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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

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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.png

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

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Answer

To find the acceleration, we need to apply Newton's second law. The total force acting on the system is given by:

Fnet=FengineFtowbarFresistanceF_{net} = F_{engine} - F_{towbar} - F_{resistance}

The total mass of both vehicles is:

mtotal=mlorry+mcar=1600extkg+900extkg=2500extkgm_{total} = m_{lorry} + m_{car} = 1600 ext{ kg} + 900 ext{ kg} = 2500 ext{ kg}

Substituting numerical values:

Fnet=1500extN600extN300extN=600extNF_{net} = 1500 ext{ N} - 600 ext{ N} - 300 ext{ N} = 600 ext{ N}

Using Newton's second law, F=maF = ma:

600=2500a600 = 2500a

Therefore, the acceleration aa is:

a = rac{600}{2500} = 0.24 ext{ m s}^{-2}

Step 2

b) the tension in the towbar

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Answer

We can determine the tension in the towbar by analyzing the forces acting on the car. The forces acting on the car are the tension in the towbar (TT), the resistance (300 N), and the subsequent equation is:

Textcos(15ext°)300=mcarimesaT ext{ cos}(15^{ ext{°}}) - 300 = m_{car} imes a

Substituting known values gives:

Textcos(15ext°)300=900imes0.24T ext{ cos}(15^{ ext{°}}) - 300 = 900 imes 0.24

Solving for TT:

Textcos(15ext°)=300+216T ext{ cos}(15^{ ext{°}}) = 300 + 216

Textcos(15ext°)=516T ext{ cos}(15^{ ext{°}}) = 516

Solving for TT using extcos(15ext°) ext{cos}(15^{ ext{°}}) results in:

T = rac{516}{ ext{cos}(15^{ ext{°}})} o T ext{ approximately equals } 534 ext{ N}

Step 3

c) find the distance moved by the car

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Answer

The car decelerates at a constant rate after the towbar breaks. The initial velocity u=6extms1u = 6 ext{ m s}^{-1} and the final velocity v=0v = 0. The only force acting on the car is the resistance, which is 300 N:

Using the equation of motion:

v2=u2+2asv^2 = u^2 + 2as

Where:

  • a = - rac{300}{900} = - rac{1}{3} ext{ m s}^{-2}

Substituting the values:

36 = - rac{2}{3} s$$ Thus: $$s = rac{36 imes 3}{2} = 54 ext{ 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

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Answer

When the towbar breaks, the vertical component of tension (TT) in the towbar is removed. This tension counteracts a portion of the weight of the car.

Without this upward force due to tension, the normal reaction force from the road on the car must increase to balance out the weight of the car. Therefore, it can be concluded that the normal reaction of the road on the car is increased.

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