Figure 4 shows two particles P and Q, of mass 3 kg and 2 kg respectively, connected by a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 7 - 2007 - Paper 1

Substituting the equations from part (a):

From the equation for Q, we have:

$T = 2g - 2a$

Substituting into the equation for P:

$2g - (2g - 2a) = 2a + 3g ext{sin}(30°)$

We know that:

$g = 9.8 ext{ m s}^{-2}$

Thus:

$3g ext{sin}(30°) = 3 \times 9.8 \times 0.5 = 14.7$

So, after simplification, we find:

$a = 0.98 ext{ m s}^{-2}$

Using the equation for Q:

$T = 2g - 2a$

Substituting the known values:

$T = 2 \times 9.8 - 2 \times 0.98$

This results in:

$T = 19.6 - 1.96 = 17.64 ext{ N}$

The inextensibility of the string is crucial in maintaining equal acceleration for both particles P and Q. This assumption is necessary to derive the equations of motion and ensure that the tensions act appropriately without any lag.

Using the formula:

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

Where:

• Initial velocity, $u = 0$
• Acceleration, $a = 0.98 ext{ m s}^{-2}$
• Distance, $s = 0.8 ext{ m}$

Thus:

$v^2 = 0 + 2 \times 0.98 \times 0.8$

So:

$v = \sqrt{1.568} \approx 1.25 ext{ m s}^{-1}$

Using the equation of motion:

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

Where:

• Initial velocity, $u = 0$
• Distance $s = 0.8 ext{ m}$
• Acceleration $a = 0.98 ext{ m s}^{-2}$

We solve:

$0.8 = 0 + \frac{1}{2} \times 0.98 imes t^2$

Thus:

$0.8 = 0.49t^2$

Solving for $t$ gives:

$t^2 = \frac{0.8}{0.49} \implies t = \sqrt{1.63265} \approx 0.51 ext{ s}$