Projectiles and variable acceleration – Year 13 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Method: treat the horizontal and vertical motions separately. Horizontally the velocity \(u\cos\theta\) is constant; vertically the acceleration is \(g=9.8\) m s\(^{-2}\) downwards, so use the suvat equations with initial velocity \(u\sin\theta\). The time \(t\) links the two motions. The ball is thrown horizontally, so the initial vertical velocity is zero. Mark scheme as shown in brackets.

   1. \(40=4.9t^2\), \(t=2.86\) s (1). \(x=15\times2.857=42.9\) m (1).
   2. \(v_y=9.8\times2.857=28.0\), speed \(=\sqrt{225+784}=31.8\) m s\(^{-1}\) (1) at \(\tan^{-1}\dfrac{28}{15}=61.8^\circ\) below the horizontal (1).

   Answer: the ball hits the sea after about 2.86 s, 42.9 m from the cliff, at about 31.8 m s\(^{-1}\) and \(61.8^\circ\) below the horizontal.

2. Answer to Question 2 (4 marks)

   Method: treat the horizontal and vertical motions separately. Horizontally the velocity \(u\cos\theta\) is constant; vertically the acceleration is \(g=9.8\) m s\(^{-2}\) downwards, so use the suvat equations with initial velocity \(u\sin\theta\). The time \(t\) links the two motions. Mark scheme as shown in brackets.

   1. The ball returns to the ground when the vertical displacement is zero, at \(t=\dfrac{2u\sin\theta}g\). \(T=\dfrac{2\times26\sin35^\circ}{9.8}=3.04\) s (2). So the ball is in the air for about 3.04 s.
   2. Use \(R=\dfrac{u^2\sin2\theta}g\) and \(H=\dfrac{(u\sin\theta)^2}{2g}\). \(R=\dfrac{26^2\sin70^\circ}{9.8}=64.8\) m (1); \(H=\dfrac{(26\sin35^\circ)^2}{19.6}=11.3\) m (1). So the range is about 64.8 m and the greatest height about 11.3 m.

3. Answer to Question 3 (4 marks)

   Method: treat the horizontal and vertical motions separately. Horizontally the velocity \(u\cos\theta\) is constant; vertically the acceleration is \(g=9.8\) m s\(^{-2}\) downwards, so use the suvat equations with initial velocity \(u\sin\theta\). The time \(t\) links the two motions. Mark scheme as shown in brackets.

   1. Write \(x\) and \(y\) in terms of \(t\), make \(t\) the subject of the \(x\) equation and substitute. \(x=20\cos30^\circ\,t\), \(y=20\sin30^\circ\,t-4.9t^2\) (1). Eliminate \(t=\dfrac{x}{20\cos30^\circ}\) to get the result (1). The path is a parabola.
   2. Substitute \(x=25\), \(\tan30^\circ=0.5774\) and \(\cos^230^\circ=0.75\). \(y=25\tan30^\circ-\dfrac{9.8\times625}{800\times0.75}=14.434-10.208=4.23\) m (2). So the ball is about 4.23 m high when it has travelled 25 m horizontally.

4. Answer to Question 4 (4 marks)

   Method: displacement is \(\int v\,dt\), but distance travelled must count every part of the motion as positive. Find when \(v=0\), integrate separately on each interval where \(v\) keeps the same sign, and add the sizes of the results. Mark scheme as shown in brackets.

   1. \((t-1)(t-5)=0\), so at rest at \(t=1\) and \(t=5\) (1).
   2. \(\left[\tfrac{t^3}3-3t^2+5t\right]_0^5=\tfrac{125}3-75+25=-\tfrac{25}3=-8.33\) m (1).
   3. From 0 to 1: \(\tfrac13-3+5=\tfrac73\) m forwards (1). From 1 to 5: \(-\tfrac{25}3-\tfrac73=-\tfrac{32}3\), i.e. \(\tfrac{32}3\) m backwards. Total distance \(=\tfrac73+\tfrac{32}3=13\) m (1).

5. Answer to Question 5 (4 marks)

   Method: when the acceleration depends on \(t\), suvat cannot be used. Differentiate to go from displacement to velocity to acceleration, and integrate (adding a constant found from the initial conditions) to go back. Mark scheme as shown in brackets.

   1. \(\mathbf v=4t\mathbf i+(3t^2-3)\mathbf j\), \(\mathbf a=4\mathbf i+6t\mathbf j\) (2).
   2. At \(t=2\): \(\mathbf v=8\mathbf i+9\mathbf j\), speed \(=\sqrt{145}=12.0\) m s\(^{-1}\) (1). \(\mathbf a=4\mathbf i+12\mathbf j\), \(|\mathbf a|=\sqrt{160}=12.6\) m s\(^{-2}\) (1).

Total : 20 marks