Question 1
Topic – ALV: 4.5
A crate of mass 200 kg is being pulled at constant speed along horizontal ground by a horizontal
rope attached to a winch. The winch is working at a constant rate of $4.5$ kW, and there is a constant
resistance to the motion of the crate of magnitude $600$ N.
(a) Find the time that it takes for the crate to move a distance of $15$ m.
1. A crate of mass 200 kg is being pulled at constant speed along horizontal ground by a horizontal
rope attached to a winch. The winch is working at a constant rate of $4.5$ kW, and there is a constant
resistance to the motion of the crate of magnitude $600$ N.
(a) Find the time that it takes for the crate to move a distance of $15$ m.
The rope breaks after the crate has moved 15 m.
(b) Find the time taken, after the rope breaks, for the crate to come to rest.
▶️Answer/Explanation
Solution: –
1(a) $4500 = \frac{600 \times 15}{t}$
$t = 2~s$
1(b) $\pm600 = 200a \Rightarrow a = \pm3$
$0 = \frac{15}{their~2} + (their – 3)t$
$t = 2.5~s$
Question 2
Topic – ALV: 4.2
2. A particle P is projected vertically upwards from horizontal ground with speed 15 m s$^{-1}$.
(a) Find the speed of P when it is 10 m above the ground.
At the same instant that P is projected, a second particle Q is dropped from a height of $18$ m above
the ground in the same vertical line as P.
(b) Find the height above the ground at which the two particles collide.
▶️Answer/Explanation
Solution: –
2(a) Use constant acceleration in an attempt to find $v$ or $v^2$
$[v^{2}=15^{2}-2g \times 10]$
Speed = $5~ms^{-1}$
2(b) $(s_{P}=)\pm(15t-\frac{1}{2}gt^{2})$, $(s_{Q}=)\pm\frac{1}{2}gt^{2}$
Use $s_{p}+s_{Q}=18$ and solve for t
So height = 10.8m
Alternative method for Question 2(b): Using relative velocity
$\pm15t$
Use $15t=18$ and solve for t
So height = 10.8m
Question 3
Topic – ALV: 4.2
3. A particle moves in a straight line starting from rest from a point O. The acceleration of the particle
at time t s after leaving O is $a~ms^{-2}$ where $a=4t^{\frac{1}{2}}$.
(a) Find the speed of the particle when $t=9$.
(b) Find the time after leaving O at which the speed (in metres per second) and the distance travelled
(in metres) are numerically equal.
▶️Answer/Explanation
Solution: –
3(a)
Attempt to integrate
$\left[(v=\frac{4}{1.5})t^{\frac{3}{2}}=\frac{8}{3}t^{\frac{3}{2}}(+c)\right]$
Substitute $t=9$ to get speed = $72 ms^{-1}$
3(b)
Attempt at integration of their v
$\left[(s=)\frac{8}{3}t^{\frac{3}{2}}=\frac{16}{15}t^{\frac{5}{2}}(+c)\right]$
Equate their v and their s and attempt to solve for t
$\left[\frac{16}{15}t^{\frac{5}{2}}=\frac{8}{3}t^{\frac{3}{2}}\Rightarrow \frac{16}{15}t-\frac{8}{3}=0\right]$
time = $\frac{5}{2}$ s
Question 4
Topic – ALV: 4.5
A toy railway locomotive of mass $0.8$ kg is towing a truck of mass $0.4$ kg on a straight horizontal
track at a constant speed of $2$ m s$^{-1}$. There is a constant resistance force of magnitude $0.2$ N on the
locomotive, but no resistance force on the truck. There is a light rigid horizontal coupling connecting
the locomotive and the truck.
(a) State the tension in the coupling.
(b) Find the power produced by the locomotive’s engine.
The power produced by the locomotive’s engine is now changed to $1.2$ W.
(c) Find the magnitude of the tension in the coupling at the instant that the locomotive begins to
accelerate.
▶️Answer/Explanation
Solution: –
4(a) Tension = 0N
4(b) Power = $[0.2 \times 2]$ = 0.4 W
4(c) Driving force = 1.2/2 [= 0.6 N]
Use of Newton’s second law for locomotive or truck or system
For locomotive: $DF – 0.2 – T = 0.8a$
For truck: $T = 0.4a$
For system: $DF – 0.2 = 1.2a$
For attempt to solve for T
$T = \frac{2}{15} N$
Question 5
Topic – ALV: 4.1
The diagram shows a block D of mass $100$ kg supported by two sloping struts AD and BD, each
attached at an angle of $45^{\circ}$ to fixed points A and B respectively on a horizontal floor. The block is also
held in place by a vertical rope CD attached to a fixed point C on a horizontal ceiling. The tension in
the rope CD is $500$ N and the block rests in equilibrium.
(a) Find the magnitude of the force in each of the struts AD and BD.
A horizontal force of magnitude $F$ N is applied to the block in a direction parallel to AB.
(b) Find the value of $F$ for which the magnitude of the force in the strut AD is zero.
▶️Answer/Explanation
Solution: –
(a) Attempt to resolve vertically
$500 + T~cos~45 + T~cos~45 – 100g = 0$
OR $500 + T_{A}~cos~45 + T_{B}~cos~45 – 100g = 0$ AND
$T_{A}(sin~45) = T_{B}(sin~45)$
$T = 354~N$
$\text{Alternative Method 1 for Question 5(a)}$
Resolve perpendicular to $T_{A}$ or $T_{B}$
$T_{A}(or~T_{B}) + 500~cos~45 = 100g~cos~45$
$T_{A} = T_{B} = 354$
$\text{Alternative Method 2 for Question 5(a): Using triangle of forces}$
Attempt Pythagoras on a right-angled triangle of forces or use of trigonometry
${T_{A}}^{2}+{T_{B}}^{2}=(100g-500)^{2}$
$OR$
$ {sin~45~or~cos~45=\frac{100g-500}{T_{A}}~or~\frac{100g-500 {T_{B}}}$
$T^{2}+T^{2}=(100g-500)^{2}$
$OR$
${T_{A}}^{2}+{T_{B}}^{2}=(100g-500)^{2}$ AND $T_{A}(sin~45)=T_{B}(sin~45)$
$OR$
$sin~45=\frac{T_{A}(or~T_{B})}{100g-500}$
$ OR $
${cos~45=\frac{T_{A}(or~T_{B})}{100g-500}}$
$T_{A}=T_{B}=354$
$\text{Alternative Method 3 for Question 5(a): Using Lami’s Theorem}$
Attempt at Lami
${100g-500}{sin~90}= {T_{A}(or~T_{B})}{sin~135}$
$T_{A}=T_{B}=354$
(b) Attempt to resolve vertically and horizontally
$T_{B}cos~45 + 500 – 100g = 0$ and
$F – T_{B}sin~45 = 0$
$F = 500$
Alternative Method 1 for Question 5(b): Resolving perpendicular
Attempt to resolve perpendicular to $T_{B}$
$F~cos~45 + 500~cos~45 = 100g~cos~45$
$F = 500$
(b) Alternative Method 2 for Question 5(b): Using Lami’s Theorem
Attempt at Lami
$\frac{100g-500}{sin~135}=\frac{F}{sin~135}=\frac{T_{B}}{sin~90}$
$F = 500$
Alternative Method 3 for Question 5(b): Using triangle of forces
Attempt use of trigonometry on right angled triangle
$tan~45=\frac{F}{100g-500}$
$F = 500$
Question 6
Topic – ALV: 4.1
A block B, of mass $2$ kg, lies on a rough inclined plane sloping at $30^{\circ}$ to the horizontal. A light rope,
inclined at an angle of $20^{\circ}$ above a line of greatest slope, is attached to B. The tension in the rope
is $T$ N. There is a friction force of $F$ N acting on B (see diagram). The coefficient of friction between
B and the plane is $\mu$.
(a) It is given that $F = 5$ and that the acceleration of B up the plane is $1.2$ ms$^{-2}$.
(i) Find the value of $T$.
(ii) Find the value of $\mu$.
(b) It is given instead that $\mu = 0.8$ and $T = 15$.
Determine whether B will move up the plane.
▶️Answer/Explanation
Solution: –
(a)(i) Attempt to resolve parallel to the plane
$Tcos~20 – 5 = 2gsin~30 = 2 \times 1.2$
$T = 18.5$
(a)(ii) Attempt resolve perpendicular to the plane
$R = 2g~cos~30 – Tsin~20$
Use of $5 = \mu R$ to get an equation in μ only
$[5 = \mu \times (2g~cos~30 – Tsin~20)]$
$\mu = 0.455$
(b) Max $F = 0.8 \times (2g~cos~30 – 15~sin~20)$
$[= 0.8 \times 12.1902… = 9.7521…]$
Net force up the plane $= 15~cos~20 – 2g~sin~30 [= 4.0953…]$
$[State~4.0953… < 9.7521…]$ hence the block does not move
[up the plane]
Alternative Method 1 for Question 6(b)
Max force down plane $= 0.8 \times (2g~cos~30 – 15~sin~20) + 2g~sin~30$
$[= 0.8 \times 12.1902 + 10… = 19.7521…]$
Force up plane $= 15~cos~20 [= 14.0953…]$
$[State~14.0953… < 19.7521…]$ hence the block does not move
[up the plane]
(b) Alternative Method 2 for Question 6(b)
$F = 15~cos~20 – 2g~sin~30 [= 4.9053…]$
Or $R = 2g~cos~30 – 15~sin~20 [= 12.1902…]$
Get $\mu = \frac{15~cos~20 – 2g~sin~30}{2g~cos~30 – 15~sin~20} [= 0.3359…]$
[State 0.3359… < 0.8,] hence the block does not move [up the plane]
Question 7
Topic – ALV: 4.4
The diagram shows a smooth track which lies in a vertical plane. The section AB is a quarter circle
of radius $1.8$ m with centre O. The section BC is a horizontal straight line of length $7.0$ m and OB is
perpendicular to BC. The section CFE is a straight line inclined at an angle of $\theta^{\circ}$ above the horizontal.
A particle P of mass $0.5$ kg is released from rest at A. Particle P collides with a particle Q of mass
$0.1$ kg which is at rest at B. Immediately after the collision, the speed of P is $4$ ms$^{-1}$ in the direction
BC. You should assume that P is moving horizontally when it collides with Q.
(a) Show that the speed of Q immediately after the collision is $10$ ms$^{-1}$.
When Q reaches C, it collides with a particle R of mass $0.4$ kg which is at rest at C. The two particles
coalesce. The combined particle comes instantaneously to rest at F. You should assume that there
is no instantaneous change in speed as the combined particle leaves C, nor when it passes through C
again as it returns down the slope.
(b) Given that the distance CF is $0.4$ m, find the value of $\theta$.
(c) Find the distance from B at which P collides with the combined particle.
▶️Answer/Explanation
Solution: –
7(a) Attempt to use conservation of energy
$[\frac{1}{2}\times0.5v^{2}=0.5g\times1.8]~or~[\frac{1}{2}\times mv^{2}=mg\times1.8$]
$v=6$
Attempt at conservation of momentum
$[0.5\times6(+0)=0.5\times4+0.1w]$
Speed of $Q(=w)=10~ms^{-1}$
7(b) Attempt at conservation of momentum
$[0.1 \times 10 = (0.1 + 0.4) \times z (\Rightarrow z=2)]$
Attempt to use conservation of energy
$\left[\frac{1}{2} \times (0.1 + 0.4) \times (their~2)^{2} = (0.1 + 0.4)gh (\Rightarrow h=0.2)\right]$
Use trigonometry to get an equation in θ and solve for
$\theta = sin^{-1}\left(\frac{their~0.2}{0.4}\right)$
$\theta = 30$
Alternative method for Question 7(b): Using constant acceleration
Attempt at conservation of momentum
$[0.1 \times 10 = 0.5 \times z (\Rightarrow z=2)]$
Attempt at use of constant acceleration
$[0^2 = (their~2)^{2} + 2 \times a \times 0.4 (\Rightarrow a=-5)]$
Use N2L to get an equation in θ leading to a positive value of
and solve for
$[(0.5)|their~a| = (0.5)g~sin~\theta]$
$\theta = 30$
7(c) Q takes 0.7s to travel from B to C
$0.4 = \frac{(their~2) + 0}{2}t \Rightarrow t = 0.4$
Distance between P moved is $(0.7 + 0.8) \times 4 = 6$
Set up equation in t using 4t, (their 2)t and their 6 and solve for
$[4t + (their~2)t = (their~1)~OR~(their~6) + 4t + (their~2)t = 7]$
Distance from B = $6\frac{2}{3}~m$
7(c) Alternative method for last 3 marks of Question 7(c)
$Time~for~P = \frac{b}{4}$
$and$
$Time~for~QR = \frac{7-b}{2}$
OR $Time~for~P = \frac{7-c}{4}$ and $Time~for~QR = \frac{c}{2}$
Attempt to form an equation from use of total time and solve for b
(or c)
$\frac{7-b}{2}+0.7+0.4+0.4=\frac{b}{4} \Rightarrow b=6\frac{2}{3}$
OR $\frac{c}{2}+0.7+0.4+0.4=\frac{7-c}{4} \Rightarrow c=\frac{1}{3}$
Distance from B = $6\frac{2}{3}~m$