \frac{3.15 \times 2}{9.8} \implies t^2 \approx 0.64 \implies t \approx 0.8 \text{ s}$

Thus, the total time from when P strikes the ground until the string becomes taut again is approximately ( t \approx 0.86 \text{ s} ).