1. [Maximum mark: 4]
A block of mass 15 kg slides down a line of greatest slope of an inclined plane. The top of the plane is at a vertical height of 1.6 m above the level of the bottom of the plane. The speed of the block at the top of the plane is 2 ms−1 and the speed of the block at the bottom of the plane is 4 ms−1. Find the work done against the resistance to motion of the block.
▶️Answer/Explanation
Solution:
Using the work-energy principle:
\[ \text{Initial KE} = \frac{1}{2} \times 15 \times 2^2 = 30 \, \text{J} \]
\[ \text{Final KE} = \frac{1}{2} \times 15 \times 4^2 = 120 \, \text{J} \]
\[ \text{PE lost} = 15 \times 10 \times 1.6 = 240 \, \text{J} \]
\[ \text{Work done against resistance} = \text{PE lost} + \text{Initial KE} – \text{Final KE} \]
\[ = 240 + 30 – 120 = 150 \, \text{J} \]
Work done = 150 J
2. [Maximum mark: 5]
The diagram shows a smooth ring \( R \), of mass \( m \, \text{kg} \), threaded on a light inextensible string. A horizontal force of magnitude 2 N acts on \( R \). The ends of the string are attached to fixed points \( A \) and \( B \) on a vertical wall. The part \( AR \) of the string makes an angle of \( 30^\circ \) with the vertical, the part \( BR \) makes an angle of \( 40^\circ \) with the vertical and the string is taut. The ring is in equilibrium. Find the tension in the string and find the value of \( m \).
▶️Answer/Explanation
Solution:
Resolving horizontally:
\[ T \sin 30^\circ + T \sin 40^\circ = 2 \]
\[ T (\sin 30^\circ + \sin 40^\circ) = 2 \implies T = \frac{2}{\sin 30^\circ + \sin 40^\circ} \approx 1.75 \, \text{N} \]
Resolving vertically:
\[ T \cos 30^\circ = T \cos 40^\circ + mg \]
\[ mg = T (\cos 30^\circ – \cos 40^\circ) \implies m = \frac{1.75 (\cos 30^\circ – \cos 40^\circ)}{10} \approx 0.0175 \, \text{kg} \]
Tension = 1.75 N, Mass \( m = 0.0175 \, \text{kg} \)
3. [Maximum mark: 5]
A block of mass 10 kg is at rest on a rough plane inclined at an angle of \( 30^\circ \) to the horizontal. A force of 120 N is applied to the block at an angle of \( 20^\circ \) above a line of greatest slope. There is a force resisting the motion of the block, and 200 J of work is done against this force when the block has moved a distance of 5 m up the plane from rest. Find the speed of the block when it has moved a distance of 5 m up the plane from rest.
▶️Answer/Explanation
Solution:
Work done by applied force:
\[ 120 \times 5 \times \cos 20^\circ \approx 563.8 \, \text{J} \]
Work done against resistance = 200 J
PE gained:
\[ 10 \times 10 \times 5 \times \sin 30^\circ = 250 \, \text{J} \]
Using the work-energy principle:
\[ 563.8 – 250 – 200 = \frac{1}{2} \times 10 \times v^2 \]
\[ 113.8 = 5v^2 \implies v = \sqrt{22.76} \approx 4.77 \, \text{ms}^{-1} \]
Speed = 4.77 ms−1
4. [Maximum mark: 7]
A particle \( P \) of mass 0.2 kg lies at rest on a rough horizontal plane. A horizontal force of 1.2 N is applied to \( P \).
(a) Given that \( P \) is in limiting equilibrium, find the coefficient of friction between \( P \) and the plane.
(b) Given instead that the coefficient of friction is 0.3, find the distance travelled by \( P \) in the third second of its motion.
▶️Answer/Explanation
(a) Solution:
In limiting equilibrium:
\[ F = \mu R \implies 1.2 = \mu \times 0.2 \times 10 \implies \mu = \frac{1.2}{2} = 0.6 \]
Coefficient of friction = 0.6
(b) Solution:
Net force:
\[ 1.2 – 0.3 \times 0.2 \times 10 = 0.6 \, \text{N} \]
Acceleration:
\[ a = \frac{0.6}{0.2} = 3 \, \text{ms}^{-2} \]
Distance in 3 seconds:
\[ s_3 = \frac{1}{2} \times 3 \times 3^2 = 13.5 \, \text{m} \]
Distance in 2 seconds:
\[ s_2 = \frac{1}{2} \times 3 \times 2^2 = 6 \, \text{m} \]
Distance in third second:
\[ s_3 – s_2 = 7.5 \, \text{m} \]
Distance = 7.5 m
5. [Maximum mark: 8]
A particle \( A \) of mass 0.5 kg is projected vertically upwards from horizontal ground with speed 25 ms−1.
(a) Find the speed of \( A \) when it reaches a height of 20 m above the ground.
(b) When \( A \) reaches a height of 20 m, it collides with a particle \( B \) of mass 0.3 kg moving downwards with speed 32.5 ms−1. In the collision, \( B \) is brought to instantaneous rest. Show that the velocity of \( A \) immediately after the collision is 4.5 ms−1 downwards.
(c) Find the time interval between \( A \) and \( B \) reaching the ground.
▶️Answer/Explanation
(a) Solution:
Using \( v^2 = u^2 + 2as \):
\[ v^2 = 25^2 – 2 \times 10 \times 20 = 225 \implies v = 15 \, \text{ms}^{-1} \]
Speed = 15 ms−1
(b) Solution:
Using conservation of momentum (upwards positive):
\[ 0.5 \times 15 + 0.3 \times (-32.5) = 0.5 \times v + 0.3 \times 0 \]
\[ 7.5 – 9.75 = 0.5v \implies v = -4.5 \, \text{ms}^{-1} \]
Velocity = 4.5 ms−1 downwards
(c) Solution:
For \( B \):
\[ 20 = \frac{1}{2} \times 10 \times t_B^2 \implies t_B = 2 \, \text{s} \]
For \( A \) (downwards positive):
\[ 20 = 4.5 t_A + \frac{1}{2} \times 10 \times t_A^2 \implies t_A = 1.6 \, \text{s} \]
Time interval:
\[ 2 – 1.6 = 0.4 \, \text{s} \]
Time interval = 0.4 s
6. [Maximum mark: 9]
A railway engine of mass 120,000 kg is towing a coach of mass 60,000 kg up a straight track inclined at an angle \( \theta \) to the horizontal where \( \sin \theta = 0.02 \). There is a light rigid coupling connecting the engine and coach. The driving force produced by the engine is 125,000 N, and there are constant resistances to motion of 22,000 N on the engine and 13,000 N on the coach.
(a) Find the acceleration of the engine and the tension in the coupling.
(b) At an instant when the engine is travelling at 30 ms−1, it comes to a section of track inclined upwards at an angle \( \phi \) to the horizontal. The power produced by the engine is now 4,500,000 W, and the engine maintains a constant speed. Assuming the resistance forces remain unchanged, find the value of \( \phi \).
▶️Answer/Explanation
(a) Solution:
For the system (engine + coach):
\[ 125000 – (120000 + 60000) \times 10 \times 0.02 – 22000 – 13000 = (120000 + 60000)a \]
\[ 125000 – 36000 – 22000 – 13000 = 180000a \implies a = \frac{54000}{180000} = 0.3 \, \text{ms}^{-2} \]
For the coach:
\[ T – 60000 \times 10 \times 0.02 – 13000 = 60000 \times 0.3 \implies T = 43000 \, \text{N} \]
Acceleration = 0.3 ms−2, Tension = 43,000 N
(b) Solution:
Driving force at constant speed:
\[ F = \frac{4500000}{30} = 150000 \, \text{N} \]
Resolving parallel to the track:
\[ 150000 – (120000 + 60000) \times 10 \times \sin \phi – 22000 – 13000 = 0 \]
\[ 150000 – 1800000 \sin \phi – 35000 = 0 \implies \sin \phi = \frac{115000}{1800000} \approx 0.0639 \implies \phi \approx 3.7^\circ \]
Angle \( \phi = 3.7^\circ \)
7. [Maximum mark: 12]
A particle \( X \) travels in a straight line. The velocity of \( X \) at time \( t \) seconds after leaving a fixed point \( O \) is denoted by \( v \, \text{ms}^{-1} \), where \( v = -0.1t^3 + 1.8t^2 – 6t + 5.6 \). The acceleration of \( X \) is zero at \( t = p \) and \( t = q \), where \( p < q \).
(a) Find the value of \( p \) and the value of \( q \).
(b) Find the velocities of \( X \) at \( t = p \) and \( t = q \), and hence sketch the velocity-time graph for the motion of \( X \) for \( 0 \leq t \leq 15 \).
(c) Find the total distance travelled by \( X \) between \( t = 0 \) and \( t = 15 \).
▶️Answer/Explanation
(a) Solution:
Differentiating \( v \) to find \( a \):
\[ a = \frac{dv}{dt} = -0.3t^2 + 3.6t – 6 \]
Setting \( a = 0 \):
\[ -0.3t^2 + 3.6t – 6 = 0 \implies t^2 – 12t + 20 = 0 \implies t = 2 \, \text{or} \, 10 \]
\( p = 2 \), \( q = 10 \)
(b) Solution:
At \( t = 2 \):
\[ v = -0.1(8) + 1.8(4) – 6(2) + 5.6 = 0 \, \text{ms}^{-1} \]
At \( t = 10 \):
\[ v = -0.1(1000) + 1.8(100) – 6(10) + 5.6 = 25.6 \, \text{ms}^{-1} \]
Graph: A curve with a minimum at \( t = 2 \) and a maximum at \( t = 10 \), crossing the \( t \)-axis at \( t = 14 \).
(c) Solution:
Integrating \( v \) to find distance:
\[ s = \int_{0}^{14} v \, dt = 176.4 \, \text{m} \]
\[ \int_{14}^{15} v \, dt = -8.025 \, \text{m} \, (\text{distance} = 8.025 \, \text{m}) \]
Total distance:
\[ 176.4 + 8.025 = 184.425 \, \text{m} \]
Total distance = 184.4 m