1. [Maximum mark: 6]
A car starts from rest and moves in a straight line with constant acceleration for a distance of 200 m, reaching a speed of 25 m s−1. The car then travels at this speed for 400 m, before decelerating uniformly to rest over a period of 5 s.
(a) Find the time for which the car is accelerating.
(b) Sketch the velocity–time graph for the motion of the car, showing the key points.
(c) Find the average speed of the car during its motion.
▶️Answer/Explanation
(a) Using \( v = u + at \) and \( s = ut + \frac{1}{2}at^2 \):
\( 25 = 0 + a t_1 \) and \( 200 = 0 + \frac{1}{2} a t_1^2 \).
Solving gives \( t_1 = 16 \, \text{s} \).
(b) The graph should show:
– A straight line from (0,0) to (16,25) for acceleration.
– A horizontal line from (16,25) to (32,25) for constant speed.
– A straight line from (32,25) to (37,0) for deceleration.
(c) Total distance = 200 + 400 + 62.5 = 662.5 m.
Total time = 16 + 16 + 5 = 37 s.
Average speed = \( \frac{662.5}{37} \approx 17.9 \, \text{m s}^{-1} \).
2. [Maximum mark: 5]
Two particles P and Q, of masses 0.5 kg and 0.3 kg respectively, are connected by a light inextensible string. The string is taut and P is vertically above Q. A force of magnitude 10 N is applied to P vertically upwards.
Find the acceleration of the particles and the tension in the string connecting them.
▶️Answer/Explanation
Solution:
For the system: \( 10 – (0.5 + 0.3)g = (0.5 + 0.3)a \).
\( 10 – 8 = 0.8a \) → \( a = 2.5 \, \text{m s}^{-2} \).
For Q: \( T – 0.3g = 0.3a \) → \( T = 0.3(10 + 2.5) = 3.75 \, \text{N} \).
Answer: Acceleration = \( 2.5 \, \text{m s}^{-2} \), Tension = \( 3.75 \, \text{N} \).
3. [Maximum mark: 5]
A crate of mass 300 kg is at rest on rough horizontal ground. The coefficient of friction between the crate and the ground is 0.5. A force of magnitude \( X \) N, acting at an angle \( \theta \) above the horizontal, is applied to the crate, where \( \sin \theta = 0.28 \).
Find the greatest value of \( X \) for which the crate remains at rest.
▶️Answer/Explanation
Solution:
Resolve vertically: \( R + X \sin \theta = 300g \).
Horizontally: \( X \cos \theta \leq \mu R \).
\( \cos \theta = \sqrt{1 – 0.28^2} = 0.96 \).
Substitute \( R = 3000 – 0.28X \):
\( 0.96X \leq 0.5(3000 – 0.28X) \).
Solve for \( X \): \( X \leq 1384.62 \, \text{N} \).
Answer: Greatest \( X \approx 1380 \, \text{N} \) (3 s.f.).
4. [Maximum mark: 6]
Three coplanar forces of magnitudes 20 N, 100 N, and \( F \) N act at a point. The directions of these forces are shown in the diagram. Given that the three forces are in equilibrium, find \( F \) and \( \theta \).
▶️Answer/Explanation
Official Mark Scheme Solution (Alternative Method):
1. Resolving forces in alternative directions:
Using the angle \( \alpha \) between the 100 N force and the resultant:
Equation 1 (Parallel to 100 N):
\( F \sin(\alpha + 20^\circ) + 20 \sin 20^\circ – 100 = 0 \)
\( F \sin(\alpha + 20^\circ) = 93.159 \)
Equation 2 (Perpendicular to 100 N):
\( F \cos(\alpha + 20^\circ) – 20 \cos 20^\circ = 0 \)
\( F \cos(\alpha + 20^\circ) = 18.793 \)
2. Solve for \( F \):
\( F = \sqrt{93.159^2 + 18.793^2} \)
\( F = 95.0 \, \text{N} \)
3. Solve for \( \alpha \):
\( \alpha = \tan^{-1}\left(\frac{93.159}{18.793}\right) – 20^\circ \)
\( \alpha = 58.6^\circ \)
Final Answer:
\( F = 95.0 \, \text{N} \) (3 s.f.), \( \theta = 58.6^\circ \) (1 d.p.)
5. [Maximum mark: 9]
Two racing cars A and B are at rest alongside each other at a point O on a straight horizontal test track. The mass of A is 1200 kg. The engine of A produces a constant driving force of 4500 N. When A arrives at a point P its speed is 25 m s−1. The distance OP is \( d \) m. The work done against the resistance force experienced by A between O and P is 75000 J.
(a) Show that \( d = 100 \).
(b) Find the mass of B.
(c) Find the steady speed which B can maintain when its engine is working at the same rate as it is at P.
▶️Answer/Explanation
(a) Net work done = KE gain:
\( 4500d – 75000 = \frac{1}{2} \times 1200 \times 25^2 \).
Solve for \( d \): \( d = 100 \, \text{m} \).
(b) For B: \( 3200 – 1200 = m_B a \). Time \( t = \frac{100}{12.5} = 8 \, \text{s} \).
\( a = \frac{25}{8} \). Solve for \( m_B \): \( m_B = 640 \, \text{kg} \).
(c) Power = \( 3200 \times 25 \). At steady speed, \( 3200 = 1200 \).
Thus, \( v = \frac{3200}{1200} \times 25 \approx 66.7 \, \text{m s}^{-1} \).
Answers: (b) 640 kg, (c) 66.7 m s−1.
6. [Maximum mark: 10]
A particle starts from a point O and moves in a straight line. The velocity \( v \) m s−1 of the particle at time \( t \) s after leaving O is given by \( v = k(3t^2 – 2t^3) \), where \( k \) is a constant.
(a) Verify that the particle returns to O when \( t = 2 \).
(b) Find \( k \) and hence find the total distance travelled in the first two seconds of motion.
▶️Answer/Explanation
(a) Integrate \( v \) to find displacement:
\( s = k(t^3 – \frac{1}{2}t^4) \). At \( t = 2 \), \( s = 0 \).
(b) Find \( t \) when \( v = 0 \): \( t = 1.5 \).
Given \( a = -13.5 \) at \( t = 1.5 \), differentiate \( v \):
\( a = k(6t – 6t^2) \). Solve for \( k = 3 \).
Total distance = \( |s(1.5)| + |s(2) – s(1.5)| = 3.375 + 3.375 = 6.75 \, \text{m} \).
Answers: \( k = 3 \), Distance = 6.75 m.
7. [Maximum mark: 9]
Two particles A and B, of masses 0.4 kg and 0.2 kg respectively, are moving down the same line of greatest slope of a smooth plane inclined at 30° to the horizontal. When the particles collide, the speeds of A and B are 3 m s−1 and 2 m s−1 respectively. In the collision, the speed of A is reduced to 2.5 m s−1.
(a) Find the speed of B immediately after the collision.
(b) Show that the speed of B immediately after it hits the barrier is 0.5 m s−1. Hence find the speed of the combined particle immediately after the second collision.
▶️Answer/Explanation
(a) Conservation of momentum:
\( 0.4 \times 3 + 0.2 \times 2 = 0.4 \times 2.5 + 0.2 \times v_B \).
Solve for \( v_B = 3 \, \text{m s}^{-1} \).
(b) After barrier, speed reduced by 90%: \( 3 \times 0.1 = 0.3 \, \text{m s}^{-1} \).
Combined mass after second collision: \( 0.6 \, \text{kg} \).
Momentum conservation: \( 0.4 \times 2.5 + 0.2 \times (-0.3) = 0.6 v \).
Solve for \( v \approx 1.57 \, \text{m s}^{-1} \).
Answers: (a) 3 m s−1, (b) 1.57 m s−1.