Two particles P and Q, of mass 0.3 kg and 0.5 kg respectively, are joined by a light horizontal rod - Edexcel - A-Level Maths Mechanics - Question 7 - 2012 - Paper 1

Using the formula for calculating speed under constant acceleration:

$v = u + at$

Where:

• u = initial speed = 0 (since the system starts from rest)
• a = acceleration = 1.25 m/s²
• t = time = 6 seconds

Now substituting the values:

$v = 0 + (1.25 \, m/s^{2})(6 \, s) = 7.5 \, m/s$

Considering particle P and applying Newton's second law:

For P:

$T - R_{P} = m_{P} a$

Substituting in known values:

$T - 1 N = (0.3 kg)(1.25 m/s^{2})$ $T - 1 = 0.375$

Thus:

$T = 1 + 0.375 = 1.375 N$

When F is removed, the system experiences a deceleration due to the resisting force. Using:

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

Where:

• u = final speed = 7.5 m/s
• a = deceleration = -\frac{R_{P}}{m_{P}} = -\frac{1 N}{0.3 kg} = -3.33 m/s² (taking initial mass only)
• t = the time until it comes to rest

To find t when speed becomes 0:

Using: $v = u + at$

Setting v = 0: $0 = 7.5 - 3.33t$ $t = \frac{7.5}{3.33} = 2.25 s$

Now substituting into the distance formula:

$s = (7.5)(2.25) + 0.5(-3.33)(2.25)^2$

Calculating each term I get: $s = 16.875 - 8.43 = 8.445 m$

For particle P during deceleration:

$T - R_{P} = m_{P} a$

Where R_{P} = 1 N. The effective mass during deceleration is calculated using:

$a = -\frac{R_{P}}{m_{P}}$ $a = -\frac{1 N}{0.3 kg} = -3.33 m/s^{2}$

Plugging into the equation: $T - 1 = 0.3(-3.33)$ $T - 1 = -1$ $T = 0.125 N$