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

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1. Answer to Question 1

   1. Weight is mass times the acceleration due to gravity. \(W=mg=6.5\times9.8=\) 63.7 N. The parcel weighs 63.7 N.
   2. Rearrange \(W=mg\) to make \(m\) the subject. \(m=\dfrac Wg=\dfrac{245}{9.8}=\) 25 kg. The crate has a mass of 25 kg.
   3. On a smooth road there is no resistance, so the resultant force is the driving force; use Newton's second law. \(F=ma=1200\times1.5=\) 1 800 N. A force of 1 800 N is needed.
   4. Force is measured in newtons (N); \(1\text{ N}=1\text{ kg m/s}^2\). Mass is the amount of matter (kg) and does not change; weight is the gravitational force on the mass (N) and depends on \(g\).

2. Answer to Question 2

   1. The forces are perpendicular, so use Pythagoras for the magnitude and trigonometry for the direction. Magnitude \(\sqrt{36+64}=\) 10 N. Direction \(\tan^{-1}\left(\tfrac86\right)=\) 53.1\(^\circ\) north of east. The resultant is 10 N on a bearing of about 037\(^\circ\).
   2. Again the forces are at right angles, so the magnitude comes from Pythagoras. \(\sqrt{144+25}=\) 13 N. The angle with the 12 N force is \(\tan^{-1}\tfrac5{12}=22.6^\circ\).
   3. Add the \(\mathbf i\) components and the \(\mathbf j\) components separately. Sum \(=2\mathbf i+6\mathbf j\) N. Magnitude \(\sqrt{4+36}=\sqrt{40}=\) 6.32 N. Angle with \(\mathbf i\): \(\tan^{-1}3=71.6^\circ\). The resultant is 6.32 N at \(71.6^\circ\) to the \(\mathbf i\) direction.

3. Answer to Question 3

   1. The resultant force is the pull minus the resistance; then apply \(F=ma\). \(F=ma\): \(50-18=8a\), so \(a=\) 4 m/s\(^2\). The box accelerates at 4 m/s\(^2\).
   2. Use \(v=u+at\) with \(u=0\). \(v=at=4\times5=\) 20 m/s. After 5 s the speed is 20 m/s.
   3. Use \(s=ut+\tfrac12at^2\) with \(u=0\). \(s=\tfrac12at^2=\tfrac12\times4\times25=\) 50 m. The box travels 50 m in 5 s.
   4. Once the pull is removed, only the 18 N resistance acts, so the box decelerates; use \(v=u+at\) with \(v=0\). Deceleration \(=\dfrac{18}8=2.25\) m/s\(^2\). \(t=\dfrac{20}{2.25}=\) 8.9 s. The box stops about 8.9 s after the pull is removed.

4. Answer to Question 4

   1. The horizontal component uses cosine and the vertical component uses sine of the angle with the horizontal. Horizontal \(20\cos30^\circ=\) 17.3 N; vertical \(20\sin30^\circ=\) 10 N (up).
   2. Use the same method; the vertical component points downwards. Horizontal \(15\cos40^\circ=\) 11.5 N; vertical \(15\sin40^\circ=\) 9.64 N (down).
   3. The force makes \(20^\circ\) with the horizontal ground, so resolve with \(\cos20^\circ\) and \(\sin20^\circ\). Horizontal \(50\cos20^\circ=\) 47.0 N; vertical \(50\sin20^\circ=\) 17.1 N.
   4. Add the two horizontal components because they act in the same direction. \(17.32+11.49=\) 28.8 N. The total horizontal force is 28.8 N.

5. Answer to Question 5

   1. Equilibrium means the resultant is zero: \(\mathbf F_3=-(\mathbf F_1+\mathbf F_2)=-(2\mathbf i+6\mathbf j)=\) \(-2\mathbf i-6\mathbf j\) N. Check: \((5-3-2)\mathbf i+(2+4-6)\mathbf j=\mathbf 0\) ✓.
   2. Use Pythagoras for the magnitude and \(\tan\) for the angle. \(|\mathbf F_3|=\sqrt{4+36}=\) 6.32 N. The force points down-left; angle below the negative \(x\)-axis \(=\tan^{-1}\left(\tfrac62\right)=71.6^\circ\).
   3. Once \(\mathbf F_3\) is removed, the resultant is equal and opposite to \(\mathbf F_3\); apply \(\mathbf F=m\mathbf a\). The resultant of the other two is \(2\mathbf i+6\mathbf j\), magnitude 6.32 N. \(a=\dfrac{6.32}4=\) 1.58 m/s\(^2\) in the direction of \(2\mathbf i+6\mathbf j\).

6. Answer to Question 6

   1. On a smooth slope only two forces act on the box. Weight \(mg=49\) N vertically down; normal reaction \(R\) perpendicular to the slope. No friction (smooth). The weight is \(5\times9.8=49\) N.
   2. There is no acceleration perpendicular to the slope, so the forces balance in that direction. Perpendicular to the slope: \(R=mg\cos30^\circ=49\times0.866=\) 42.4 N. The normal reaction is 42.4 N.
   3. The only force along the slope is the component of the weight, \(mg\sin30^\circ\). Along the slope: \(mg\sin30^\circ=ma\), so \(a=g\sin30^\circ=\) 4.9 m/s\(^2\). The acceleration does not depend on the mass of the box.
   4. Use \(v=u+at\) with \(u=0\). \(v=4.9\times3=\) 14.7 m/s. After 3 s the box moves at 14.7 m/s.

7. Answer to Question 7

   1. Take upwards as positive; the resultant force is \(R-mg\). \(R-mg=ma\): \(R=60(9.8+1.2)=\) 660 N. The floor pushes with 660 N.
   2. Take downwards as positive; the resultant force is \(mg-R\). \(mg-R=ma\): \(R=60(9.8-0.8)=\) 540 N. The floor pushes with 540 N.
   3. Constant speed means zero acceleration, so the forces balance. No acceleration: \(R=mg=\) 588 N. The floor pushes with 588 N.
   4. Accelerating up the floor pushes harder (you feel heavier, 660 N); accelerating down you feel lighter (540 N); at constant speed you feel your normal weight.

8. Answer to Question 8

   1. Treat the car and caravan as one body: the tension is internal and cancels. Whole system: \(3500-600-400=2000a\), so \(a=\) 1.25 m/s\(^2\). The acceleration is 1.25 m/s\(^2\).
   2. Now consider the caravan alone: the only horizontal forces on it are the tension and its resistance. Caravan alone: \(T-400=800\times1.25\), so \(T=\) 1 400 N. The tension in the tow bar is 1 400 N.
   3. Take the direction of motion as positive and consider the whole system first. Forces opposing motion: \(600+400+2000=3000\) N, so \(a=\dfrac{-3000}{2000}=-1.5\) m/s\(^2\) (deceleration 1.5). Caravan: With tension \(T\) forwards on the caravan: \(T-400=800\times(-1.5)\), so \(T=-800\) N, meaning the bar pushes (the tow bar is in thrust, not tension): 800 N thrust.

9. Answer to Question 9

   1. The heavier particle moves down and the lighter one moves up, with the same acceleration \(a\) because the string is inextensible. Heavier particle (down): \(5g-T=5a\). Lighter particle (up): \(T-3g=3a\).
   2. Adding the equations eliminates \(T\). Add: \(2g=8a\), \(a=\dfrac{2\times9.8}8=\) 2.45 m/s\(^2\). The particles accelerate at 2.45 m/s\(^2\).
   3. Substitute \(a\) into the equation for the lighter particle. \(T=3(g+a)=3\times12.25=\) 36.75 N. The tension is 36.75 N.
   4. Use \(v=at\) and \(s=\tfrac12at^2\) from rest. \(v=2.45\times2=\) 4.9 m/s; distance \(=\tfrac12\times2.45\times4=\) 4.9 m. Each particle moves 4.9 m in 2 s and reaches 4.9 m/s.

10. Answer to Question 10

    1. On horizontal ground the normal reaction balances the weight. \(R=mg=196\) N. \(F_{max}=\mu R=0.3\times196=\) 58.8 N. The maximum friction is 58.8 N.
    2. Compare the push with the maximum available friction. 50 N is less than 58.8 N, so friction balances the push: the box does not move.
    3. Friction is now limiting and equal to 58.8 N; apply \(F=ma\) to the resultant. \(80-58.8=20a\), so \(a=\) 1.06 m/s\(^2\). The box accelerates at 1.06 m/s\(^2\).
    4. Use \(s=\tfrac12at^2\) from rest. \(5=\tfrac12\times1.06t^2\), \(t^2=9.43\), \(t=\) 3.07 s. The box takes about 3.07 s to travel 5 m.

11. Answer to Question 11

    1. Each cable has vertical component \(T\sin\theta\); in equilibrium the upward forces balance the 200 N weight. Resolving vertically: the vertical components of both tensions balance the weight, \(2T\sin\theta=200\).
    2. Substitute \(\sin30^\circ=\tfrac12\). \(T=\dfrac{200}{2\sin30^\circ}=\) 200 N.
    3. Substitute \(\sin20^\circ=0.342\). \(T=\dfrac{200}{2\sin20^\circ}=\dfrac{100}{0.342}=\) 292.4 N. The tension rises to 292.4 N.
    4. \(\sin\theta\le1\), so \(T=\dfrac{100}{\sin\theta}\ge100\) N. As \(\theta\to0\) the vertical component of each tension shrinks, so a bigger tension is needed. This is why cables are never pulled completely flat.

12. Answer to Question 12

    1. Take upwards as positive; the resultant force is \(T-mg\). \(T-250g=250\times0.4\), so \(T=2450+100=\) 2 550 N. The tension is 2 550 N.
    2. At constant speed the acceleration is zero, so the forces balance. \(T=mg=\) 2 450 N. The tension is 2 450 N.
    3. Use \(v^2=u^2+2as\) with \(u=2\), \(v=0\), \(s=5\), then Newton's second law. \(0=4-2a\times5\), so the deceleration is \(a=0.4\) m/s\(^2\). Then \(250g-T=250\times0.4\) gives \(T=2450-100=\) 2 350 N. The deceleration is 0.4 m/s\(^2\) and the tension is 2 350 N.
    4. Apply Newton's second law with \(T=2600\) N. \(2600-2450=250a\), so \(a=\) 0.6 m/s\(^2\). The greatest safe acceleration is 0.6 m/s\(^2\).

13. Answer to Question 13

    1. By symmetry the sideways components cancel and only the components along the bisector remain. The resultant lies along the bisector, at \(30^\circ\) to each rope: \(2\times5\cos30^\circ=\) 8.66 kN. The combined pull is 8.66 kN.
    2. Subtract the water resistance from the forward pull. \(8.66-6=\) 2.66 kN. The net force is 2.66 kN forwards.
    3. Convert to SI units (2 660 N and 40 000 kg) and use \(a=\dfrac Fm\). \(a=\dfrac{2660}{40000}=\) 0.0665 m/s\(^2\). The ship accelerates at about 0.0665 m/s\(^2\).
    4. Each rope now makes \(45^\circ\) with the direction of motion. \(2\times5\cos45^\circ=7.07\) kN, smaller than 8.66 kN. Pulling at a wider angle wastes some of the force sideways. The net force would only be 1.07 kN.

14. Answer to Question 14

    1. Use \(v^2=u^2-2as\) with \(u=25\), \(v=0\), \(s=50\). \(0=625-2a\times50\), so \(a=\) 6.25 m/s\(^2\). The deceleration is 6.25 m/s\(^2\).
    2. Use Newton's second law. \(F=ma=900\times6.25=\) 5 625 N. The braking force is 5 625 N.
    3. Use \(v=u-at\) with \(v=0\). \(t=\dfrac{25}{6.25}=\) 4 s. The car takes 4 s to stop.
    4. Find the new deceleration from \(F=ma\), then use \(v^2=u^2-2as\). \(a=\dfrac{3600}{900}=4\) m/s\(^2\). \(s=\dfrac{625}{2\times4}=\) 78.1 m. On the wet road the stopping distance increases to about 78.1 m.

15. Answer to Question 15

    1. Treat the whole train as one body: the coupling forces are internal. Total mass 70 t = 70 000 kg, total resistance 35 000 N. \(60000-35000=70000a\), \(a=\) 0.357 m/s\(^2\). The train accelerates at 0.357 m/s\(^2\).
    2. Consider the three trucks together: the only external horizontal forces are the coupling tension and their 15 000 N resistance. Trucks 1-3 (30 000 kg): \(T_1-15000=30000\times0.357\), so \(T_1=15000+10714=\) 25 714 N. The tension is about 25 700 N.
    3. Consider truck 3 alone. Truck 3 alone: \(T_3-5000=10000\times0.357\), \(T_3=\) 8 571 N. The tension in the last coupling is about 8 570 N.
    4. Deceleration of the whole train: \(\dfrac{20000+35000}{70000}=0.786\) m/s\(^2\). Trucks 1-3: the trucks need a retarding force of \(30000\times0.786=23\,571\) N but resistance gives only 15 000 N, so the coupling must push backwards on the trucks with \(23\,571-15\,000=8\,571\) N. Thrust in the coupling = 8 571 N.

16. Answer to Question 16

    1. Take the direction of motion as positive for each particle; the tension is the same throughout the string. \(B\): \(6g-T=6a\). \(A\): \(T-F=4a\) where \(F=0.25\times4g=9.8\) N.
    2. Add the two equations to eliminate \(T\). Add: \(6g-9.8=10a\), so \(a=\dfrac{58.8-9.8}{10}=\) 4.9 m/s\(^2\). \(T=6g-6a=58.8-29.4=\) 29.4 N. The acceleration is 4.9 m/s\(^2\) and the tension is 29.4 N.
    3. Use \(v^2=u^2+2as\) with \(u=0\), \(s=1.2\). \(v^2=2\times4.9\times1.2=11.76\), so \(v=\) 3.43 m/s. \(B\) hits the floor at 3.43 m/s.
    4. Once the string is slack, only friction acts horizontally on \(A\). Deceleration of \(A\) \(=\dfrac{9.8}4=2.45\) m/s\(^2\). Distance \(=\dfrac{v^2}{2a}=\dfrac{11.76}{4.9}=\) 2.4 m. \(A\) slides another 2.4 m before stopping.

17. Answer to Question 17

    1. In equilibrium the horizontal components of the tensions balance, and the vertical components support the weight \(50g=490\) N. Horizontally: \(T_1\cos40^\circ=T_2\cos25^\circ\). Vertically: \(T_1\sin40^\circ+T_2\sin25^\circ=490\).
    2. From the first, \(T_2=T_1\dfrac{\cos40^\circ}{\cos25^\circ}\). Substitute: \(T_1\left[\sin40^\circ+\cos40^\circ\tan25^\circ\right]=490\). The bracket equals \(\dfrac{\sin(40^\circ+25^\circ)}{\cos25^\circ}=\dfrac{\sin65^\circ}{\cos25^\circ}=1\) (because \(\sin65^\circ=\cos25^\circ\)). So \(T_1=\) 490 N.
    3. Use the horizontal equation. \(T_2=490\times\dfrac{0.7660}{0.9063}=\) 414 N. The tension in rope 2 is 414 N.
    4. Rope 1 carries 490 N, the larger tension, and is steeper so it takes more of the weight.

18. Answer to Question 18

    1. Resolve perpendicular to the slope, then use \(F_{max}=\mu R\). \(R=8g\cos30^\circ=67.9\) N. \(F_{max}=0.4\times67.9=\) 27.2 N. Maximum friction is 27.2 N.
    2. Compare the component of the weight down the slope with the maximum friction. Component of weight down the slope \(=8g\sin30^\circ=39.2\) N, which exceeds 27.2 N, so the block slides. \(8a=39.2-27.2=12.0\), \(a=\) 1.51 m/s\(^2\). The block slides down with acceleration 1.51 m/s\(^2\).
    3. When the block moves up, both the weight component and friction act down the slope. \(80-39.2-27.2=8a\), so \(a=\dfrac{13.64}8=\) 1.71 m/s\(^2\). The block accelerates up the slope at 1.71 m/s\(^2\).
    4. On the point of moving up, the tension balances the weight component plus limiting friction. \(T=39.2+27.2=\) 66.4 N. The smallest tension is 66.4 N.

19. Answer to Question 19

    1. Take upwards as positive and apply \(T-mg=ma\) to the lift (total 800 kg) in each phase. \(T-mg=ma\): phase 1: \(800(9.8+1.5)=\) 9 040 N; phase 2: \(800\times9.8=\) 7 840 N; phase 3: \(800(9.8-1.2)=\) 6 880 N.
    2. Apply \(R-mg=ma\) to the 70 kg passenger with the same accelerations. Phase 1: \(70(9.8+1.5)=\) 791 N; phase 2: 686 N; phase 3: \(70(9.8-1.2)=\) 602 N. The passenger feels heaviest in phase 1 and lightest in phase 3.
    3. Phase 1, when the tension is greatest. The tension of 9 040 N is the largest of the three phases.

20. Answer to Question 20

    1. In equilibrium the sum of the \(\mathbf i\) components and the sum of the \(\mathbf j\) components must both be zero. \(\mathbf i\): \(3k-4+1=0\) gives \(k=1\). \(\mathbf j\): \(2+(k+1)-4=0\) gives \(k=1\). Both agree: \(k=1\).
    2. With \(\mathbf F_3\) removed, the resultant is \(\mathbf F_1+\mathbf F_2\); use \(\mathbf F=m\mathbf a\). \(\mathbf F_1+\mathbf F_2=(3-4)\mathbf i+(2+2)\mathbf j=-\mathbf i+4\mathbf j\) N. \(\mathbf a=\dfrac12(-\mathbf i+4\mathbf j)=\) \((-0.5\mathbf i+2\mathbf j)\) m/s\(^2\). Magnitude \(\sqrt{0.25+4}=\) 2.06 m/s\(^2\).
    3. Sketch the vector \(-0.5\mathbf i+2\mathbf j\): it points left and up. The acceleration is in the second quadrant: angle \(=180^\circ-\tan^{-1}\left(\tfrac{2}{0.5}\right)=180^\circ-76.0^\circ=\) 104.0\(^\circ\) from the positive \(x\)-axis. The acceleration makes an angle of \(104.0^\circ\) with the positive \(x\)-axis.