A moving car hit a parked van and the two vehicles became locked together and skidded to a stop. From the skid marks it was estimated that just after the collision the common velocity of the two vehicles was 8 ms⁻¹. The masses of the van and car were 1 500 kg and 1 000 kg respectively.

Step 1: Identify the forces acting on each particle.
For particle A (mass $4m$) on the inclined plane:

• Weight component parallel to the plane (downwards): $W_{A, \text{parallel}} = 4mg \sin(30^\circ)$
• Tension (up the plane): $T$ For particle B (mass $m$) hanging freely:
• Weight (downwards): $W_B = mg$
• Tension (upwards): $T$

Step 2: Apply Newton's second law for each particle.
Since the string is inextensible and the pulley is smooth, both particles will have the same magnitude of acceleration, $a$. Given that particle A accelerates down the plane, particle B must accelerate upwards.

For particle A (down the plane is positive direction):

$$4mg \sin(30^\circ) - T = 4ma$$

Since $\sin(30^\circ) = \frac{1}{2}$:

$$4mg \left(\frac{1}{2}\right) - T = 4ma$$ $$2mg - T = 4ma \quad \text{(Equation 1)}$$

For particle B (upwards is positive direction):

$$T - mg = ma \quad \text{(Equation 2)}$$