SchoolNinjapsst! google school.ninja not schoolspaceninja 💀
BREAKING
Student finds x, x refuses to elaborate★Paper 1 still non-calculator, nation in shock★Local teen shows working, examiner weeps with joy★Student finds x, x refuses to elaborate★Paper 1 still non-calculator, nation in shock★Local teen shows working, examiner weeps with joy★
Also involves
Forces and Newton’s laws ·
Resolving forces
A truck of mass 750 kg is moving with constant speed v m s–1 down a straight road inclined at an angle θ to the horizontal, where sinθ. The resistance to motion of the truck is modelled as a constant force of magnitude 1200 N. The engine of the truck is working at a constant rate of 9 kW.
(a)
Find the value of v.
[4]
Reveal mark scheme
▾
(b)
On another occasion the truck is moving up the same straight road. The resistance to motion of the truck from non-gravitational forces is modelled as a constant force of magnitude 1200 N. The engine of the truck is working at a constant rate of 9 kW.
Find the acceleration of the truck at the instant when it is moving with speed 4.5 m s–1.
A particle, of mass 0.3 kg, is projected from a point O on horizontal ground with speed u. The particle is projected at an angle α above the horizontal, where tan = 2, and moves freely under gravity. When the particle has moved a horizontal distance x from O, its height above the ground is y.
(a)
Show that
[5]
Reveal mark scheme
▾
(b)
The particle hits the ground at the point A, where OA = 36m.
Find u, the speed of projection.
[2]
Reveal mark scheme
▾
(c)
Find the minimum kinetic energy of the particle as it moves between O and A.
[3]
Reveal mark scheme
▾
(d)
The point B lies on the path of the particle. The direction of motion of the particle at B is perpendicular to the initial direction of motion of the particle.
Also involves
Forces and Newton’s laws ·
Resolving forces
A truck of mass 900 kg is towing a trailer of mass 150 kg up an inclined straight road with constant speed 15 m s–1. The trailer is attached to the truck by a light inextensible towbar which is parallel to the road. The road is inclined at an angle θ to the horizontal, where sin θ = . The resistance to motion of the truck from non-gravitational forces has constant magnitude 200 N and the resistance to motion of the trailer from non-gravitational forces has constant magnitude 50 N.
(a)
Find the rate at which the engine of the truck is working.
[5]
Reveal mark scheme
▾
(b)
When the truck and trailer are moving up the road at 15m s–1 the towbar breaks, and the trailer is no longer attached to the truck. The rate at which the engine of the truck is working is unchanged. The resistance to motion of the truck from non-gravitational forces and the resistance to motion of the trailer from non-gravitational forces are still forces of constant magnitudes 200 N and 50 N respectively.
Find the acceleration of the truck at the instant after the towbar breaks.
[3]
Reveal mark scheme
▾
(c)
Use the work-energy principle to find out how much further up the road the trailer travels before coming to instantaneous rest.
A car of mass 800kg is moving on a straight road which is inclined at an angle to the horizontal, where sin θ. The resistance to the motion of the car from non-gravitational forces is modelled as a constant force of magnitude R newtons. When the car is moving up the road at a constant speed of 12.5ms, the engine of the car is working at a constant rate of 3 P watts. When the car is moving down the road at a constant speed of 12.5ms, the engine of the car is working at a constant rate of P watts.
(a)
Find
Reveal mark scheme
▾
(a)(i)
the value of P,
(a)(ii)
the value of R.
[6]
(b)
When the car is moving up the road at 12.5ms the engine is switched off and the car comes to rest, without braking, in a distance d metres. The resistance to the motion of the car from non-gravitational forces is still modelled as a constant force of magnitude R newtons.
Use the work-energy principle to find the value of d.
Also involves
Forces and Newton’s laws ·
Resolving forces
A van of mass 600 kg is moving up a straight road inclined at an angle θ to the horizontal, where sin θ. The resistance to motion of the van from non-gravitational forces has constant magnitude R newtons. When the van is moving at a constant speed of 20 m s–1, the van’s engine is working at a constant rate of 25 kW.
(a)
Find the value of R.
[4]
Reveal mark scheme
▾
(b)
The power developed by the van’s engine is now increased to 30 kW. The resistance to motion from non-gravitational forces is unchanged. At the instant when the van is moving up the road at 20 m s–1, the acceleration of the van is a m s–2.
A truck of mass 1800 kg is towing a trailer of mass 800 kg up a straight road which is inclined to the horizontal at an angle , where sin . The truck is connected to the trailer by a light inextensible rope which is parallel to the direction of motion of the truck. The resistances to motion of the truck and the trailer from non-gravitational forces are modelled as constant forces of magnitudes 300 N and 200 N respectively. The truck is moving at constant speed v m s–1 and the engine of the truck is working at a rate of 40 kW.
Reveal mark scheme
▾
(a)
Find the value of v.
[5]
Reveal mark scheme
▾
(b)
As the truck is moving up the road the rope breaks.
Find the acceleration of the truck immediately after the rope breaks.
A particle P of mass 2 kg is released from rest at a point A on a rough inclined plane and slides down a line of greatest slope. The plane is inclined at 30° to the horizontal. The point B is 5 m from A on the line of greatest slope through A, as shown in Figure 3.
(a)
Find the potential energy lost by P as it moves from A to B.
[2]
Reveal mark scheme
▾
(b)(i)
The speed of P as it reaches B is 4 m s–1.
Use the work-energy principle to find the magnitude of the constant frictional force acting on P as it moves from A to B.
(b)(ii)
Find the coefficient of friction between P and the plane.
[7]
(c)
The particle P is now placed at A and projected down the plane towards B with speed 3 m s–1. Given that the frictional force remains constant,
The points A and B are 10 m apart on a line of greatest slope of a fixed rough inclined plane, with A above B. The plane is inclined at 25° to the horizontal. A particle P of mass 5 kg is released from rest at A and slides down the slope. As P passes B, it is moving with speed 7 m s–1.
(a)
Find, using the work-energy principle, the work done against friction as P moves from A to B.
[4]
Reveal mark scheme
▾
(b)
Find the coefficient of friction between the particle and the plane.
A caravan of mass 600 kg is towed by a car of mass 900 kg along a straight horizontal road. The towbar joining the car to the caravan is modelled as a light rod parallel to the road. The total resistance to motion of the car is modelled as having magnitude 300 N. The total resistance to motion of the caravan is modelled as having magnitude 150 N. At a given instant the car and the caravan are moving with speed 20 m s–1 and acceleration 0.2 m s–2.
Reveal mark scheme
▾
(a)
Find the power being developed by the car’s engine at this instant.
A ball of mass 0.2 kg is projected vertically upwards from a point O with speed 20 m s–1. The non-gravitational resistance acting on the ball is modelled as a force of constant magnitude 1.24 N and the ball is modelled as a particle. Find, using the work-energy principle, the speed of the ball when it first reaches the point which is 8 m vertically above O.
A particle P of mass 3 kg moves from point A to point B up a line of greatest slope of a fixed rough plane. The plane is inclined at 20 to the horizontal. The coefficient of friction between P and the plane is 0.4 Given that AB = 15 m and that the speed of P at A is 20 m s–1, find
(a)
the work done against friction as P moves from A to B,
A ball is projected from a point A which is 8 m above horizontal ground as shown in Figure 4. The ball is projected with speed u m s–1 at an angle above the horizontal. The ball moves freely under gravity and hits the ground at the point B. The speed of the ball immediately before it hits the ground is 2u m s–1.
(a)
By considering energy, find the value of u.
[5]
Reveal mark scheme
▾
(b)
The time taken for the ball to move from A to B is 2 seconds. Find
[4]
Reveal mark scheme
▾
(c)
the minimum speed of the ball on its path from A to B.
Also involves
Forces and Newton’s laws ·
Resolving forces
A lorry of mass 1800 kg travels along a straight horizontal road. The lorry’s engine is working at a constant rate of 30 kW. When the lorry’s speed is 20 m s−1, its acceleration is 0.4 m s−2. The magnitude of the resistance to the motion of the lorry is R newtons.
(a)
Find the value of R.
[4]
Reveal mark scheme
▾
(b)
The lorry now travels up a straight road which is inclined at an angle to the horizontal, where sin α = . The magnitude of the non-gravitational resistance to motion is R newtons. The lorry travels at a constant speed of 20 m s−1.
Find the new rate of working of the lorry’s engine.
The point A lies on a rough plane inclined at an angle to the horizontal, where sin θ = A particle P is projected from A, up a line of greatest slope of the plane, with speed U m s−1. The mass of P is 2 kg and the coefficient of friction between P and the plane is . The particle comes to instantaneous rest at the point B on the plane, where AB =1 5. m. It then moves back down the plane to A.
Reveal mark scheme
▾
(a)
Find the work done against friction as P moves from A to B.
[4]
Reveal mark scheme
▾
(b)
Use the work-energy principle to find the value of U.
A car of mass 1200 kg pulls a trailer of mass 400 kg up a straight road which is inclined to the horizontal at an angle , where sin . The trailer is attached to the car by a light inextensible towbar which is parallel to the road. The car’s engine works at a constant rate of 60 kW. The non-gravitational resistances to motion are constant and of magnitude 1000 N on the car and 200 N on the trailer. At a given instant, the car is moving at 10 m s–1. Find
Reveal mark scheme
▾
(a)
the acceleration of the car at this instant,
[5]
Reveal mark scheme
▾
(b)
the tension in the towbar at this instant.
[4]
Reveal mark scheme
▾
(c)
The towbar breaks when the car is moving at 12 m s–1.
Find, using the work-energy principle, the further distance that the trailer travels before coming instantaneously to rest.
A car of mass 1000 kg moves with constant speed V m s1 up a straight road inclined at an angle to the horizontal, where sin = . The engine of the car is working at a rate of 12 kW. The resistance to motion from non-gravitational forces has magnitude 500 N. Find the value of V.
A ball of mass 0.5 kg is moving with velocity 12i m s–1 when it is struck by a bat. The impulse received by the ball is (–4 i + 7 j) N s. By modelling the ball as a particle, find
Reveal mark scheme
▾
(a)
the speed of the ball immediately after the impact,
[4]
Reveal mark scheme
▾
(b)
the angle, in degrees, between the velocity of the ball immediately after the impact and the vector i,
[2]
Reveal mark scheme
▾
(c)
the kinetic energy gained by the ball as a result of the impact.
A particle P of mass 0.5 kg is projected from a point A up a line of greatest slope AB of a fixed plane. The plane is inclined at 30° to the horizontal and AB = 2 m with B above A, as shown in Figure 2. The particle P passes through B with speed 5 m s–1. The plane is smooth from A to B.
Reveal mark scheme
▾
(a)
Find the speed of projection.
[4]
Reveal mark scheme
▾
(b)
The particle P comes to instantaneous rest at the point C on the plane, where C is above B and BC = 1.5 m. From B to C the plane is rough and the coefficient of friction between P and the plane is . By using the work-energy principle,
A cyclist starts from rest and moves along a straight horizontal road. The combined mass of the cyclist and his cycle is 120 kg. The resistance to motion is modelled as a constant force of magnitude 32 N. The rate at which the cyclist works is 384 W. The cyclist accelerates until he reaches a constant speed of v m s–1. Find
Reveal mark scheme
▾
(a)
the value of v,
[3]
Reveal mark scheme
▾
(b)
the acceleration of the cyclist at the instant when the speed is 9 m s–1.
A particle of mass 2 kg is moving with velocity (5i + j) m s–1 when it receives an impulse of (–6 i + 8 j) N s. Find the kinetic energy of the particle immediately after receiving the impulse.
A box of mass 30 kg is held at rest at point A on a rough inclined plane. The plane is inclined at 20° to the horizontal. Point B is 50 m from A up a line of greatest slope of the plane, as shown in Figure 1. The box is dragged from A to B by a force acting parallel to AB and then held at rest at B. The coefficient of friction between the box and the plane is . Friction is the only non-gravitational resistive force acting on the box. Modelling the box as a particle,
Reveal mark scheme
▾
(a)
find the work done in dragging the box from A to B.
[6]
Reveal mark scheme
▾
(b)
The box is released from rest at the point B and slides down the slope. Using the work-energy principle, or otherwise,
A particle P of mass 0.6 kg is released from rest and slides down a line of greatest slope of a rough plane. The plane is inclined at 30° to the horizontal. When P has moved 12 m, its speed is 4 m s–1. Given that friction is the only non-gravitational resistive force acting on P, find
Reveal mark scheme
▾
(a)
the work done against friction as the speed of P increases from 0 m s–1 to 4 m s–1,
[4]
Reveal mark scheme
▾
(b)
the coefficient of friction between the particle and the plane.
Also involves
Forces and Newton’s laws ·
Resolving forces
A car of mass 750 kg is moving up a straight road inclined at an angle θ to the horizontal, where sin θ = The resistance to motion of the car from non-gravitational forces has constant magnitude R newtons. The power developed by the car’s engine is 15 kW and the car is moving at a constant speed of 20 m s–1.
Reveal mark scheme
▾
(a)
Show that R = 260.
[4]
Reveal mark scheme
▾
(b)
The power developed by the car’s engine is now increased to 18 kW. The magnitude of the resistance to motion from non-gravitational forces remains at 260 N. At the instant when the car is moving up the road at 20 m s–1 the car’s acceleration is a m s–2.
[ In this question i and j are perpendicular unit vectors in a horizontal plane. ] A ball of mass 0.5 kg is moving with velocity (10i + 24j) m s–1 when it is struck by a bat. Immediately after the impact the ball is moving with velocity 20i m s–1. Find
Reveal mark scheme
▾
(a)
the magnitude of the impulse of the bat on the ball,
[4]
Reveal mark scheme
▾
(b)
the size of the angle between the vector i and the impulse exerted by the bat on the ball,
[2]
Reveal mark scheme
▾
(c)
the kinetic energy lost by the ball in the impact.
A particle of mass 0.5 kg is projected vertically upwards from ground level with a speed of 20 m s–1. It comes to instantaneous rest at a height of 10 m above the ground. As the particle moves it is subject to air resistance of constant magnitude R newtons. Using the work-energy principle, or otherwise, find the value of R.
Also involves
Forces and Newton’s laws ·
Resolving forces
A cyclist and her bicycle have a total mass of 70 kg. She cycles along a straight horizontal road with constant speed 3.5 m s–1. She is working at a constant rate of 490 W.
(a)
Find the magnitude of the resistance to motion.
[4]
Reveal mark scheme
▾
(b)
The cyclist now cycles down a straight road which is inclined at an angle θ to the
gravitational resistance to motion is modelled as 40U newtons. She is now working at a constant rate of 24 W.
Also involves
Weight and motion under gravity ·
Resolving forces
A truck of mass of 300 kg moves along a straight horizontal road with a constant speed of 10 m s–1. The resistance to motion of the truck has magnitude 120 N.
(a)
Find the rate at which the engine of the truck is working.
[2]
Reveal mark scheme
▾
(b)
On another occasion the truck moves at a constant speed up a hill inclined at θ to the horizontal, where sin θ . The resistance to motion of the truck from non-gravitational forces remains of magnitude 120 N. The rate at which the engine works is the same as in part (a).
A particle P of mass 2 kg is projected up a rough plane with initial speed 14 m s–1, from a point X on the plane, as shown in Figure 4. The particle moves up the plane along the line of greatest slope through X and comes to instantaneous rest at the point Y. The plane is inclined at an angle α to the horizontal, where tan α = . The coefficient of friction between the particle and the plane is .
(a)
Use the work-energy principle to show that XY = 25 m.
[7]
Reveal mark scheme
▾
(b)
After reaching Y, the particle P slides back down the plane.
A lorry of mass 2000 kg is moving down a straight road inclined at angle to the horizontal, where sin=αsin = . The resistance to motion is modelled as a constant force of magnitude 1600 N. The lorry is moving at a constant speed of 14 m s–1. Find, in kW, the rate at which the lorry’s engine is working.
A package of mass 3.5 kg is sliding down a ramp. The package is modelled as a particle and the ramp as a rough plane inclined at an angle of 20° to the horizontal. The package slides down a line of greatest slope of the plane from a point A to a point B, where AB = 14 m. At A the package has speed 12 m s–1 and at B the package has speed 8 m s–1, as shown in Figure 1. Find
Reveal mark scheme
▾
(a)
the total energy lost by the package in travelling from A to B,
[5]
(b)
the coefficient of friction between the package and the ramp.
A parcel of mass 2.5 kg is moving in a straight line on a smooth horizontal floor. Initially the parcel is moving with speed 8 m s–1. The parcel is brought to rest in a distance of 20 m by a constant horizontal force of magnitude R newtons. Modelling the parcel as a particle, find
Reveal mark scheme
▾
(a)
the kinetic energy lost by the parcel in coming to rest,
A car of mass 1000 kg is moving at a constant speed of 16 m s–1 up a straight road inclined at an angle θ to the horizontal. The rate of working of the engine of the car is 20 kW and the resistance to motion from non-gravitational forces is modelled as a constant force of magnitude 550 N.
(a)
[5]
Reveal mark scheme
▾
(b)
When the car is travelling up the road at 16 m s–1, the engine is switched off. The car comes to rest, without braking, having moved a distance y metres from the point where the engine was switched off. The resistance to motion from non-gravitational forces is again modelled as a constant force of magnitude 550 N.
Find the value of y.
[4]
Reveal mark scheme
▾
Report a problem with a question on this page
Memes
Certified Maths memes. Peer-reviewed by nobody.
Laugh now. Cry in the exam hall later. Then come back and do a past paper.
Learn the formula
Draw a vague triangle and hope
😰
Me with 4 minutes left on Paper 3
Using a calculator🧠
Using the formula sheet🧠
Doing it in your head🧠
Writing ‘by inspection’ and walking out🧠
Maths enlightenment, ranked
Mum 💖
did you do your maths homework sweetie?? 🥰
yes I solved for x
proud of you!!!
📱 Screen time this week · TikTok6h 7m
x was never found
Maths starter pack
📐Protractor you've never used🧮Calculator older than you✍️“Show your working” trauma🥷This website