Newton's Second Law (Cambridge (CIE) IGCSE Physics): Revision Note

A car salesperson says that their best car has a mass of 900 kg and can accelerate from 0 to 27 m/s in 3 seconds.

Calculate:

a) the acceleration of the car in the first 3 seconds. [2]

b) the force required to produce this acceleration. [2]

Answer:

Part (a)

Step 1: List the known quantities

• Initial velocity, $0\mathrm{m}/\mathrm{s}$ 

• Final velocity, $27\mathrm{m}/\mathrm{s}$ 

• Time, $∆t=3\mathrm{s}$

Step 2: State the equation for acceleration, in terms of change in velocity

$a=\frac{∆v}{∆t}$

$a=\frac{27-0}{3}$ [1 mark]

$a=9\mathrm{m}/{\mathrm{s}}^2$ [1 mark]

Part (b)

Step 1: List the known quantities

• Mass of the car, $m=900\mathrm{kg}$ 

• Acceleration, $a=9\mathrm{m}/{\mathrm{s}}^2$ 

Step 2: Identify which law of motion to apply

• The question involves quantities of force, mass and acceleration, so Newton's second law is required:

$F=ma$

Step 3: Calculate the force required to accelerate the car

$F=900\times 9$ [1 mark]

$F=8100\mathrm{N}$ [1 mark]

Three shopping trolleys, A, B and C, are being pushed using the same force. This force causes each trolley to accelerate.

State which trolley would have the smallest acceleration. Explain your answer.

[3]

Answer:

Step 1: Identify which law of motion to apply

• The question involves quantities of force and acceleration, and the image shows trolleys of different masses, so Newton's second law is required:

$F=ma$

Step 2: Re-arrange the equation to make acceleration the subject

$a=\frac{F}{m}$

Step 3: Explain the inverse proportionality between acceleration and mass  

• Acceleration is inversely proportional to mass [1 mark]

• This means that for the same amount of force, a large mass will experience a small acceleration [1 mark]

• Therefore, trolley C will have the smallest acceleration because it has the largest mass [1 mark]

Remember that in Newton's second law, F is the resultant force. An object can have several forces acting on it, but the object can only have one acceleration. So you must substitute the resultant force into $F=ma$ to calculate the correct acceleration.