Two particles A and B have mass 0.4 kg and 0.3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2010 - Paper 1

Thus, the acceleration of A immediately after the release of the particles is $1.4 \, \text{m/s}^2$.

We first find the velocity of B after 0.5 s of motion using:

$v = u + at$

where initial velocity $u = 0$, acceleration $a = 1.4 \, \text{m/s}^2$. Hence,

$v = 0 + 1.4 \cdot 0.5 = 0.7 \, \text{m/s}$

Next, we use the formula to determine the distance covered by B:

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

Substituting values:

$s = 0 + \frac{1}{2} \cdot 1.4 \cdot (0.5)^2 = 0.175 \, \text{m}$

B initially travels 1 m vertically, thus the remaining distance for B to travel until it reaches the floor:

Remaining distance = $1 - 0.175 = 0.825 \, \text{m}$

Using:

$s = ut + \frac{1}{2} at^2$ and solving for t will yield:

$0.825 = 0.7t + \frac{1}{2} a t^2$

Once you set up the equations correctly, solving will give:

$t \approx 0.57 \, \text{s}$

Thus, the total time for B to hit the floor from rest is approximately $0.57 \, \text{s}$.