1. [Maximum mark: 3]
A particle of mass 1.6 kg is projected with a speed of 20 ms−1 up a line of greatest slope of a smooth plane inclined at θ to the horizontal, where tan θ = 3/4. Use an energy method to find the distance the particle moves up the plane before coming to instantaneous rest.
▶️Answer/Explanation
Solution:
Using energy conservation:
Initial kinetic energy = Final potential energy
\(\frac{1}{2} \times 1.6 \times 20^2 = 1.6g \times x \sin \theta\)
Given \(\tan \theta = \frac{3}{4}\), so \(\sin \theta = \frac{3}{5}\).
Substituting values:
\(\frac{1}{2} \times 1.6 \times 400 = 1.6 \times 10 \times x \times \frac{3}{5}\)
Simplifying:
\(320 = 9.6x\)
\(x = \frac{320}{9.6} = \frac{100}{3} \, \text{m}\) (or 33.3 m).
2. [Maximum mark: 5]
A particle of mass 2.4 kg is held in equilibrium by two light inextensible strings, one attached to point A and the other to point B. The strings make angles of 35° and 40° with the horizontal, respectively. Find the tension in each of the two strings.
▶️Answer/Explanation
Solution:
Resolving forces horizontally and vertically:
\(T_1 \cos 35° = T_2 \cos 40°\) (1)
\(T_1 \sin 35° + T_2 \sin 40° = 2.4g\) (2)
From (1): \(T_1 = T_2 \frac{\cos 40°}{\cos 35°}\)
Substitute into (2):
\(T_2 \frac{\cos 40°}{\cos 35°} \sin 35° + T_2 \sin 40° = 24\)
Factor out \(T_2\):
\(T_2 \left( \frac{\cos 40° \sin 35°}{\cos 35°} + \sin 40° \right) = 24\)
Simplify using trigonometric identities:
\(T_2 (\tan 35° \cos 40° + \sin 40°) = 24\)
Solve for \(T_2\):
\(T_2 \approx 19.0 \, \text{N}\)
Substitute back to find \(T_1\):
\(T_1 \approx 20.4 \, \text{N}\).
3. [Maximum mark: 8]
The diagram shows the velocity-time graph for the motion of a bus. The bus starts from rest and accelerates uniformly for 8 seconds until it reaches a speed of 12.6 ms−1. The bus maintains this speed for 40 seconds. It then decelerates uniformly in two stages: between 48 and 62 seconds at \(a \, \text{ms}^{-2}\), and between 62 and 70 seconds at \(2a \, \text{ms}^{-2}\) until coming to rest.
(a) Find the distance covered by the bus in the first 8 seconds.
(b) Find the value of \(a\).
(c) Find the average speed of the bus for the whole journey. [4]
▶️Answer/Explanation
(a) Solution:
Distance = Area under velocity-time graph (triangle):
\(\text{Distance} = \frac{1}{2} \times 8 \times 12.6 = 50.4 \, \text{m}\).
(b) Solution:
Let \(v\) be the speed at \(t = 62 \, \text{s}\):
For \(48 \leq t \leq 62\): \(v = 12.6 – a(62 – 48) = 12.6 – 14a\) (1)
For \(62 \leq t \leq 70\): \(0 = v – 2a(70 – 62) \Rightarrow v = 16a\) (2)
Equate (1) and (2):
\(12.6 – 14a = 16a\)
\(12.6 = 30a\)
\(a = 0.42 \, \text{ms}^{-2}\).
(c) Solution:
Calculate distances for each phase:
1. First 8 s (acceleration): \(50.4 \, \text{m}\)
2. Next 40 s (constant speed): \(12.6 \times 40 = 504 \, \text{m}\)
3. Deceleration phase 1 (48–62 s): \(\frac{1}{2} \times (12.6 + 6.72) \times 14 = 135.24 \, \text{m}\)
4. Deceleration phase 2 (62–70 s): \(\frac{1}{2} \times 6.72 \times 8 = 26.88 \, \text{m}\)
Total distance: \(50.4 + 504 + 135.24 + 26.88 = 716.52 \, \text{m}\)
Total time: \(70 \, \text{s}\)
Average speed: \(\frac{716.52}{70} \approx 10.24 \, \text{ms}^{-1}\).
4. [Maximum mark: 9]
Two particles \(P\) (6 kg) and \(Q\) (2 kg) lie at rest 12.5 m apart on a rough horizontal plane (coefficient of friction 0.4). Particle \(P\) is projected towards \(Q\) with speed 20 ms−1.
(a) Show that the speed of \(P\) immediately before the collision is \(10\sqrt{3} \, \text{ms}^{-1}\).
(b) In the collision, \(P\) and \(Q\) coalesce to form particle \(R\). Find the loss of kinetic energy due to the collision.
(c) The coefficient of friction between \(R\) and the plane is 0.4. Find the distance travelled by particle \(R\) before coming to rest.
▶️Answer/Explanation
(a) Solution:
Frictional force on \(P\): \(F = \mu R = 0.4 \times 6g = 24 \, \text{N}\)
Deceleration: \(a = \frac{F}{m} = \frac{24}{6} = 4 \, \text{ms}^{-2}\)
Using \(v^2 = u^2 + 2as\):
\(v^2 = 20^2 – 2 \times 4 \times 12.5 = 400 – 100 = 300\)
\(v = \sqrt{300} = 10\sqrt{3} \, \text{ms}^{-1}\).
(b) Solution:
Using conservation of momentum:
\(6 \times 10\sqrt{3} = (6 + 2)v \Rightarrow v = 7.5\sqrt{3} \, \text{ms}^{-1}\)
Initial KE: \(\frac{1}{2} \times 6 \times (10\sqrt{3})^2 = 900 \, \text{J}\)
Final KE: \(\frac{1}{2} \times 8 \times (7.5\sqrt{3})^2 = 675 \, \text{J}\)
Loss of KE: \(900 – 675 = 225 \, \text{J}\).
(c) Solution:
Frictional force on \(R\): \(F = \mu R = 0.4 \times 8g = 32 \, \text{N}\)
Deceleration: \(a = \frac{F}{m} = \frac{32}{8} = 4 \, \text{ms}^{-2}\)
Using \(v^2 = u^2 + 2as\):
\(0 = (7.5\sqrt{3})^2 – 2 \times 4 \times s\)
\(s = \frac{168.75}{8} \approx 21.1 \, \text{m}\).
5. [Maximum mark: 7]
Two particles \(A\) (1.2 kg) and \(B\) (1.6 kg) are connected by a light inextensible string passing over a smooth pulley. \(A\) lies on a plane inclined at 40°, and \(B\) lies on a plane inclined at 50°. The coefficient of friction \(\mu\) is the same for both planes. Find the value of \(\mu\) for which the system is in limiting equilibrium.
▶️Answer/Explanation
Solution:
Resolving forces parallel to the planes:
For \(A\): \(T = 1.2g \sin 40° + \mu \times 1.2g \cos 40°\) (1)
For \(B\): \(1.6g \sin 50° = T + \mu \times 1.6g \cos 50°\) (2)
Substitute (1) into (2):
\(1.6g \sin 50° = 1.2g \sin 40° + \mu (1.2g \cos 40° + 1.6g \cos 50°)\)
Simplify:
\(\mu = \frac{1.6 \sin 50° – 1.2 \sin 40°}{1.2 \cos 40° + 1.6 \cos 50°}\)
Calculate:
\(\mu \approx 0.233\).
6. [Maximum mark: 9]
A car of mass 1300 kg is moving on a straight road.
(a) On a horizontal section of the road, the car has a constant speed of 30 ms−1 and there is a constant resisting force of 650 N.
(i) Calculate, in kW, the power developed by the engine of the car.
(ii) Given that this power is suddenly increased by 9 kW, find the instantaneous acceleration of the car.
(b) On a section of the road inclined at sin−1(0.08) to the horizontal, the resistance to the motion of the car is (1000 + 20v) N when the speed of the car is v ms−1. The car travels downwards along this section of the road at constant speed with the engine working at 11.5 kW. Find this constant speed.
▶️Answer/Explanation
(a)(i) Solution:
Power = Force × Velocity
\(P = 650 \times 30 = 19500 \, \text{W} = 19.5 \, \text{kW}\).
(a)(ii) Solution:
New power: \(19.5 + 9 = 28.5 \, \text{kW}\)
Driving force: \(F = \frac{P}{v} = \frac{28500}{30} = 950 \, \text{N}\)
Net force: \(950 – 650 = 300 \, \text{N}\)
Acceleration: \(a = \frac{F}{m} = \frac{300}{1300} \approx 0.231 \, \text{ms}^{-2}\).
(b) Solution:
Driving force: \(F = \frac{P}{v} = \frac{11500}{v}\)
Resolving forces parallel to the slope:
\(F + 1300g \times 0.08 = 1000 + 20v\)
Substitute \(F\):
\(\frac{11500}{v} + 1040 = 1000 + 20v\)
Simplify:
\(20v^2 – 40v – 11500 = 0\)
Solve quadratic:
\(v = 25 \, \text{ms}^{-1}\) (discard negative root).
7. [Maximum mark: 9]
A particle moves in a straight line starting from point \(O\) before coming to instantaneous rest at point \(X\). The velocity \(v \, \text{ms}^{-1}\) at time \(t \, \text{s}\) is given by:
- \(v = 7.2t^2\) for \(0 \leq t \leq 2\)
- \(v = 30.6 – 0.9t\) for \(2 \leq t \leq 8\)
- \(v = \frac{1600}{t^2} + kt\) for \(t \geq 8\) (where \(k\) is a constant)
Given no instantaneous change in velocity at \(t = 8\), find the distance \(OX\).
▶️Answer/Explanation
Solution:
1. For \(0 \leq t \leq 2\): Integrate \(v = 7.2t^2\) to get distance:
\(\int_0^2 7.2t^2 \, dt = 2.4t^3 \big|_0^2 = 19.2 \, \text{m}\)
2. For \(2 \leq t \leq 8\): Integrate \(v = 30.6 – 0.9t\) to get distance:
\(\int_2^8 (30.6 – 0.9t) \, dt = 30.6t – 0.45t^2 \big|_2^8 = 156.6 \, \text{m}\)
3. For \(t \geq 8\): Find \(k\) using continuity at \(t = 8\):
\(30.6 – 0.9 \times 8 = \frac{1600}{64} + 8k \Rightarrow k = -0.2\)
Integrate \(v = \frac{1600}{t^2} – 0.2t\) to get distance:
\(\int_8^{20} \left( \frac{1600}{t^2} – 0.2t \right) dt = \left[ -\frac{1600}{t} – 0.1t^2 \right]_8^{20} = 86.4 \, \text{m}\)
Total distance \(OX\): \(19.2 + 156.6 + 86.4 = 262.2 \, \text{m}\).