CIE AS/A level Mathematics- Mechanics Paper 4 Prediction - 2025
CIE AS/A level Mathematics- Mechanics Paper 4 Prediction- 2025
To excel in A level Math Exam, consistent practice with CIE AS & A Level Revision resources is key. CIE AS/A level Mathematics- Mechanics Paper 4 Prediction will guide you for exam pattern.
IITian Academy offers a vast collection of questions that can aid your understanding of specific topics and solidify your concepts. By practicing regularly and focusing on these key areas, you’ll be well-prepared for the A level Math exam
Question 1
A block is at rest on a rough horizontal plane. The coefficient of friction between the block and the plane is 1.25.
(i) State, giving a reason for your answer, whether the minimum vertical force required to move the block is greater or less than the minimum horizontal force required to move the block.
A horizontal force of continuously increasing magnitude P N and fixed direction is applied to the block.
(ii) Given that the weight of the block is 60 N, find the value of P when the acceleration of the block is 4 m s−2.
Answer/Explanation
Ans:
(i) Less than
F = 1.25W so W< F
(ii) [P – 60 × 1.25 = 6 × 4]
P = 99
Question 2
A particle of mass 0.4 kg is released from rest at a height of 1.8 m above the surface of the water in a
tank. There is no instantaneous change of speed when the particle enters the water. The water exerts an upward force of 5.6 N on the particle when it is in the water.
(i) Find the velocity of the particle at the instant when it reaches the surface of the water.
(ii) Find the time that it takes from the instant when the particle enters the water until it comes to instantaneous rest in the water. You may assume that the tank is deep enough so that the particle does not reach the bottom of the tank.
(iii) Sketch a velocity-time graph for the motion of the particle from the instant at which it is released
until it comes to instantaneous rest in the water.
▶️Answer/Explanation
(i) Velocity just before entering the water:
The potential energy lost by the particle is converted into kinetic energy just before entering the water. The potential energy lost is given by \( mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.
\( \text{Potential energy lost} = mgh \)
This potential energy is converted into kinetic energy just before entering the water:
\( \frac{1}{2}m\nu^2 = mgh \)
Solving for \( \nu \):
\( \nu = \sqrt{2gh} \)
Substitute \( g = 9.8 \, \text{m/s}^2 \) and \( h = 1.8 \, \text{m} \):
\( \nu = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 1.8 \, \text{m}} \)
\( \nu = 6 \, \text{m/s} \)
(ii) Time taken to come to rest in the water:
Once in the water, the net force acting on the particle is the difference between the buoyant force (\(5.6 \, \text{N}\)) and its weight (\(mg\)). The net force is then used to find the deceleration.
\( F_{\text{net}} = 5.6 \, \text{N} – 0.4 \, \text{kg} \times 9.8 \, \text{m/s}^2 \)
Calculate \( F_{\text{net}} \):
\( F_{\text{net}} = -3.6 \, \text{N} \)
\( a = \frac{F_{\text{net}}}{m} = -4 \, \text{m/s}^2 \)
Now, we’ll use the kinematic equation \( \nu = u + at \), where \( u \) is the initial velocity (velocity just before entering water), \( a \) is the deceleration, and \( t \) is the time.
\( \nu = 0 \) (since the particle comes to rest)
Solve for \( t \):
\( 0 = \sqrt{2gh} + at \)
\( t = 1.5 \, \text{s} \)
(iii) Velocity-time graph:
The graph starts at the origin (0,0) with a positive gradient, indicating a linear increase in velocity during free fall. After entering the water, there’s a second straight line with a negative gradient, indicating a linear decrease in velocity until the particle comes to rest. The graph finishes at (2.1, 0), showing that the velocity becomes zero at \( t = 2.1 \) seconds.
Question 3
A block of mass 5kg is being pulled along a rough horizontal floor by a force of magnitude X N acting at \(30^o\) above the horizontal (see diagram). The block starts from rest and travels 2m in the first 5s of its motion.
(a) Find the acceleration of the block
(b) Given that the coefficient of friction between the block and the floor is 0.4, find X.
The block is now placed on a part of the floor where the coefficient of friction between the block and the floor has a different value. The value of X is changed to 25, and the block is now in limiting equilibrium.
(c) Find the value of the coefficient of friction between the block and this part of the floor.
▶️Answer/Explanation
(a) Use the kinematic equation for uniformly accelerated motion:
\( s = ut + \frac{1}{2}at^2 \)
Given that the block starts from rest (\(u = 0\)), travels \(2 \, \text{m}\), and \(t = 5 \, \text{s}\):
\(2 = \frac{1}{2}a(5^2) \)
Solve for \(a\):
\(a = \frac{2}{\frac{1}{2} \times 5^2} = \frac{2}{\frac{1}{2} \times 25} = \frac{2}{\frac{25}{2}} = \frac{4}{25} \, \text{m/s}^2 \)
\( a = 0.16 \, \text{m/s}^2 \)
(b) Finding \(X\) with Coefficient of Friction \(0.4\):
In the horizontal direction:
\( X \cos(30^\circ) – f = ma \)
The frictional force (\(f\)) is given by \(f = \mu mg\), where \(\mu = 0.4\).
\( X \cos(30^\circ) – 0.4mg = m \cdot \frac{4}{25} \)
Assuming \(g = 9.8 \, \text{m/s}^2\):
\( X \cos(30^\circ) – 0.4 \times 5 \times 9.8 = 5 \cdot \frac{4}{25} \)
Solve for \(X\):
\( X = 19.5 \, \text{N} \)
(c) Finding Coefficient of Friction with \(X = 25\):
In limiting equilibrium, the frictional force is at its maximum, so \(f_{\text{max}} = \mu_{\text{max}} mg\).
In the horizontal direction:
\( X \cos(30^\circ) – f_{\text{max}} = 0 \)
Substituting \(X = 25\) and solve for \(\mu_{\text{max}}\):
\( R = 5g – 25 \sin(30^\circ) = 37.5 \, \text{N} \)
\( F = 25 \cos(30^\circ) = \frac{25 \sqrt{3}}{2} \, \text{N} \)
\( \mu = \frac{F}{R} = \frac{\frac{25 \sqrt{3}}{2}}{37.5} = 0.577 \)
Question 4
Two particles P and Q of masses kg and kg respectively are free to move in a horizontal straight line on a smooth horizontal plane. P is projected towards Q with speed \(0.5ms^{-1}\). At the same instant Q is projected towards P with speed \(1ms^{-1}\). Q comes to rest in the resulting collision.
Find the speed of P after the collision.
▶️Answer/Explanation
The conservation of linear momentum equation is:
\( m_Pv_P + m_Qv_Q = (m_P + m_Q)v_f \)
Substitute the given values:
\( 0.2 \times 0.5 + 0.3 \times (-1) = (0.3-0.2) \times v_f \)
Now, let’s solve for \(v_f\):
\( -0.1 = 0.1 \times v_f \)
Solving for \(v_f\), we get:
\( v_f = -1 \, \text{m/s} \)
So, the correct speed of \(P\) after the collision is \(1 \, \text{m/s}\).
Question 5
A uniform solid cone has weight 5 N and base radius 0.1 m. AB is a diameter of the base of the cone.
The cone is held in equilibrium, with A in contact with a rough horizontal surface and AB vertical, by a force applied at B. This force has magnitude 3 N and acts parallel to the axis of the cone (see
diagram). Calculate the height of the cone.
▶️Answer/Explanation
Vertical forces:
In the vertical direction, the weight of the cone (\( W \)) and the normal force (\( N \)) must balance each other.
\( N = W \)
\( N = 5 \, \text{N} \)
Torque (Moment) about point A:
The force applied at B creates a torque about point A. The torque (\( \tau \)) is given by the product of the force (\( F \)) and the perpendicular distance (\( r \)) from the line of action of the force to point A. The perpendicular distance is the radius of the base (\( r \)).
\( \tau = F \cdot r \)
\( \tau = 3 \, \text{N} \cdot 0.1 \, \text{m} = 0.3 \, \text{Nm} \)
Torque due to the weight of the cone:
The weight of the cone creates a torque about point A. The weight acts vertically downward at the center of mass, which is located at \( \frac{1}{3} \) of the height of the cone (\( h \)). The perpendicular distance (\( d’ \)) from the line of action of the weight to point A is \( \frac{2}{3} \) of the height.
\( \tau’ = W \cdot d’ \)
\( \tau’ = 5 \, \text{N} \cdot \frac{2}{3}h \)
Since the cone is in equilibrium, the torques about point A must balance:
\( \tau = \tau’ \)
\( 0.3 \, \text{Nm} = 5 \, \text{N} \cdot \frac{2}{3}h \)
\( h = \frac{0.3}{5} \times \frac{3}{2} \, \text{m} \)
\( h = 0.03 \, \text{m} \)
So, the height of the cone is \(0.03 \, \text{m}\) or \(3 \, \text{cm}\).
Question 6
A tractor of mass 3700 kg is travelling along a straight horizontal road at a constant speed of 12 \(m s^{−1}\). The total resistance to motion is 1150 N.
(i) Find the power output of the tractor’s engine. The tractor comes to a hill inclined at 4Å above the horizontal. The power output is increased to 25 kW and the resistance to motion is unchanged.
(ii) Find the deceleration of the tractor at the instant it begins to climb the hill.
(iii) Find the constant speed that the tractor could maintain on the hill when working at this power.
▶️Answer/Explanation
(i) The power output of the tractor’s engine on the horizontal road is correctly calculated as:
\(P = \text{Resistance} \times \text{Velocity} = 1150 \, \text{N} \times 12 \, \text{m/s} = 13,800 \, \text{W}\)
So, the power output on the horizontal road is 13,800 W, which is equivalent to 13.8 kW.
(ii) The deceleration (negative acceleration) is correctly calculated using the work-energy principle:
\(F_{\text{driving}} = \frac{25000 \, \text{W}}{12 \, \text{m/s}}\)
\(a = \frac{F_{\text{driving}} – \text{Resistance} – \text{Weight component parallel to the hill}}{m} = \frac{25000/12 – 1150 – 3700g \sin(4^\circ)}{3700}\)
Calculating this gives \(a \approx -0.445 \, \text{m/s}^2\), indicating deceleration.
(iii) The constant speed on the hill is correctly calculated using the power formula:
\(\frac{25000 \, \text{W}}{v} = 1150 + 3700g \sin(4^\circ)\)
\(v = \frac{25000 \, \text{W}}{1150 + 3700g \sin(4^\circ)} \approx 6.70 \, \text{m/s}\)
So, the constant speed that the tractor can maintain on the hill when working at 25 kW is approximately 6.70 m/s.
Question 7
An elevator moves vertically, supported by a cable. The diagram shows a velocity-time graph which models the motion of the elevator. The graph consists of 7 straight line segments.
The elevator accelerates upwards from rest to a speed of \(2ms^{-1}\) over a period of 1.5s and then travels at this speed for before decelerating to rest over a period of 1 s.
The elevator then remains at rest for 6 s. before accelerating to a speed of \( Vms^{-1}\) downwards over a period of 2s. The elevator travels at this speed for a period of 5s, before decelerating to rest over a period of 1.5s.
(a) Find the acceleration of the elevator during the first 1.5s.
(b) Given that the elevator starts and finishes its journey on the ground floor, find V.
(c) The combined weight of the elevator and passengers on its upward journey is 1500kg. Assuming that there is no resistance to motion, find the tension in the elevator cable on its upward journey when the elevator is decelerating.
▶️Answer/Explanation
(a) Acceleration during the first \(1.5 \, \text{s}\):
The formula for acceleration is \(a = \frac{\Delta v}{\Delta t}\), where \(\Delta v\) is the change in velocity and \(\Delta t\) is the change in time.
In the given velocity-time graph, during the first \(1.5 \, \text{s}\), the elevator accelerates from rest to a speed of \(2 \, \text{m/s}\). Therefore, the change in velocity \(\Delta v\) is \(2 \, \text{m/s}\), and the change in time \(\Delta t\) is \(1.5 \, \text{s}\). Plug these values into the formula to find the acceleration \(a\):
\( a = \frac{2 \, \text{m/s}}{1.5 \, \text{s}} = \frac{4}{3} \, \text{m/s}^2 \)
(b) Find \(V\):
For this part, we consider the downward journey. The equation of motion for constant acceleration is given by:
\( s = ut + \frac{1}{2} a t^2 \)
For the first segment (acceleration), \(s_1 = \frac{1}{2} a t_1^2\) where \(a = \frac{4}{3} \, \text{m/s}^2\) and \(t_1 = 1.5 \, \text{s}\).
For the second segment (constant velocity), \(s_2 = V t_2\) where \(V\) is the constant velocity and \(t_2 = 1 \, \text{s}\).
For the third segment (deceleration), \(s_3 = \frac{1}{2} a t_3^2\) where \(a = -2 \, \text{m/s}^2\) and \(t_3 = 1.5 \, \text{s}\)
The total displacement is given by \(s_1 + s_2 + s_3\) and is set equal to zero since the elevator starts and finishes at the same position. Substituting the known values into the equation and solving for \(V\) gives:
\( \frac{1}{2} \left(\frac{4}{3}\right) \times (1.5)^2 + V \times 1 + \frac{1}{2} \times (-2) \times (1.5)^2 = 0 \)
Solving this equation yields \(V = 1.70 \, \text{m/s}\) (rounded to three significant figures).
(c) Tension during upward journey when decelerating:
Use Newton’s second law \(\Sigma F = ma\). The net force is \(T – mg\), where \(T\) is the tension, \(m\) is the mass of the elevator and passengers (\(1500 \, \text{kg}\)), \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)), and \(a\) is the acceleration during deceleration (\(-2 \, \text{m/s}^2\)).
\( T – mg = ma \)
\( T – 1500 \times 9.8 = 1500 \times (-2) \)
Solving this equation gives \(T = 12000 \, \text{N}\).
Question 8
The diagram shows a velocity-time graph which models the motion of a cyclist. The graph consists
of five straight line segments. The cyclist accelerates from rest to a speed of 5\( ms^{-1}\) over a period of 10 s, and then travels at this speed for a further 20 s. The cyclist then descends a hill, accelerating to speed V\( m s^{-1}\)over a period of 10 s. This speed is maintained for a further 30 s. The cyclist thendecelerates to rest over a period of 20 s.
(i) Find the acceleration of the cyclist during the first 10 seconds.
(ii) Show that the total distance travelled by the cyclist in the 90 seconds of motion may be expressed
as(45V + 150) m. Hence find V, given that the total distance travelled by the cyclist is 465 m. (iii) The combined mass of the cyclist and the bicycle is 80 kg. The cyclist experiences a constant resistance to motion of 20 N. Use an energy method to find the vertical distance which the cyclist descends during the downhill section from t = 30 to t = 40, assuming that the cyclist does no work during this time.
▶️Answer/Explanation
(i) Acceleration during the first 10 seconds:
The cyclist accelerates from rest to a speed of \(5 \, \text{m/s}\) over a period of \(10 \, \text{s}\). Acceleration (\(a\)) is given by the formula \(a = \frac{\Delta v}{\Delta t}\), where \(\Delta v\) is the change in velocity and \(\Delta t\) is the time interval.
In this case, the change in velocity (\(\Delta v\)) is \(5 \, \text{m/s}\) (initial velocity is \(0 \, \text{m/s}\) as the cyclist starts from rest), and the time interval (\(\Delta t\)) is \(10 \, \text{s}\).
So, \(a = \frac{5 \, \text{m/s}}{10 \, \text{s}} = 0.5 \, \text{m/s}^2\).
(ii) Total distance travelled by the cyclist:
The total distance (\(S\)) can be found by adding the areas under each segment of the velocity-time graph.
– First segment (10 seconds): Area = \(\frac{1}{2} \times 10 \times 5 = 25 \, \text{m}\)
– Second segment (20 seconds):Area = \(10 \times 5 = 50 \, \text{m}\)
– Third segment (10 seconds):Area = \(\frac{1}{2} \times 10 \times (5 + V)\)
– Fourth segment (30 seconds):Area = \(10 \times V\)
– Fifth segment (20 seconds):Area = \(\frac{1}{2} \times 20 \times V = 10V\)
The total distance is \(S = 25 + 50 + \frac{1}{2} \times 10 \times (5 + V) + 10V = 150 + 45V\).
Given that \(S = 465 \, \text{m}\), we can set up the equation \(150 + 45V = 465\) to solve for \(V\).
\(
\begin{aligned}
150 + 45V &= 465 \\
45V &= 315 \\
V &= 7 \, \text{m/s}
\end{aligned}
\)
(iii) Vertical distance descended during the downhill section:**
The energy method involves equating the potential energy lost to the work done against resistance. The potential energy lost is \(mgh\), where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the vertical distance.
The work done against resistance is the force multiplied by the distance, \(20 \, \text{N} \times h\).
Setting these equal, we get:
\(mgh = 20 \times h\)
Substitute \(m = 80 \, \text{kg}\), \(g = 9.8 \, \text{m/s}^2\), and solve for \(h\):
\(
\begin{aligned}
& 80 \times 9.8 \times h = 20 \times h \\
& 784h = 20h \\
& h = \frac{20}{784} \\
& h \approx 0.0255 \, \text{m} \, (\text{or} \, 2.55 \, \text{cm})
\end{aligned}
\)
This represents the vertical distance descended during the downhill section.