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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
Question 7
Figure 4 shows a lorry of mass 1600 kg towing a car of mass 900 kg along a straight horizontal road. The two vehicles are joined by a light towbar which is at an ang... show full transcript
Worked Solution & Example Answer: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
Step 1
(a) the acceleration of the lorry and the car
Answer
To find the acceleration of the lorry and the car, we start by applying Newton's second law:
The total force acting on the system (lorry + car) is given by:
Total Force = Force from engine - Resistance from lorry - Resistance from car
Substituting in the values:
Total Force = 1500 N - 600 N - 300 N = 600 N
Next, we find the total mass:
Total mass = mass of lorry + mass of car = 1600 kg + 900 kg = 2500 kg
Using Newton's second law, we have:
a = \frac{F}{m} = \frac{600}{2500} = 0.24 , m/s^2
Step 2
(b) the tension in the towbar.
Answer
To find the tension in the towbar, we can examine the forces acting on the car. The equations can be set for the forces in the direction of the motion:
Consider the forces acting on the car:
( T \cos(15°) - 300 = \frac{900}{a} )
Substituting the known acceleration:
( T \cos(15°) - 300 = \frac{900}{0.24} )
Solving for T:
( T = 534 \text{ N} )
Step 3
(c) find the distance moved by the car from the moment the towbar breaks to the moment when the car comes to rest.
Answer
When the towbar breaks, the forces acting on the car change. The deceleration can be calculated:
Let the deceleration of the car be a, then:
( F = ma ) Where F is the resisting force:
Taking the resistance as 300 N:
( 300 = \frac{900}{s} \Rightarrow s = \frac{300}{900} = \frac{1}{3} ) m/s²
Using the kinematic equation:
( v^2 = u^2 + 2as )
With final velocity (v = 0), initial velocity (u = 6 m/s), we can find s:
( 0 = 6^2 + 2(-\frac{1}{3})s \Rightarrow s = 54 , m)
Step 4
(d) State whether, when the towbar breaks, the normal reaction of the road on the car is increased, decreased or remains constant. Give a reason for your answer.
Answer
When the towbar breaks, the vertical component of the tension T is removed. This tension was contributing to the upward normal force acting on the car. Hence, the normal reaction of the road on the car will increase since there is no longer a counteracting vertical force from the towbar.