Forces and Newton's laws (connected particles, pulleys, friction) – A-level Maths exam questions with mark scheme

1CoreCalculator4 marks

A lift of mass 600 kg is raised by a vertical cable. A woman of mass 80 kg stands in the lift. The lift moves upwards with constant acceleration $0.5\ \mathrm{m\,s^{-2}}$. Air resistance is ignored.

(a) Find the tension in the cable. [2]

(b) Find the magnitude of the force exerted on the woman by the floor of the lift. [2]

(a) N2L for lift and woman together: $T-680g=680\times0.5$ (or N2L for the lift alone, $T-600g-R=600\times0.5$, with their $R$)M1
(a) $T=680\times10.3=7004$, so $T=7000$ N (2 s.f.) or $7.00\times10^{3}$ N (3 s.f.); do not accept 7004 N (more than 3 s.f. after using $g=9.8$)A1
(b) N2L for woman: $R-80g=80\times0.5$M1
(b) $R=824$ N (accept 820 N)A1
Answer: (a) 7000 N (b) 824 N
Examiner tip: Each N2L equation must have the correct mass on the right: 680 kg for the whole system, 80 kg for the woman alone.

2CoreCalculator4 marks

[In this question $\mathbf{i}$ and $\mathbf{j}$ are perpendicular horizontal unit vectors.]

Two forces, $\mathbf{F}_1=(3\mathbf{i}-2\mathbf{j})$ N and $\mathbf{F}_2=(p\mathbf{i}+5\mathbf{j})$ N, act on a particle of mass 2 kg. These are the only forces acting on the particle. The particle moves with acceleration $(4\mathbf{i}+q\mathbf{j})\ \mathrm{m\,s^{-2}}$.

(a) Find the value of $p$ and the value of $q$. [3]

(b) Find the magnitude of the acceleration. [1]

(a) $\mathbf{F}_1+\mathbf{F}_2=2(4\mathbf{i}+q\mathbf{j})$, i.e. $(3+p)\mathbf{i}+3\mathbf{j}=8\mathbf{i}+2q\mathbf{j}$M1
(a) $p=5$A1
(a) $q=1.5$A1
(b) $\sqrt{4^2+1.5^2}=$ awrt $4.27\ \mathrm{m\,s^{-2}}$ (accept exact $\frac{\sqrt{73}}{2}$)B1
Answer: (a) $p=5$, $q=1.5$; (b) $4.27\ \mathrm{m\,s^{-2}}$
Examiner tip: Write $\mathbf{F}=m\mathbf{a}$ as a vector equation and compare $\mathbf{i}$ and $\mathbf{j}$ components separately.

3HardCalculator10 marks

A particle $A$ of mass 3 kg rests on a rough horizontal table. The coefficient of friction between $A$ and the table is 0.2. $A$ is attached to one end of a light inextensible string which passes over a small smooth pulley fixed at the edge of the table. The other end of the string is attached to a particle $B$ of mass 2 kg, which hangs freely 0.5 m above the floor, as shown. The string is taut and the part from $A$ to the pulley is horizontal.

The system is released from rest.

(a) Find the acceleration of $B$. [3]

(b) Find the tension in the string. [2]

When $B$ hits the floor it does not rebound. $A$ does not reach the pulley.

(d) Find the further distance that $A$ moves before it comes to rest. [3]

(a) N2L for $B$: $2g-T=2a$M1
(a) N2L for $A$ with $F=0.2\times3g$: $T-0.6g=3a$M1
(a) $a=0.28g=2.744$, awrt $2.7\ \mathrm{m\,s^{-2}}$A1
(b) substitutes their $a$ into either equationM1
(b) $T=$ awrt 14 N (14.1 N)A1
(c) $v^2=0^2+2\times2.744\times0.5$M1
(c) awrt $1.7\ \mathrm{m\,s^{-1}}$ (1.66)A1
(d) deceleration of $A$ from $0.2\times3g=3a'$, $a'=0.2g=1.96$M1
(d) $0=v^2-2\times1.96\times s$ with their $v$M1
(d) $s=0.7$ mA1
Answer: (a) $2.7\ \mathrm{m\,s^{-2}}$; (b) 14 N; (c) $1.7\ \mathrm{m\,s^{-1}}$; (d) 0.7 m
Examiner tip: Once $B$ lands the string goes slack, so the tension disappears and only friction acts on $A$ – don't reuse $T$ in part (d).

4StandardCalculator5 marks

A box of mass 5 kg is placed on a rough plane inclined at $30^\circ$ to the horizontal. The coefficient of friction between the box and the plane is 0.4. The box is released from rest. The box is modelled as a particle.

Show that the box slides down the plane and find its acceleration. [5]

resolves perpendicular to the plane: $R=5g\cos30^\circ$ $(=42.4)$M1
uses $F_{\max}=0.4R$ $(=16.97)$M1
compares with component of weight down the plane: $5g\sin30^\circ=24.5>16.97$, so the box slides (or equivalent: $\tan30^\circ=0.577>\mu=0.4$)A1
N2L down the plane: $5g\sin30^\circ-0.4R=5a$M1
$a=1.5$ or $1.51\ \mathrm{m\,s^{-2}}$ (2 or 3 s.f. only, since $g=9.8$ was used)A1
Answer: $24.5\ \text{N}>F_{\max}=17.0$ N so it slides; $a\approx1.5\ \mathrm{m\,s^{-2}}$
Examiner tip: The normal reaction on a slope is $mg\cos\theta$, not $mg$ – using $5g$ for $R$ loses most of the marks.

5StandardCalculator7 marks

A car of mass 1200 kg tows a trailer of mass 400 kg along a straight horizontal road using a light rigid towbar. The resistances to motion of the car and the trailer are constant and of magnitude 300 N and 100 N respectively. The engine of the car produces a constant driving force of 2000 N.

(a) Find the acceleration of the car and trailer. [2]

(b) Find the tension in the towbar. [2]

When the speed is $12\ \mathrm{m\,s^{-1}}$ the towbar breaks. The resistance to the motion of the trailer is unchanged.

(a) N2L whole system: $2000-300-100=1600a$M1
(a) $a=1\ \mathrm{m\,s^{-2}}$A1
(b) N2L for trailer: $T-100=400\times1$ (or for car $2000-300-T=1200\times1$)M1
(b) $T=500$ NA1
(c) $-100=400a'$ giving deceleration $0.25\ \mathrm{m\,s^{-2}}$M1
(c) $0^2=12^2-2\times0.25\times s$M1
(c) 288 mA1
Answer: (a) $1\ \mathrm{m\,s^{-2}}$; (b) 500 N; (c) 288 m
Examiner tip: Use the trailer alone for the tension – it has the fewest forces, so there's less to get wrong.

6HardCalculator6 marks

A block of mass 4 kg is pulled along rough horizontal ground by a force of magnitude 30 N. The force acts at an angle of $20^\circ$ above the horizontal. The block accelerates at $2\ \mathrm{m\,s^{-2}}$. The block is modelled as a particle.

(a) Find the coefficient of friction between the block and the ground. [5]

(b) State how you have used the modelling assumption that the block is a particle. [1]

(a) resolves vertically: $R+30\sin20^\circ=4g$M1
(a) $R=4g-30\sin20^\circ$ $(=28.94)$A1
(a) N2L horizontally: $30\cos20^\circ-F=4\times2$M1
(a) uses $F=\mu R$ and solves for $\mu$M1
(a) $\mu=0.70$ or $0.698$ (2 or 3 s.f. only, since $g=9.8$ was used)A1
(b) all the forces (weight, reaction, friction, pull) act at a single point, so the block does not rotate or topple. Allow 'the forces act at the same point' or 'the block does not rotate/tip over'. Do not accept 'the block has no size/dimensions' alone, or anything about air resistance or the groundB1
Answer: (a) $\mu\approx0.70$; (b) all the forces act at a single point, so the block does not rotate
Examiner tip: The upward component of the pull reduces the normal reaction – $R$ is not $4g$ here.

7StretchCalculator11 marks

A particle $A$ of mass 4 kg lies on a rough plane inclined at an angle $\alpha$ to the horizontal, where $\tan\alpha=\frac34$. The coefficient of friction between $A$ and the plane is $\frac14$. $A$ is attached to a light inextensible string which passes over a small smooth pulley at the top of the plane. The other end of the string is attached to a particle $B$ of mass $m$ kg hanging freely below the pulley. The part of the string from $A$ to the pulley lies along a line of greatest slope of the plane.

(a) Given that $A$ is on the point of sliding up the plane, find the value of $m$. [4]

Particle $B$ is replaced by a particle of mass 5 kg and the system is released from rest.

(b) Find the acceleration of $A$ and the tension in the string. [4]

(a) $R=4g\cos\alpha=3.2g$ and $F=\frac14R=0.8g$B1
(a) resolves along the plane for $A$ with friction acting down the plane: $T=4g\sin\alpha+F$M1
(a) $T=2.4g+0.8g=3.2g$ and $T=mg$A1
(a) $m=3.2$A1
(b) N2L for $B$: $5g-T=5a$M1
(b) N2L for $A$: $T-2.4g-0.8g=4a$M1
(b) $a=0.2g=1.96\ \mathrm{m\,s^{-2}}$ (accept 2.0)A1
(b) $T=4g=39.2$ N (accept 39 N)A1
(c) angle between the two parts of the string is $90^\circ-\alpha$ $(=53.1^\circ)$M1
(c) $2T\cos\left(\frac{90^\circ-\alpha}{2}\right)$, or cosine rule $\sqrt{2T^2-2T^2\cos(90^\circ+\alpha)}$, or resolves both tensions horizontally and vertically and combines componentsM1
(c) 70 N or 70.1 N (2 or 3 s.f.)A1
Answer: (a) $m=3.2$; (b) $1.96\ \mathrm{m\,s^{-2}}$, 39.2 N; (c) 70 N
Examiner tip: 'On the point of sliding up' tells you the direction of friction (down the slope) – decide this before writing any equation.

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