Figure 4 shows a lorry of mass 1600 kg towing a car of mass 900 kg along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 7 - 2005 - Paper 1

To find the tension, we can analyze the forces acting on the car. The forces include the tension in the towbar, the gravitational force, and the resistance to motion:

For the car moving on a slope, we can resolve the components:

Let T be the tension in the towbar. The component of the tension acting against the resistance (along the road) can be expressed as:

$T imes ext{cos}(15°) - 300 ext{ N} = m imes a$

Substitute values into the equation for the car:

$T imes 0.9659 - 300 = 900 imes 0.24$

Solving this equation gives:

$T imes 0.9659 = 300 + 216$

Thus,

$T = \frac{516}{0.9659} = 534 ext{ N}$

When the towbar breaks, we need to analyze the deceleration of the car. The only force acting on the car is the resistance of 300 N:

Using Newton’s second law:

$F = m imes a$

Thus, the deceleration 'a' of the car can be calculated as follows:

$300 = 900 imes a$

Hence,

$a = \frac{300}{900} = \frac{1}{3} ext{ m s}^{-2}$

Using the equation of motion to find the distance covered before coming to rest:

Using:

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

Where:

• v = final velocity = 0 (car comes to rest)
• u = initial velocity = 6 m/s
• a = -1/3 m/s² (negative because it is deceleration)

Substituting the values:

$0 = 6^2 + 2 \left(-\frac{1}{3}\right) s$

This simplifies to:

$0 = 36 - \frac{2}{3} s$

Reorganizing gives:

$s = \frac{36 \times 3}{2} = 54 ext{ m}$

When the towbar breaks, the vertical component of the tension T is removed. Since T was acting at an angle, it contributed a vertical force component on the car. Therefore, with the breaking of the towbar:

• The normal reaction from the ground on the car will increase as the upward component of tension is gone. Hence, the normal force will adjust to maintain equilibrium.