1. [Maximum mark: 3]
A particle of mass 0.6 kg is projected with a speed of 4 m s−1 down a line of greatest slope of a smooth plane inclined at 10° to the horizontal. Use an energy method to find the speed of the particle after it has moved 15 m down the plane.
▶️Answer/Explanation
Solution:
Initial KE = \(\frac{1}{2} \times 0.6 \times 4^2 = 4.8\) J
Final KE = \(\frac{1}{2} \times 0.6 \times v^2\)
PE loss = \(0.6 \times g \times 15 \sin 10° = 15.628\) J
Using energy conservation:
\(15.628 + 4.8 = \frac{1}{2} \times 0.6 \times v^2\)
Solving for \(v\):
\(v = 8.25 \, \text{m s}^{-1}\)
2. [Maximum mark: 6]
Coplanar forces of magnitudes 34 N, 30 N, and 26 N act at a point in the directions shown in the diagram. Given that \(\sin \alpha = \frac{5}{13}\) and \(\sin \beta = \frac{8}{17}\), find the magnitude and direction of the resultant of the three forces.
▶️Answer/Explanation
Solution:
Resolve horizontally:
\(X = 30 – 34 \times \frac{8}{17} – 26 \times \frac{5}{13} = 4\) N
Resolve vertically:
\(Y = 34 \times \frac{15}{17} – 26 \times \frac{12}{13} = 6\) N
$[R] = \sqrt{X^2 + Y^2}$
$[\beta] = \tan^{-1} \left( \frac{Y}{X} \right) \text{ or } [\beta] = \tan^{-1} \left( \frac{X}{Y} \right)$
3. [Maximum mark: 6]
A ring of mass 0.3 kg is threaded on a horizontal rough rod. The coefficient of friction between the ring and the rod is 0.8. A force of magnitude 8 N acts on the ring at an angle of 10° above the horizontal. Find the time taken for the ring to move, from rest, 0.6 m along the rod.
▶️Answer/Explanation
Solution:
Resolve vertically:
\(R + 8 \sin 10° = 0.3g \implies R = 1.61\) N
Friction:
\(F = \mu R = 0.8 \times 1.61 = 1.29\) N
Resolve horizontally:
\(8 \cos 10° – F = 0.3a \implies a = 21.97 \, \text{m s}^{-2}\)
Using \(s = ut + \frac{1}{2}at^2\):
\(0.6 = 0 + \frac{1}{2} \times 21.97 \times t^2\)
Solving for \(t\):
\(t = 0.234\) s
4. [Maximum mark: 6]
A particle of mass 12 kg is stationary on a rough plane inclined at 25° to the horizontal. A pulling force of magnitude \(P\) N acts at an angle of 8° above the line of greatest slope. The coefficient of friction is 0.3. Find the greatest possible value of \(P\).
▶️Answer/Explanation
Solution:
Resolve parallel to the plane:
\(P \cos 8° = F + 12g \sin 25°\)
Resolve perpendicular to the plane:
\(R + P \sin 8° = 12g \cos 25°\)
Friction:
\(F = 0.3R\)
Substitute \(F\) and solve for \(P\):
\(P \cos 8° = 0.3(12g \cos 25° – P \sin 8°) + 12g \sin 25°\)
Solving numerically:
\(P = 80.8\) N
5. [Maximum mark: 11]
A car of mass 1250 kg pulls a caravan of mass 800 kg along a straight road. The resistances are 440 N (car) and 280 N (caravan). The system is connected by a light rigid tow-bar.
(a) (i) On a horizontal road at constant speed 30 m s−1, calculate the power developed by the car’s engine.
(ii) If the power is suddenly decreased by 8 kW, find the instantaneous deceleration and the tension in the tow-bar.
(b) The car and caravan now travel along a part of the road inclined at sin^−1 0.06 to the horizontal. The car and caravan travel up the incline at constant speed with the engine of the car working at 28 kW
(i)Find this constant speed.
(ii) Find the increase in the potential energy of the caravan in one minute.
▶️Answer/Explanation
Solution:
(a)(i)
Power = Total resistance force × speed = \((440 + 280) \times 30 = 21.6\) kW
(a)(ii)
New power = 21.6 kW – 8 kW = 13.6 kW
New driving force = \(\frac{13600}{30} = 453.33\) N
Deceleration:
\(453.33 – 720 = 2050a \implies a = -0.13 \, \text{m s}^{-2}\)
Tension:
\(T – 280 = 800a \implies T = 176\) N
(b)(i)
Driving force = Total resistance + Component of weight = \(720 + 2050g \times 0.06 = 1950\) N
Speed = \(\frac{28000}{1950} = 14.4 \, \text{m s}^{-1}\)
(b)(ii)
Height gained in 60 s = \(14.4 \times 60 \times 0.06 = 51.84\) m
PE increase = \(800g \times 51.84 = 414,000\) J (414 kJ)
6. [Maximum mark: 8]
Particle A is projected vertically upwards from the ground at 30 m s−1. Simultaneously, particle B is released from rest 15 m above A. The masses are in a 2:1 ratio. They collide and coalesce. Find the difference between the two possible times at which the combined particle C hits the ground.
▶️Answer/Explanation
Solution:
Time to collision:
\(s_A + s_B = 15 \implies 30t – 5t^2 + 5t^2 = 15 \implies t = 0.5\) s
Velocities at collision:
\(v_A = 30 – 10 \times 0.5 = 25\) m s−1 (upwards)
\(v_B = -10 \times 0.5 = -5\) m s−1 (downwards)
Case 1: \(m_A = 2m\), \(m_B = m\)
Momentum: \(2m \times 25 + m \times (-5) = 3m v \implies v = 15\) m s−1
Time to ground: Solve \(-13.75 = 15t – 5t^2\) → \(t = 3.74\) s
Case 2: \(m_A = m\), \(m_B = 2m\)
Momentum: \(m \times 25 + 2m \times (-5) = 3m v \implies v = 5\) m s−1
Time to ground: Solve \(-13.75 = 5t – 5t^2\) → \(t = 2.23\) s
Difference: \(3.74 – 2.23 = 1.50\) s
7. [Maximum mark: 10]
A particle P moving in a straight line starts from rest at a point O and comes to rest 16 s later . Its acceleration \(a \, \text{m s}^{-2}\) at time \(t\) s is given by:
- \(a = 6 + 4t\) for \(0 \leq t < 2\)
- \(a = 14\) for \(2 \leq t < 4\)
- \(a = 16 – 2t\) for \(4 \leq t \leq 16\)
There is no sudden change in velocity at any instant
(a) Find the times when the velocity is 55 m s−1.
(b) Complete the sketch of the velocity-time diagram.
(c) Find the distance travelled by P when it is decelerating..
▶️Answer/Explanation
Solution:
(a)
Integrate acceleration to find velocity:
Stage 1: \(v = 6t + 2t^2\) → At \(t = 2\), \(v = 20\) m s−1
Stage 2: \(v = 14t – 8\) → At \(t = 4\), \(v = 48\) m s−1
Stage 3: \(v = 16t – t^2\) → Solve \(55 = 16t – t^2\) → \(t = 5\) s and \(t = 11\) s
(b)
Graph: Positive quadratic (0–2 s), linear (2–4 s), negative quadratic (4–16 s) with maximum at \(t = 8\) s.
(c)
Deceleration occurs for \(8 \leq t \leq 16\). Integrate \(v\):
Distance = \(\int_{8}^{16} (16t – t^2) \, dt = 341.\overline{3}\) m