Two particles A and B have masses 2m and 3m respectively - Edexcel - A-Level Maths Mechanics - Question 5 - 2014 - Paper 1

In the time it takes for B to stop, we can find the distance A travels. The velocity of B just before hitting the plane can be found from:

Using the equations of motion:

$v^2 = u^2 + 2as \rightarrow (0)^2 = u^2 + 2(-\frac{g}{5})(1.5)$

So,

$u^2 = 2 \times \frac{g}{5} \times 1.5 = \frac{3g}{5}$

At the time of impact, A would have moved downward as well. Let the distance travelled by A before the string becomes taut be (d). Using a similar analysis:

After B hits the plane, A descends according to:

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

From the situation:

$d + 1.5 = 0.6 \implies d = 0.6 - 1.5 = 0.3 m$

Thus the distance travelled by A is 0.3 m.

The impulse experienced by B can be calculated as:

Impulse is given by:

$J = m(v - u)$

Where (m = 3m) and the final velocity (v = 0 ) (B comes to rest), and thus:

$J = 3m(0 - u) = -3mu$

From the earlier calculation:

The downward speed of B just before impact was (u = 0.6g ). Thus:

The magnitude of the impulse is:

$|J| = 3m(0.6g) = 3m(0.6 \times 9.8) = 3 imes 0.5 imes 0.6 imes 9.8 \approx 3.6\ (Ns)$

Therefore, the magnitude of the impulse on B due to the impact with the plane is (3.6\ (Ns)).