Forces in two dimensions and friction – Year 13 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Method: resolve both forces east and north, add the components, then use Pythagoras and the inverse tangent. Mark scheme: 1 mark for each component, 1 for the magnitude, 1 for the direction.

   East component \(=15+10\cos50^\circ=21.43\) N (1); north component \(=10\sin50^\circ=7.66\) N (1). Magnitude \(=\sqrt{21.43^2+7.66^2}=22.8\) N (1). Direction \(\tan^{-1}\dfrac{7.66}{21.43}=19.7^\circ\) north of east (1).

   Answer: the resultant is about 22.8 N on a bearing \(19.7^\circ\) north of east.

2. Answer to Question 2 (4 marks)

   Method: the picture is in equilibrium, so the horizontal components of the tensions balance each other and the vertical components together balance the weight. Mark scheme: 1 mark for each equation, 1 mark for each tension.

   Horizontal: \(T_1\cos35^\circ=T_2\cos55^\circ\) (1). Vertical: \(T_1\sin35^\circ+T_2\sin55^\circ=60\) (1). Since \(\cos55^\circ=\sin35^\circ\), the first equation gives \(T_1=T_2\tan35^\circ\). Substituting, \(T_2(\tan35^\circ\sin35^\circ+\sin55^\circ)=60\), so \(1.2208T_2=60\) and \(T_2=49.1\) N (1), \(T_1=34.4\) N (1). Check: \(34.41\sin35^\circ+49.15\sin55^\circ=19.74+40.26=60\).

   Answer: the steeper string carries more of the weight: \(T_2\approx49.1\) N and \(T_1\approx34.4\) N.

3. Answer to Question 3 (5 marks)

   Method: friction can take any value up to the limiting value \(\mu R\). The object moves only if the applied force is greater than \(\mu R\); once it moves, friction equals \(\mu R\) and acts against the motion. Mark scheme as shown in brackets.

   1. The vertical component of the tension, \(40\sin30^\circ=20\) N, lifts the crate a little, so \(R\) is less than the weight. \(R=12\times9.8-40\sin30^\circ=117.6-20=97.6\) N (2). So \(R=97.6\) N.
   2. Compare the horizontal pull with the limiting friction. \(F_{max}=0.3\times97.6=29.3\) N (1). Horizontal pull \(40\cos30^\circ=34.6\) N \(\gt 29.3\) N, so the crate moves (1). So the crate moves and friction is 29.3 N.
   3. Apply \(F=ma\) horizontally. \(a=\dfrac{34.64-29.28}{12}=0.447\) m s\(^{-2}\) (1). So the crate accelerates at about 0.447 m s\(^{-2}\).

4. Answer to Question 4 (4 marks)

   Method: on a slope, resolve parallel and perpendicular to the plane. Perpendicular: \(R=mg\cos\alpha\) (plus any other perpendicular components). Parallel: use \(F=ma\), with friction \(F=\mu R\) acting against the motion or the likely motion. Mark scheme: 2 marks for each part.

   1. The block stays at rest only if \(\tan\alpha\le\mu\), because then \(mg\sin\alpha\le\mu mg\cos\alpha\). \(\tan35^\circ=0.700\gt 0.4\), so friction cannot hold it (2). So the block slides down the plane.
   2. Friction is limiting and acts up the slope; \(F=ma\) along the slope gives \(mg\sin35^\circ-0.4mg\cos35^\circ=ma\) and the mass cancels. \(a=g(\sin35^\circ-0.4\cos35^\circ)=9.8(0.5736-0.3277)=2.41\) m s\(^{-2}\) (2). So the block accelerates at about 2.41 m s\(^{-2}\).

5. Answer to Question 5 (3 marks)

   Method: write \(F=ma\) for each particle in its own direction of motion; the string is light and inextensible, so the tension and the acceleration are the same for both. Add the equations to eliminate \(T\). Mark scheme: 1 mark for the equations, 1 for \(a\), 1 for \(T\).

   Hanging: \(2g-T=2a\). Table: \(T=3a\) (1). So \(2g=5a\) and \(a=3.92\) m s\(^{-2}\) (1). \(T=3\times3.92=11.8\) N (1).

   Answer: the system accelerates at 3.92 m s\(^{-2}\) and the tension is about 11.8 N.

Total : 20 marks