1. [Maximum mark: 3]
A particle is projected vertically upwards from horizontal ground with a speed of \( u \, \text{ms}^{-1} \). The particle has height \( s \) above the ground at times 3 seconds and 4 seconds after projection. Find the value of \( u \) and the value of \( s \).
▶️Answer/Explanation
Solution:
Using the equation of motion \( s = ut – \frac{1}{2}gt^2 \), where \( g = 10 \, \text{ms}^{-2} \):
At \( t = 3 \): \( s = 3u – 5 \times 9 \) → \( s = 3u – 45 \) (1)
At \( t = 4 \): \( s = 4u – 5 \times 16 \) → \( s = 4u – 80 \) (2)
Equating (1) and (2): \( 3u – 45 = 4u – 80 \) → \( u = 35 \).
Substituting \( u = 35 \) into (1): \( s = 3(35) – 45 = 60 \).
Answer: \( u = 35 \), \( s = 60 \).
2. [Maximum mark: 5]
A machine for driving a nail into a block of wood causes a hammerhead to drop vertically onto the top of a nail. The mass of the hammerhead is 1.2 kg and the mass of the nail is 0.004 kg. The hammerhead hits the nail with speed \( v \, \text{ms}^{-1} \) and remains in contact with the nail after the impact. The combined hammerhead and nail move immediately after the impact with speed \( 40 \, \text{ms}^{-1} \).
(a) Calculate \( v \), giving your answer as an exact fraction. [2]
(b) The nail is driven 4 cm into the wood. Find the constant force resisting the motion. [3]
▶️Answer/Explanation
(a) Solution:
Using conservation of momentum:
\( 1.2v = (1.2 + 0.004) \times 40 \) → \( v = \frac{1.204 \times 40}{1.2} = \frac{602}{15} \).
Answer: \( v = \frac{602}{15} \).
(b) Solution:
Using \( v^2 = u^2 + 2as \):
\( 0 = 40^2 + 2 \times 0.04 \times a \) → \( a = -20000 \, \text{ms}^{-2} \).
Using Newton’s Second Law:
\( -R + (1.2 + 0.004)g = (1.2 + 0.004) \times (-20000) \).
Solving: \( R = 24100 \, \text{N} \).
Answer: \( R = 24100 \, \text{N} \).
3. [Maximum mark: 6]
A block of mass 8 kg slides down a rough plane inclined at 30° to the horizontal, starting from rest. The coefficient of friction between the block and the plane is \( \mu \). The block accelerates uniformly down the plane at \( 2.4 \, \text{ms}^{-2} \).
(a) Draw a diagram showing the forces acting on the block. [1]
(b) Find the value of \( \mu \). [4]
(c) Find the speed of the block after it has moved 3 m down the plane. [1]
▶️Answer/Explanation
(a) Solution:
The diagram should show:
– Weight \( 8g \) vertically downward.
– Normal reaction \( R \) perpendicular to the plane.
– Frictional force \( F \) opposing motion up the plane.
(b) Solution:
Resolving perpendicular to the plane: \( R = 8g \cos 30° = 40\sqrt{3} \).
Resolving parallel to the plane: \( 8g \sin 30° – F = 8 \times 2.4 \).
Using \( F = \mu R \): \( 40 – \mu \times 40\sqrt{3} = 19.2 \).
Solving: \( \mu = 0.3 \).
Answer: \( \mu = 0.3 \).
(c) Solution:
Using \( v^2 = u^2 + 2as \):
\( v^2 = 0 + 2 \times 2.4 \times 3 \) → \( v = \sqrt{14.4} = 3.79 \, \text{ms}^{-1} \).
Answer: \( v = 3.79 \, \text{ms}^{-1} \).
4. [Maximum mark: 7]
A car has mass 1600 kg.
(a) The car is moving along a straight horizontal road at a constant speed of \( 24 \, \text{ms}^{-1} \) and is subject to a constant resistance of magnitude 480 N. Find, in kW, the rate at which the engine of the car is working. [2]
(b) The car now moves down a hill inclined at an angle of \( \theta \) to the horizontal, where \( \sin \theta = 0.09 \). The engine of the car is working at a constant rate of 12 kW. The speed of the car is \( 24 \, \text{ms}^{-1} \) at the top of the hill. Ten seconds later the car has travelled 280 m down the hill and has speed \( 32 \, \text{ms}^{-1} \). Given that the resistance is not constant, use an energy method to find the total work done against the resistance during the ten seconds. [5]
▶️Answer/Explanation
(a) Solution:
Power \( P = Fv = 480 \times 24 = 11520 \, \text{W} = 11.52 \, \text{kW} \).
Answer: \( 11.52 \, \text{kW} \).
(b) Solution:
Change in KE: \( \frac{1}{2} \times 1600 \times (32^2 – 24^2) = 358400 \, \text{J} \).
Change in PE: \( 1600g \times 280 \times 0.09 = 403200 \, \text{J} \).
Work done by engine: \( 12000 \times 10 = 120000 \, \text{J} \).
Using energy conservation:
\( 120000 + 403200 = 358400 + \text{Work done against resistance} \).
Solving: Work done against resistance \( = 164800 \, \text{J} \).
Answer: \( 164800 \, \text{J} \).
5. [Maximum mark: 8]
A light string AB is fixed at A and has a particle of weight 80 N attached at B. A horizontal force of magnitude \( P \) N is applied at B such that the string makes an angle \( \theta \) to the vertical.
(a) It is given that \( P = 32 \) and the system is in equilibrium. Find the tension in the string and the value of \( \theta \). [4]
(b) It is given instead that the tension in the string is 120 N and that the particle attached at B still has weight 80 N. Find the value of \( P \) and the value of \( \theta \). [4]
▶️Answer/Explanation
(a) Solution:
Resolving horizontally: \( T \sin \theta = 32 \).
Resolving vertically: \( T \cos \theta = 80 \).
Dividing: \( \tan \theta = \frac{32}{80} \) → \( \theta = 21.8° \).
Substituting: \( T = \frac{80}{\cos 21.8°} = 86.2 \, \text{N} \).
Answer: \( T = 86.2 \, \text{N} \), \( \theta = 21.8° \).
(b) Solution:
Resolving vertically: \( 120 \cos \theta = 80 \) → \( \cos \theta = \frac{2}{3} \) → \( \theta = 48.2° \).
Resolving horizontally: \( P = 120 \sin 48.2° = 89.4 \, \text{N} \).
Answer: \( P = 89.4 \, \text{N} \), \( \theta = 48.2° \).
6. [Maximum mark: 8]
A particle moves in a straight line. At time \( t \) s, the acceleration, \( a \, \text{ms}^{-2} \), of the particle is given by \( a = 36 – 6t \). The velocity of the particle is \( 27 \, \text{ms}^{-1} \) when \( t = 2 \).
(a) Find the values of \( t \) when the particle is at instantaneous rest. [4]
(b) Find the total distance the particle travels during the first 12 seconds. [4]
▶️Answer/Explanation
(a) Solution:
Integrate \( a \) to find \( v \): \( v = 36t – 3t^2 + c \).
At \( t = 2 \), \( v = 27 \): \( 27 = 72 – 12 + c \) → \( c = -33 \).
Set \( v = 0 \): \( 36t – 3t^2 – 33 = 0 \) → \( t = 1 \) or \( t = 11 \).
Answer: \( t = 1 \), \( t = 11 \).
(b) Solution:
Integrate \( v \) to find \( s \): \( s = 18t^2 – t^3 – 33t \).
Evaluate \( s \) at \( t = 0, 1, 11, 12 \):
\( s(0) = 0 \), \( s(1) = -16 \), \( s(11) = 484 \), \( s(12) = 468 \).
Total distance \( = | -16 – 0 | + | 484 – (-16) | + | 468 – 484 | = 532 \, \text{m} \).
Answer: \( 532 \, \text{m} \).
7. [Maximum mark: 13]
Particles A and B, of masses 2.4 kg and 3.3 kg respectively, are connected by a light inextensible string that passes over a smooth pulley fixed to the top of a rough plane. The plane makes an angle of \( \theta \) with horizontal ground. Particle A is on the plane, and the string between A and the pulley is parallel to a line of greatest slope of the plane. Particle B hangs vertically below the pulley and is 1 m above the ground. The coefficient of friction between the plane and A is \( \mu \).
(a) It is given that \( \theta = 30° \) and the system is in equilibrium with A on the point of moving directly up the plane. Show that \( \mu = 1.01 \) correct to 3 significant figures. [5]
(b) It is given instead that \( \theta = 20° \) and \( \mu = 1.01 \). The system is released from rest with the string taut. Find the total distance travelled by A before coming to instantaneous rest. You may assume that A does not reach the pulley and that B remains at rest after it hits the ground. [8]
▶️Answer/Explanation
(a) Solution:
For equilibrium:
For B: \( T = 3.3g \).
For A: \( T = 2.4g \sin 30° + F \).
Using \( F = \mu R \) and \( R = 2.4g \cos 30° \):
\( 3.3g = 2.4g \times 0.5 + \mu \times 2.4g \times \cos 30° \).
Solving: \( \mu = \frac{3.3 – 1.2}{2.4 \cos 30°} = 1.01 \).
Answer: \( \mu = 1.01 \).
(b) Solution:
For motion before B hits the ground:
Using Newton’s Second Law for the system:
\( 3.3g – 2.4g \sin 20° – \mu \times 2.4g \cos 20° = (2.4 + 3.3)a \).
Solving: \( a = 0.353 \, \text{ms}^{-2} \).
Velocity when B hits the ground: \( v^2 = 2 \times 0.353 \times 1 \) → \( v = 0.841 \, \text{ms}^{-1} \).
For motion after B hits the ground:
Deceleration of A: \( a = -12.9 \, \text{ms}^{-2} \).
Distance travelled: \( s = \frac{0 – 0.841^2}{2 \times (-12.9)} = 0.0274 \, \text{m} \).
Total distance: \( 1 + 0.0274 = 1.03 \, \text{m} \).
Answer: \( 1.03 \, \text{m} \).