1. [Maximum mark: 4]
Particles P of mass 0.4 kg and Q of mass 0.5 kg are free to move on a smooth horizontal plane. P and Q are moving directly towards each other with speeds 2.5 m s−1 and 1.5 m s−1 respectively. After P and Q collide, the speed of Q is twice the speed of P.
Find the two possible values of the speed of P after the collision.
▶️Answer/Explanation
Solution:
Using conservation of momentum:
\(0.4 \times 2.5 – 0.5 \times 1.5 = 0.4v + 0.5 \times 2v\)
Simplifying:
\(1 – 0.75 = 0.4v + v\)
\(0.25 = 1.4v\)
\(v = 0.179 \, \text{m s}^{-1}\) (first possible value)
Alternatively, considering direction reversal:
\(0.4 \times 2.5 – 0.5 \times 1.5 = -0.4v + 0.5 \times 2v\)
Simplifying:
\(0.25 = 0.6v\)
\(v = 0.417 \, \text{m s}^{-1}\) (second possible value)
2. [Maximum mark: 5]
A cyclist is travelling along a straight horizontal road. She is working at a constant rate of 150 W. At an instant when her speed is 4 m s−1, her acceleration is 0.25 m s−2. The resistance to motion is 20 N.
(a) Find the total mass of the cyclist and her bicycle.
(b) The cyclist comes to a straight hill inclined at an angle θ above the horizontal. She ascends the hill at constant speed 3 m s−1. She continues to work at the same rate as before and the resistance force is unchanged. Find the value of θ.
▶️Answer/Explanation
(a)
Using power formula \(P = Fv\):
\(150 = F \times 4\) → \(F = 37.5 \, \text{N}\)
Using Newton’s second law:
\(37.5 – 20 = m \times 0.25\)
\(m = 70 \, \text{kg}\)
(b)
Resolving forces up the hill:
\(\frac{150}{3} – 20 – 70g \sin \theta = 0\)
\(50 – 20 = 700 \sin \theta\)
\(\sin \theta = \frac{30}{700}\)
\(\theta = 2.5^\circ\)
3. [Maximum mark: 6]
Four coplanar forces act at a point. The magnitudes of the forces are 20 N, 30 N, 40 N and F N. The directions of the forces are as shown in the diagram, where \(\sin \alpha^\circ = 0.28\) and \(\sin \beta^\circ = 0.6\). Given that the forces are in equilibrium, find F and θ.
▶️Answer/Explanation
Solution:
Resolving vertically:
\(F \sin \theta + 20 \sin 60^\circ – 30 \times 0.28 – 40 \times 0.6 = 0\)
\(F \sin \theta = 15.08\)
Resolving horizontally:
\(F \cos \theta + 40 \times 0.8 – 30 \times 0.96 – 20 \cos 60^\circ = 0\)
\(F \cos \theta = 6.8\)
Dividing the two equations:
\(\tan \theta = \frac{15.08}{6.8}\) → \(\theta = 65.7^\circ\)
\(F = \sqrt{15.08^2 + 6.8^2} = 16.5 \, \text{N}\)
4. [Maximum mark: 6]
A particle is projected vertically upwards with speed \(u \, \text{m s}^{-1}\) from a point on horizontal ground. After 2 seconds, the height of the particle above the ground is 24 m.
(a) Show that \(u = 22\).
(b) The height of the particle above the ground is more than \(h \, \text{m}\) for a period of 3.6 s. Find \(h\).
▶️Answer/Explanation
(a)
Using \(s = ut – \frac{1}{2}gt^2\):
\(24 = u \times 2 – \frac{1}{2} \times 10 \times 2^2\)
\(24 = 2u – 20\) → \(u = 22\) (shown)
(b)
Time to maximum height: \(t = \frac{22}{10} = 2.2 \, \text{s}\)
Maximum height: \(s = 22 \times 2.2 – \frac{1}{2} \times 10 \times 2.2^2 = 24.2 \, \text{m}\)
Time descending: \(3.6 – (2.2 – t_1) = 1.8 \, \text{s}\)
Distance descended: \(s = \frac{1}{2} \times 10 \times 1.8^2 = 16.2 \, \text{m}\)
\(h = 24.2 – 16.2 = 8 \, \text{m}\)
5. [Maximum mark: 9]
A car of mass 1400 kg is towing a trailer of mass 500 kg down a straight hill inclined at an angle of 5° to the horizontal. The car and trailer are connected by a light rigid tow-bar. At the top of the hill the speed of the car and trailer is 20 m s−1 and at the bottom of the hill their speed is 30 m s−1.
(a) It is given that as the car and trailer descend the hill, the engine of the car does 150,000 J of work, and there are no resistance forces. Find the length of the hill.
(b) It is given instead that there is a resistance force of 100 N on the trailer, the length of the hill is 200 m, and the acceleration of the car and trailer is constant. Find the tension in the tow-bar between the car and trailer.
▶️Answer/Explanation
(a)
Change in KE: \(\frac{1}{2} \times 1900 \times (30^2 – 20^2) = 475,000 \, \text{J}\)
Loss in PE: \(1900 \times 10 \times s \times \sin 5^\circ = 1655.95s \, \text{J}\)
Energy equation: \(1655.95s + 150,000 = 475,000\)
\(s = 196 \, \text{m}\)
(b)
Using \(v^2 = u^2 + 2as\):
\(30^2 = 20^2 + 2a \times 200\) → \(a = 1.25 \, \text{m s}^{-2}\)
For trailer: \(T + 500g \sin 5^\circ – 100 = 500 \times 1.25\)
\(T = 289 \, \text{N}\)
6. [Maximum mark: 10]
A particle moves in a straight line and passes through the point A at time \(t = 0\). The velocity of the particle at time \(t\) s after leaving A is \(v \, \text{m s}^{-1}\), where \(v = 2t^2 – 5t + 3\).
(a) Find the times at which the particle is instantaneously at rest. Hence or otherwise find the minimum velocity of the particle.
(b) Sketch the velocity-time graph for the first 3 seconds of motion.
(c) Find the distance travelled between the two times when the particle is instantaneously at rest.
▶️Answer/Explanation
(a)
Solving \(v = 0\):
\(2t^2 – 5t + 3 = 0\) → \((2t – 3)(t – 1) = 0\)
\(t = 1 \, \text{s}\) or \(t = 1.5 \, \text{s}\)
Minimum velocity at \(t = 1.25 \, \text{s}\):
\(v = 2(1.25)^2 – 5(1.25) + 3 = -0.125 \, \text{m s}^{-1}\)
(b)
Graph should show:
– Parabola opening upwards
– Roots at \(t = 1\) and \(t = 1.5\)
– Minimum at \((1.25, -0.125)\)
– Points at \((0, 3)\) and \((3, 6)\)
(c)
Distance = \(\int_{1}^{1.5} (2t^2 – 5t + 3) dt = \left[\frac{2}{3}t^3 – \frac{5}{2}t^2 + 3t\right]_{1}^{1.5} = 0.0417 \, \text{m}\)
7. [Maximum mark: 10]
A particle P of mass 0.3 kg rests on a rough plane inclined at an angle θ to the horizontal, where \(\sin \theta = \frac{7}{25}\). A horizontal force of magnitude 4 N, acting in the vertical plane containing a line of greatest slope of the plane, is applied to P. The particle is on the point of sliding up the plane.
(a) Show that the coefficient of friction between the particle and the plane is \(\frac{3}{4}\).
(b) The force acting horizontally is replaced by a force of magnitude 4 N acting up the plane parallel to a line of greatest slope. Find the acceleration of P.
(c) Starting with P at rest, the force of 4 N parallel to the plane acts for 3 seconds and is then removed. Find the total distance travelled until P comes to instantaneous rest.
▶️Answer/Explanation
(a)
Resolving perpendicular: \(R = 0.3g \cos \theta + 4 \sin \theta = 4 \, \text{N}\)
Resolving parallel: \(F = 4 \cos \theta – 0.3g \sin \theta = 3 \, \text{N}\)
\(\mu = \frac{F}{R} = \frac{3}{4}\) (shown)
(b)
Using Newton’s second law:
\(4 – \mu R – 0.3g \sin \theta = 0.3a\)
\(4 – 2.16 – 0.84 = 0.3a\) → \(a = \frac{10}{3} \, \text{m s}^{-2}\)
(c)
Distance in first 3s: \(s_1 = \frac{1}{2} \times \frac{10}{3} \times 3^2 = 15 \, \text{m}\)
Final velocity: \(v = \frac{10}{3} \times 3 = 10 \, \text{m s}^{-1}\)
Deceleration after force removed: \(a = -10 \, \text{m s}^{-2}\)
Distance to stop: \(s_2 = \frac{0 – 10^2}{2 \times (-10)} = 5 \, \text{m}\)
Total distance: \(15 + 5 = 20 \, \text{m}\)