Projectiles and moments – A-level Maths exam questions with mark scheme

1CoreCalculator4 marks

A uniform beam $AB$ has length 6 m and mass 20 kg. It rests horizontally on two supports, one at $A$ and one at $B$. A particle $P$ of mass 15 kg is placed on the beam at a distance 2 m from $A$, as shown.

Find the magnitude of the reaction on the beam at each support. [4]

takes moments about $A$ (or $B$): $R_B\times6=20g\times3+15g\times2$M1
$R_B=15g=147$ N (accept 150 N or $15g$)A1
resolves vertically $R_A+R_B=35g$ (or second moments equation)M1
$R_A=20g=196$ N (accept 200 N or $20g$)A1
Answer: $R_A=196$ N, $R_B=147$ N
Examiner tip: Take moments about a support so that its unknown reaction disappears from the equation.

2CoreCalculator6 marks

A stone is thrown horizontally with speed $12\ \mathrm{m\,s^{-1}}$ from the top of a vertical cliff. The point of projection is 30 m above the sea. The stone is modelled as a particle moving freely under gravity.

(a) Find the time taken for the stone to reach the sea. [2]

(b) Find the horizontal distance from the foot of the cliff to the point where the stone enters the sea. [2]

(a) vertical motion: $30=\frac12\times9.8\,t^2$M1
(a) $t=$ awrt 2.5 s (2.47)A1
(b) $x=12\times$ their $t$M1
(b) awrt 30 m (29.7)A1
(c) vertical component $v_y=9.8t$ $(=24.2)$ or $v_y^2=2\times9.8\times30$, and $\sqrt{12^2+v_y^2}$M1
(c) awrt $27\ \mathrm{m\,s^{-1}}$ (27.1)A1
Answer: (a) 2.5 s; (b) 30 m; (c) $27\ \mathrm{m\,s^{-1}}$
Examiner tip: The horizontal velocity stays 12 throughout – the speed at impact combines it with the vertical velocity, not just one or the other.

3StandardCalculator8 marks

A uniform plank $AB$ has length 5 m and mass 30 kg. It rests horizontally on two supports at $C$ and $D$, where $AC=1$ m and $DB=1.5$ m, as shown. A man of mass 60 kg stands on the plank at its midpoint. The man is modelled as a particle.

(a) Find the magnitude of the reaction on the plank at $C$ and at $D$. [4]

The man now walks along the plank towards $B$.

(b) Find the distance of the man from $B$ when the plank is about to tilt. [3]

(a) moments about $C$ (or $D$), e.g. $R_D\times2.5=30g\times1.5+60g\times1.5$M1
(a) $R_D=54g=529$ N (accept 530 N)A1
(a) resolves vertically $R_C+R_D=90g$ (or second moments equation)M1
(a) $R_C=36g=353$ N (accept 350 N)A1
(b) about to tilt about $D$, so $R_C=0$B1
(b) moments about $D$: $60g\times x=30g\times1$ (giving $x=0.5$ m beyond $D$)M1
(b) 1 m from $B$A1
(c) the weight of the plank acts at its midpoint (2.5 m from $A$) / centre of mass at the midpointB1
Answer: (a) $R_C=36g\approx353$ N, $R_D=54g\approx529$ N; (b) 1 m from $B$; (c) weight of plank taken to act at its midpoint
Examiner tip: 'About to tilt' means the reaction at the other support is zero – set $R_C=0$ before taking moments about $D$.

4StandardCalculator7 marks

A ball is kicked from a point $O$ on horizontal ground with speed $25\ \mathrm{m\,s^{-1}}$ at an angle $\alpha$ above the horizontal, where $\tan\alpha=\frac34$. The ball is modelled as a particle moving freely under gravity.

(a) Find the greatest height above the ground reached by the ball. [2]

(b) Find the horizontal distance from $O$ to the point where the ball first hits the ground. [4]

(a) $0^2=(25\sin\alpha)^2-2\times9.8\times h$ with $25\sin\alpha=15$M1
(a) awrt 11 m (11.5)A1
(b) vertical: $0=15t-4.9t^2$M1
(b) $t=\frac{15}{4.9}=3.06$ sA1
(b) horizontal: $x=25\cos\alpha\times t=20t$M1
(b) awrt 61 m (61.2)A1
(c) $20\ \mathrm{m\,s^{-1}}$B1
Answer: (a) 11.5 m; (b) 61 m; (c) $20\ \mathrm{m\,s^{-1}}$
Examiner tip: With $\tan\alpha=\frac34$ use the 3-4-5 triangle: $\sin\alpha=0.6$, $\cos\alpha=0.8$ – exact components avoid rounding errors.

5HardCalculator8 marks

A ball is projected from a point $O$ on horizontal ground with speed $U\ \mathrm{m\,s^{-1}}$ at an angle of $40^\circ$ above the horizontal. The ball just passes over the top of a vertical wall of height 3 m. The wall is at a horizontal distance of 20 m from $O$. The ball is modelled as a particle moving freely under gravity.

(a) Find the value of $U$. [5]

(b) Find the distance beyond the wall at which the ball first hits the ground. [3]

(a) horizontal: $20=U\cos40^\circ\,t$M1
(a) vertical: $3=U\sin40^\circ\,t-\frac12(9.8)t^2$M1
(a) both equations correctA1
(a) eliminates $t$, e.g. $3=20\tan40^\circ-\dfrac{4.9\times400}{U^2\cos^2 40^\circ}$, and solves for $U^2$ $(=242.3)$dM1
(a) $U=$ 15.6 (awrt 16)A1
(b) time of flight $\frac{2U\sin40^\circ}{9.8}$ (=2.04 s) or range $\frac{U^2\sin80^\circ}{9.8}$M1
(b) horizontal distance $U\cos40^\circ\times$ their time (or their range), then subtracts 20M1
(b) awrt 4.4 m (4.35)A1
Answer: (a) $U\approx15.6$; (b) 4.4 m
Examiner tip: Keep $U^2$ unrounded into part (b) – using $U=16$ puts the landing point about 5.7 m beyond the wall instead of 4.4 m.

6HardCalculator6 marks

A uniform ladder $AB$, of length 4 m and mass 12 kg, rests with its end $A$ on rough horizontal ground and its end $B$ against a smooth vertical wall. The ladder makes an angle $\theta$ with the ground and lies in a vertical plane perpendicular to the wall. The coefficient of friction between the ladder and the ground is 0.4.

The ladder is in limiting equilibrium.

(a) Show that $\tan\theta=\frac54$. [5]

(b) Hence find the size of angle $\theta$. [1]

(a) resolves vertically: $R=12g$M1
(a) resolves horizontally: $F=S$ (wall reaction) with $F=0.4R$, so $S=4.8g$M1
(a) moments about $A$: $S\times4\sin\theta=12g\times2\cos\theta$ (dimensionally correct, all terms)M1
(a) correct equation, e.g. $4.8g\times4\sin\theta=24g\cos\theta$A1
(a) $\tan\theta=\frac{24}{19.2}=\frac54$ (given answer – no errors seen)A1
(b) $\theta=$ awrt $51.3^\circ$B1
Answer: (b) $\theta\approx51.3^\circ$
Examiner tip: The wall is smooth, so its reaction is horizontal only – all the friction is at the ground.

7HardCalculator7 marks

A non-uniform rod $AB$ has length 3 m and mass 8 kg. The rod hangs in equilibrium in a horizontal position, supported by two vertical light strings attached to the rod at $A$ and at $C$, where $AC=2.4$ m, as shown. The tension in the string at $C$ is three times the tension in the string at $A$.

(a) Find the distance of the centre of mass of the rod from $A$. [4]

A particle of mass $m$ kg is now attached to the rod at $A$. The rod remains horizontal and in equilibrium, and the tensions in the two strings are now equal.

(b) Find the value of $m$. [3]

(a) resolves vertically: $T+3T=8g$M1
(a) $T_A=2g$ (and $T_C=6g$)A1
(a) moments about $A$: $3T\times2.4=8g\times x$M1
(a) $x=1.8$ mA1
(b) moments about $A$ (particle has no moment): $T'\times2.4=8g\times1.8$, so $T'=6g$M1
(b) resolves vertically: $2T'=(8+m)g$M1
(b) $m=4$A1
Answer: (a) 1.8 m from $A$; (b) $m=4$
Examiner tip: Taking moments about $A$ in (b) removes both the new particle and the string at $A$ – choose the pivot that kills the most unknowns.

8HardCalculator10 marks

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

A particle $P$ is projected from a point $A$, which is 2.5 m above horizontal ground, with velocity $(8\mathbf{i}+10\mathbf{j})\ \mathrm{m\,s^{-1}}$. $P$ moves freely under gravity until it hits the ground.

(a) Find the greatest height of $P$ above the ground. [2]

(b) Find the time taken for $P$ to reach the ground. [3]

(c) Find the speed of $P$ as it hits the ground and the angle its direction of motion makes with the horizontal at this instant. [4]

(d) State one limitation of the model that could affect the accuracy of your answers. [1]

(a) $0^2=10^2-2\times9.8\,s$ and adds 2.5M1
(a) awrt 7.6 mA1
(b) $-2.5=10t-4.9t^2$M1
(b) solves $4.9t^2-10t-2.5=0$M1
(b) $t=$ awrt 2.3 s (2.27)A1
(c) vertical velocity $10-9.8t$ $(=-12.2)$ or $v_y^2=10^2+2\times9.8\times2.5$, then speed $=\sqrt{8^2+v_y^2}$M1
(c) $\sqrt{213}=14.6\ \mathrm{m\,s^{-1}}$ (awrt 15)A1
(c) angle using $\tan^{-1}\dfrac{|v_y|}{8}$ (or $\cos^{-1}\dfrac{8}{\text{speed}}$)M1
(c) awrt $57^\circ$ (56.8°) below the horizontalA1
(d) e.g. air resistance ignored / $P$ treated as a particle (no size or spin) / $g$ taken as constant (allow any valid limitation; ignore extra correct ones)B1
Answer: (a) 7.6 m; (b) 2.3 s (2.27); (c) $14.6\ \mathrm{m\,s^{-1}}$ at $56.8^\circ$ below the horizontal; (d) e.g. air resistance is ignored
Examiner tip: In (c) the angle must be described – 'below the horizontal' – and found from $\tan^{-1}\frac{|v_y|}{8}$, not from the launch velocity.

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