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Question 1

Topic-4.1 – Forces and Motion

Two particles, of masses 1.8 kg and 1.2 kg, are connected by a light inextensible string that passes over
a fixed smooth pulley. The particles hang vertically. The system is released from rest.

Find the magnitude of the acceleration of the particles and find the tension in the string.

▶️Answer/Explanation

Solution :-

Use of Newton’s second law for either particle or system

T – 1.2g = 1.2a

1.8g – T = 1.8a

1.8g – 1.2g = (1.2 + 1.8)a

For attempt to solve for T

$ a = 2 {ms}^{-2} $

T = 14.4 N

Question 2

(a) Topic-4.2 – Energy, Work and Power

(b) Topic-4.2 – Energy, Work and Power

A particle of mass 7.5kg, starting from rest at A, slides down an inclined plane AB. The point B is
12.5 metres vertically below the level of A, as shown in the diagram.

(a) Given that the plane is smooth, use an energy method to find the speed of the particle at B.

(b) It is given instead that the plane is rough and the particle reaches B with a speed of 8 ms⁻¹. The
plane is 25 m long and the constant frictional force has magnitude F N.

Find the value of F.

▶️Answer/Explanation

Solution :-

(a)

KE = $\frac{1}{2} \times 7.5 \times v^{2}$

PE = 7.5 $\times g \times 12.5 [= 937.5]$

$v = 15.8 \text{ ms}^{-1}$

(b)

$KE_{B}=0.5\times7.5\times8^{2}[=240]$

$7.5\times g\times12.5=0.5\times7.5\times8^{2}+F\times25$

$F=27.9$

$ALTERNATIVE FOR 2(b)$

$8^{2}=0^{2}+2a\times25\Rightarrow a=1.28$

$7.5g\times\frac{12.5}{25}-F=7.5\times a$

$F=27.9$

Question 3

Topic-4.1 – Forces and Motion

Coplanar forces of magnitudes 52 N, 39 N and P N act at a point in the directions shown in the diagram.
The system is in equilibrium.

Find the values of P and 0.

▶️Answer/Explanation

Solution :-

Resolving in any direction to get an equation

$Pcos\theta=39$ $Psin\theta=52$

$P=\sqrt{39^{2}+52^{2}}$

$\theta=tan^{-1}(\frac{52}{39})$

$P=65$ $\theta=53.1$

Question 4

(a) Topic-4.3 – Kinematics – Motion in a Straight Line

(b) Topic-4.3 – Kinematics – Motion in a Straight Line

A bus travels between two stops, A and B. The bus starts from rest at A and accelerates at a constant rate
of $a$ m s$^{-2}$ until it reaches a speed of 16 m s$^{-1}$. It then travels at this constant speed before decelerating
at a constant rate of 0.75$a$ m s$^{-2}$, coming to rest at B. The total time for the journey is 240 s.

(a) Sketch the velocity-time graph for the bus’s journey from A to B.

(b) Find an expression, in terms of a, for the length of time that the bus is travelling with constant
speed.

▶️Answer/Explanation

Solution :-

(a)

(b)

$t_{1}=\frac{16}{a}, \quad t_{2}=\frac{64}{3a}$

$T=240-(\frac{16}{a}+\frac{64}{3a})$

(c)

$3000=\frac{1}{2}\times16\times(T+240)$ [T = 135]

$3000=\frac{1}{2}\times16\times\left[240+240-(\frac{16}{a}+\frac{64}{3a})\right]$

$135=240-(\frac{16}{a}+\frac{64}{3a})$

$a=\frac{16}{45}$

Question 5

(a) Topic-4.3 – Kinematics – Motion in a Straight Line

(b) Topic-4.3 – Kinematics – Motion in a Straight Line

A particle, A, is projected vertically upwards from a point O with a speed of 80 m s⁻¹. One second later
a second particle, B, with the same mass as A, is projected vertically upwards from O with a speed of
100 m s⁻¹. At time T s after the first particle is projected, the two particles collide and coalesce to form
a particle C.

(a) Show that T = 3.5.

(b) Find the height above O at which the particles collide.

▶️Answer/Explanation

Solution :-

(a)

$s_{A}=80T-\frac{1}{2}gT^{2}$

$s_{B}=100(T-1)-\frac{1}{2}g(T-1)^{2}$

Two correct expressions for the displacement of both particles at time $T$

$100(T-1)-5(T-1)^{2}=80T-5T^{2}$

Leading to $T=3.5$

(b)

$[s=80\times3.5-\frac{1}{2}g\times3.5^{2}]=218.75m$

(c)

$v_{A}=80-g\times3.5[=45]$

$v_{B}=100-g\times2.5[=75]$

$45m+75m=2mv$

$v=60$

$-218.75=60t-\frac{1}{2}g\times t^{2}$

$t=14.9+3.5=18.4s$

Question 6

Topic-4.1 – Forces and Motion

A particle of mass 1.2 kg is placed on a rough plane which is inclined at an angle θ to the horizontal,
where $sin~\theta=\frac{3}{5}$. The particle is kept in equilibrium by a horizontal force of magnitude P N acting in a
vertical plane containing a line of greatest slope (see diagram). The coefficient of friction between the
particle and the plane is 0.15.

Find the least possible value of P.

▶️Answer/Explanation

Solution :-

Attempt at resolving perpendicular to the plane to get an equation

$R=1.2g\times\frac{24}{25}+P\times\frac{7}{25}$

Attempt at resolving parallel to the plane to get an equation

$F+P\times\frac{24}{25}=1.2g\times\frac{7}{25}$

Use of $F=0.15R$ to get an equation in P only

Solve to get $P=1.63$

Question 7

(a) Topic-4.2 – Energy, Work and Power

(b) Topic-4.2 – Energy, Work and Power

A car has mass 1200 kg. When the car is travelling at a speed of v ms⁻¹, there is a resistive force of
magnitude kv N. The maximum power of the car’s engine is 92.16 kW.

(a) The car travels along a straight level road.

(i) The car has a greatest possible constant speed of 48 ms⁻¹. Show that k = 40.

(ii) At an instant when its speed is $45 \text{ ms}^{-1}$, find the greatest possible acceleration of the car.

(b) The car now travels at a constant speed up a hill inclined at an angle of $\sin^{-1}(0.15)$ to the horizontal.
Find the greatest possible speed of the car going up the hill.

▶️Answer/Explanation

Solution :-

7(a)(i)

$Power = k \times 48^2 = 92160 \Longrightarrow k = 40$

(a)(ii)

$[DF=] \frac{92160}{45}[=2048]$

$2048-40 \times 45 = 1200a$

$a = \frac{248}{1200} = \frac{31}{150} ms^{-2}$

(b)

$DF = 40v + 1200g \times 0.15$

$\frac{92160}{v} = 40v + 1200g \times 0.15$

$40v^{2} + 1800v – 92160 [= 0]$

$v = 30.5 ms^{-1}$

Question 8

(a) Topic-4.2 – Energy, Work and Power

(b) Topic-4.2 – Energy, Work and Power

A particle P moves in a straight line, passing through a point O with velocity $4.2 \text{ ms}^{-1}$. At time $t$ s after P passes O, the acceleration, $a \text{ ms}^{-2}$, of P is given by $a = 0.6t – 2.7$.
Find the distance P travels between the times at which it is at instantaneous rest.

▶️Answer/Explanation

Solution :-

$v=0.3t^{2}-2.7t+c$ $[c=4.2]$

$0.3t^{2}-2.7t+4.2[=0]$

$[(t-2)(t-7)=0\Rightarrow]t=2,7$

Attempt to integrate v

$s=0.1t^{3}-1.35t^{2}+4.2t[+c]$

For use of their positive t limits in their cubic expression for s

Total distance $=6.25m$

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