1. [Maximum mark: 4]
A crane is used to raise a block of mass 600 kg vertically upwards at a constant speed through a height of 15 m. There is a resistance to the motion of the block, which the crane does 10 000 J of work to overcome.
(a) Find the total work done by the crane.
(b) Given that the average power exerted by the crane is 12.5 kW, find the total time for which the block is in motion.
▶️Answer/Explanation
(a) \[ \text{Potential energy} = 600 \times 10 \times 15 = 90,000 \, \text{J} \] \[ \text{Total work done} = 90,000 + 10,000 = 100,000 \, \text{J} \]
(b) \[ \text{Work done} = \text{Power} \times \text{Time} \] \[ 100,000 = 12,500 \times t \] \[ t = 8 \, \text{s} \]
2. [Maximum mark: 5]
A particle \( P \) is projected vertically upwards from horizontal ground with speed \( u \, \text{ms}^{-1} \). \( P \) reaches a maximum height of 20 m above the ground.
(a) Find the value of \( u \).
(b) Find the total time for which \( P \) is at least 15 m above the ground.
▶️Answer/Explanation
(a) \[ 0 = u^2 – 2 \times 10 \times 20 \] \[ u = 20 \, \text{ms}^{-1} \]
(b) \[ 15 = 20t – \frac{1}{2} \times 10 \times t^2 \] \[ t = 1 \, \text{s or} \, t = 3 \, \text{s} \] \[ \text{Total time} = 3 – 1 = 2 \, \text{s} \]
3. [Maximum mark: 5]
A car of mass \( m \, \text{kg} \) is towing a trailer of mass 300 kg down a straight hill inclined at \( 3^\circ \) to the horizontal at a constant speed. There are resistance forces on the car and on the trailer, and the total work done against the resistance forces in a distance of 50 m is 40,000 J. The engine of the car is doing no work and the tow-bar is light and rigid.
(a) Find the value of \( m \).
(b) The resistance force on the trailer is 200 N. Find the tension in the tow-bar between the car and the trailer.
▶️Answer/Explanation
(a) \[ \text{PE lost} = (m + 300) \times 10 \times 50 \sin 3^\circ \] \[ (m + 300) \times 10 \times 50 \sin 3^\circ – 40,000 = 0 \] \[ m = 1230 \, \text{kg (to 3 sf)} \]
(b) \[ T + 300 \times 10 \sin 3^\circ – 200 = 0 \] \[ T = 43.0 \, \text{N (to 3 sf)} \]
4. [Maximum mark: 6]
The total mass of a cyclist and her bicycle is 70 kg. The cyclist is riding with constant power of 180 W up a straight hill inclined at an angle \( \theta \) to the horizontal, where \( \sin \theta = 0.05 \). At an instant when the cyclist’s speed is 6 ms\(^{-1}\), her acceleration is \(-0.2 \, \text{ms}^{-2}\). There is a constant resistance to motion of magnitude \( F \, \text{N} \).
(a) Find the value of \( F \).
(b) Find the steady speed that the cyclist could maintain up the hill when working at this power.
▶️Answer/Explanation
(a) \[ \text{Driving force} = \frac{180}{6} = 30 \, \text{N} \] \[ 30 – F – 70 \times 10 \times 0.05 = 70 \times (-0.2) \] \[ F = 9 \, \text{N} \]
(b) \[ \frac{180}{v} – 9 – 70 \times 10 \times 0.05 = 0 \] \[ v = 4.09 \, \text{ms}^{-1} \]
5. [Maximum mark: 7]
Four coplanar forces act at a point. The magnitudes of the forces are 10 N, \( F \, \text{N} \), \( G \, \text{N} \), and \( 2F \, \text{N} \). The directions of the forces are as shown in the diagram.
(a) Given that the forces are in equilibrium, find the values of \( F \) and \( G \).
(b) Given instead that \( F = 3 \), find the value of \( G \) for which the resultant of the forces is perpendicular to the 10 N force.
▶️Answer/Explanation
(a) Resolving vertically: \[ G \sin 60^\circ + 2F \sin 40^\circ – 10 = 0 \] Resolving horizontally: \[ F + G \cos 60^\circ – 2F \cos 40^\circ = 0 \] Solving simultaneously: \[ F = 4.53, \, G = 4.82 \]
(b) \[ G \sin 60^\circ + 2 \times 3 \sin 40^\circ – 10 = 0 \] \[ G = 7.09 \, \text{(to 3 sf)} \]
6. [Maximum mark: 11]
A cyclist starts from rest at a fixed point \( O \) and moves in a straight line, before coming to rest \( k \) seconds later. The acceleration of the cyclist at time \( t \) s after leaving \( O \) is \( a \, \text{ms}^{-2} \), where \( a = 2t^{-\frac{1}{2}} – \frac{3}{5}t^{\frac{1}{2}} \) for \( 0 < t \leq k \).
(a) Find the value of \( k \).
(b) Find the maximum speed of the cyclist.
(c) Find an expression for the displacement from \( O \) in terms of \( t \). Hence find the total distance travelled by the cyclist from the time at which she reaches her maximum speed until she comes to rest.
▶️Answer/Explanation
(a) Integrate \( a \) to find \( v \): \[ v = 4t^{\frac{1}{2}} – \frac{2}{5}t^{\frac{3}{2}} \] Set \( v = 0 \) and solve: \[ k = 10 \, \text{s} \]
(b) Set \( a = 0 \) to find \( t \): \[ t = \frac{10}{3} \, \text{s} \] Substitute into \( v \): \[ \text{Maximum speed} = 4.87 \, \text{ms}^{-1} \]
(c) Integrate \( v \) to find \( s \): \[ s = \frac{8}{3}t^{\frac{3}{2}} – \frac{4}{25}t^{\frac{5}{2}} \] Distance from \( t = \frac{10}{3} \) to \( t = 10 \): \[ \text{Distance} = 20.7 \, \text{m} \]
7. [Maximum mark: 12]
A bead, \( A \), of mass 0.1 kg is threaded on a long straight rigid wire which is inclined at \( \sin^{-1}\left(\frac{7}{25}\right) \) to the horizontal. \( A \) is released from rest and moves down the wire. The coefficient of friction between \( A \) and the wire is \( \mu \). When \( A \) has travelled 0.45 m down the wire, its speed is 0.6 ms\(^{-1}\).
(a) Show that \( \mu = 0.25 \).
(b) Another bead, \( B \), of mass 0.5 kg is also threaded on the wire. At the point where \( A \) has travelled 0.45 m down the wire, it hits \( B \) which is instantaneously at rest on the wire. \( A \) is brought to instantaneous rest in the collision. The coefficient of friction between \( B \) and the wire is 0.275. Find the time from when the collision occurs until \( A \) collides with \( B \) again.
▶️Answer/Explanation
(a) Use \( v^2 = u^2 + 2as \): \[ 0.6^2 = 0 + 2a \times 0.45 \] \[ a = 0.4 \, \text{ms}^{-2} \] Resolve forces: \[ 0.1 \times 10 \times \frac{7}{25} – \mu \times 0.1 \times 10 \times \frac{24}{25} = 0.1 \times 0.4 \] \[ \mu = 0.25 \]
(b) Conservation of momentum: \[ 0.1 \times 0.6 = 0.5v \] \[ v = 0.12 \, \text{ms}^{-1} \] For \( B \): \[ 0.5 \times 10 \times \frac{7}{25} – 0.275 \times 0.5 \times 10 \times \frac{24}{25} = 0.5a \] \[ a = 0.16 \, \text{ms}^{-2} \] Solve \( s_A = s_B \): \[ t = 1 \, \text{s} \]