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

After B strikes the plane, it comes to rest while A continues moving. Using the equations of motion for A, we have:

Let ( u ) be the initial velocity of A when B strikes the plane, which can be equated as follows:

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

Deriving travel distance using (a = \frac{g}{5}):

We have (v = 0) at the time of impact, so substituting:

( v = 2 \times \frac{g}{5} \times 1.5 = 0.6g ) leads us to total distance:

$s = 2 \times 1.5 \times \frac{g}{5} = 0.6 \text{ m}.$

When B strikes the plane, it experiences an impulse given by:

$J = m(v - u)$

Here, (u = \text{velocity before impact}) and (v = 0) after impact:

Given that ( m = 3m ), substituting yields:

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

To find the magnitude: substituting (u = 0.6g), we have:

$|J| = 3m(0.6g) = 3m(0.6 \times 9.81) = 3m \times 5.88\approx 3.64 ext{ N·s}.$ Thus, the answer is ( 3.64 ext{ N·s} ).