1. [Maximum mark: 4]
Two particles P and Q, of masses 0.3 kg and 0.2 kg respectively, are at rest on a smooth horizontal plane. P is projected at a speed of 4 ms−1 directly towards Q. After P and Q collide, Q begins to move with a speed of 3 ms−1.
(a) Find the speed of P after the collision.
(b) After the collision, Q moves directly towards a third particle R, of mass m kg, which is at rest on the plane. The two particles Q and R coalesce on impact and move with a speed of 2 ms−1. Find m.
▶️Answer/Explanation
(a) Using conservation of momentum:
0.3 × 4 + 0 = 0.3v + 0.2 × 3
Speed of P after collision = 2 ms−1.
(b) Using conservation of momentum:
0.2 × 3 + 0 = (0.2 + m) × 2
m = 0.1 kg.
2. [Maximum mark: 5]
A particle P is projected vertically upwards from horizontal ground. P reaches a maximum height of 45 m. After reaching the ground, P comes to rest without rebounding.
(a) Find the speed at which P was projected.
(b) Find the total time for which the speed of P is at least 10 ms−1.
▶️Answer/Explanation
(a) Using the equation of motion:
0 = u2 − 2 × 10 × 45
Speed = 30 ms−1.
(b) Time to reach 10 ms−1 upwards:
10 = 30 − 10t ⇒ t = 2 s.
Total time = 2 × 2 s = 4 s.
3. [Maximum mark: 5]
The displacement of a particle moving in a straight line is s metres at time t seconds after leaving a fixed point O. The particle starts from rest and passes through points P, Q, and R, at times t = 5, t = 10, and t = 15 respectively, and returns to O at time t = 20. The distances OP, OQ, and OR are 50 m, 150 m, and 200 m respectively.
(a) Find the speed of the particle between t = 5 and t = 10.
(b) Find the acceleration of the particle between t = 0 and t = 5, given that it is constant.
(c) Find the average speed of the particle during its motion.
▶️Answer/Explanation
(a) Speed = 20 ms−1.
(b) Using v = u + at:
20 = 0 + a × 5 ⇒ a = 4 ms−2.
(c) Total distance = 50 + 100 + 50 + 200 = 400 m.
Total time = 20 s.
Average speed = 20 ms−1.
4. [Maximum mark: 8]
The diagram shows a block of mass 10 kg suspended below a horizontal ceiling by two strings AC and BC, of lengths 0.8 m and 0.6 m respectively, attached to fixed points on the ceiling. Angle ACB = 90°. There is a horizontal force of magnitude F N acting on the block. The block is in equilibrium.
(a) In the case where F = 20, find the tensions in each of the strings.
(b) Find the greatest value of F for which the block remains in equilibrium in the position shown.
▶️Answer/Explanation
(a) Resolving forces:
TA × 0.8 − TB × 0.6 − 20 = 0 (horizontal)
TA × 0.6 + TB × 0.8 − 100 = 0 (vertical)
Solving gives TA = 76 N, TB = 68 N.
(b) When TB = 0:
TA × 0.6 = 100 ⇒ TA = 500/3 N.
TA × 0.8 = F ⇒ F = 400/3 N ≈ 133 N.
5. [Maximum mark: 8]
A cyclist is riding along a straight horizontal road. The total mass of the cyclist and her bicycle is 70 kg. At an instant when the cyclist’s speed is 4 ms−1, her acceleration is 0.3 ms−2. There is a constant resistance to motion of magnitude 30 N.
(a) Find the power developed by the cyclist.
(b) The cyclist comes to the top of a hill inclined at 5° to the horizontal. The cyclist stops pedalling and freewheels down the hill. The magnitude of the resistance force remains at 30 N. Over a distance of d m, the speed of the cyclist increases from 6 ms−1 to 12 ms−1.
(i) Find the change in kinetic energy.
(ii) Use an energy method to find d.
▶️Answer/Explanation
(a) Using Newton’s Second Law:
F − 30 = 70 × 0.3 ⇒ F = 51 N.
Power = F × v = 51 × 4 = 204 W.
(b)(i) Change in KE = ½ × 70 × (122 − 62) = 3780 J.
(b)(ii) Work-energy equation:
70 × 10 × d × sin5° − 30 × d = 3780 ⇒ d ≈ 122 m.
6. [Maximum mark: 10]
Two particles P and Q, of masses 0.3 kg and 0.2 kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley at B which is attached to two inclined planes. P lies on a smooth plane AB inclined at 60° to the horizontal. Q lies on a plane BC inclined at 30° to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes.
(a) It is given that the plane BC is smooth and that the particles are released from rest. Find the tension in the string and the magnitude of the acceleration of the particles.
(b) It is given instead that the plane BC is rough. A force of magnitude 3 N is applied to Q directly up the plane along a line of greatest slope of the plane. Find the least value of the coefficient of friction between Q and the plane BC for which the particles remain at rest.
▶️Answer/Explanation
(a) Using Newton’s Second Law:
For P: T − 0.3g sin60° = 0.3a.
For Q: 0.2g sin30° − T = 0.2a.
Solving gives a ≈ 7.20 ms−2, T ≈ 0.439 N.
(b) For equilibrium:
For P: T = 0.3g sin60° ≈ 2.598 N.
For Q: T + 3 − 0.2g sin30° − F = 0 ⇒ F ≈ 4.598 N.
R = 0.2g cos30° ≈ 1.732 N.
Coefficient of friction μ = F/R ≈ 0.345.
7. [Maximum mark: 10]
A particle P moves in a straight line through a point O. The velocity v ms−1 of P, at time t s after passing O, is given by v = 9/4 + b/(t + 1)2 − ct2, where b and c are positive constants. At t = 5, the velocity of P is zero and its acceleration is −13/12 ms−2.
(a) Show that b = 9 and find the value of c.
(b) Given that the velocity of P is zero only at t = 5, find the distance travelled in the first 10 seconds of motion.
▶️Answer/Explanation
(a) At t = 5, v = 0:
0 = 9/4 + b/36 − 25c.
Differentiating v gives a = −2b/(t + 1)3 − 2ct.
At t = 5, a = −13/12:
−13/12 = −2b/216 − 10c.
Solving gives b = 9, c = 0.1.
(b) Integrating v:
s = ∫(9/4 + 9/(t + 1)2 − 0.1t2) dt = 9t/4 − 9/(t + 1) − t3/30 + C.
Distance from 0 to 5: 5.583 m.
Distance from 5 to 10: 11.651 m.
Total distance ≈ 31.8 m.