Forces and Newton’s laws – Year 12 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   The forces are at right angles, so the magnitude comes from Pythagoras and the angle from \(\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}\).

   \(\sqrt{81+144}=15\) N (2). Angle \(=\tan^{-1}\left(\tfrac9{12}\right)=36.9^\circ\) (2).

   The resultant is 15 N at \(36.9^\circ\) to the 12 N force (opposite side 9 N, adjacent side 12 N).

   The resultant lies between the two forces, closer to the larger 12 N force.

   A labelled sketch of the right-angled triangle of forces earns credit even if the arithmetic slips.

2. Answer to Question 2 (4 marks)

   1. The resultant force is the pull minus the resistance; apply \(F=ma\). \(30-12=5a\), \(a=3.6\) m/s\(^2\) (2) The block accelerates at 3.6 m/s\(^2\). The resultant force on the block is 18 N.
   2. Use \(s=ut+\tfrac12at^2\) with \(u=0\) and \(t=4\). \(s=\tfrac12\times3.6\times16=28.8\) m (2) The block travels 28.8 m in 4 s.

   Draw a force diagram first: 30 N forward, 12 N backward, and the weight \(5g\) balanced by the normal reaction. Using 30 N alone in \(F=ma\) gives the wrong answer 6 m/s\(^2\).

3. Answer to Question 3 (4 marks)

   The 4 kg particle moves down and the 2 kg particle moves up, with the same acceleration; write \(F=ma\) for each.

   \(4g-T=4a\) and \(T-2g=2a\) (1). Add: \(2g=6a\), \(a=3.27\) m/s\(^2\) (1). \(T=2(g+a)=26.1\) N (2).

   Substituting into \(T-2g=2a\) gives \(T=2(9.8+3.27)=26.1\) N. So \(a=3.27\) m/s\(^2\) and \(T=26.1\) N.

   As expected, the tension lies between the two weights, \(2g=19.6\) N and \(4g=39.2\) N.

   Here \(a=\tfrac g3\) exactly, and \(9.8\div3=3.27\) to 3 s.f.

4. Answer to Question 4 (4 marks)

   On a smooth slope, resolve perpendicular to the slope (no acceleration) and along the slope (Newton's second law).

   \(R=3g\cos40^\circ=22.5\) N (2). \(a=g\sin40^\circ=6.30\) m/s\(^2\) (2).

   So the normal reaction is 22.5 N and the block accelerates down the slope at 6.30 m/s\(^2\), whatever its mass.

   The normal reaction is less than the weight \(3g=29.4\) N because it only balances the component of the weight perpendicular to the slope.

5. Answer to Question 5 (4 marks)

   1. Treat lorry and trailer as one body; the tension is internal, and the total resistance is 2 000 N. \(8000-2000=4000a\), \(a=1.5\) m/s\(^2\) (2) The acceleration is 1.5 m/s\(^2\).
   2. Consider the trailer alone: the tension pulls it forward and its resistance acts backwards. Trailer: \(T-500=1000\times1.5\), \(T=2000\) N (2) The tension in the tow-bar is 2 000 N.

   Check with the lorry alone, where the tension acts backwards: \(8000-1500-2000=4500=3000\times1.5\) ✓.

Total : 20 marks