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A-Level · Mechanics

M8 Mechanics: Projectiles and Applications Depth

EDEXCEL 9MA0 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Key results: projectile motion and combining kinematics with forces

Original content written for Revision Library.

Unless a question states otherwise, take g = 9.8 m/s^2 and model every object as a particle moving freely under gravity once released, so that gravity is the only force acting during the flight itself. For a particle projected from a point on horizontal ground with speed U at angle alpha above the horizontal, resolve the initial velocity into a horizontal component U cos(alpha), which is constant throughout the flight, and a vertical component U sin(alpha), which reduces uniformly under gravity; the horizontal and vertical motions can then be treated independently using the suvat equations, and combined using Pythagoras' theorem and trigonometry to find the resultant speed and direction at any instant. Two standard results you may quote without proof, unless a question asks you to derive them, are the time of flight on level ground, T = 2Usin(alpha)/g, and the range on level ground, R = U^2 sin(2alpha)/g; the greatest height reached is H = U^2 sin^2(alpha)/(2g). Several questions in this worksheet combine projectile motion with forces: where a particle or block is first accelerated by gravity and/or friction while in contact with a surface (for example sliding down a rough inclined plane, or being pulled by a connected particle over a pulley), use Newton's second law, F = ma, resolving forces parallel and perpendicular to the relevant surface, to find its speed at the point where it becomes a projectile, and then apply projectile motion to the remainder of its path. Vectors are given in terms of perpendicular unit vectors i and j; a position vector r(t) for a projectile can be written in terms of t by integrating, or by using the suvat equations componentwise, with constant acceleration -gj.

A ball is projected from ground level with speed 18 m/s at an angle of 55 degrees to the horizontal, and lands on the same horizontal ground. Take g = 9.8 m/s^2. You may quote the result R = U² sin(2alpha)/g without proof.

Find the horizontal range of the ball, giving your answer to 3 significant figures.

(Total for Question 1 is 3 marks)

A particle is projected from ground level with speed U m/s at an angle of 50 degrees to the horizontal, and lands on the same horizontal ground at a horizontal distance of 40 m from the point of projection. Take g = 9.8 m/s^2. You may quote the result R = U² sin(2alpha)/g without proof.

Find the value of U, giving your answer to 3 significant figures.

(Total for Question 2 is 3 marks)

A particle is projected from a point O on horizontal ground with speed 22 m/s at an angle of 42 degrees above the horizontal. The particle moves freely under gravity. Take g = 9.8 m/s^2.

(a)Find the greatest height above the ground reached by the particle, giving your answer to 3 significant figures.(3)

(b)Find the time taken for the particle to reach this greatest height, giving your answer to 3 significant figures.(2)

(Total for Question 3 is 5 marks)

A ball is thrown horizontally with speed 17 m/s from a point at the top of a vertical cliff. The point of projection is 122.5 m vertically above the sea, which may be modelled as a horizontal plane. The ball is modelled as a particle moving freely under gravity. Take g = 9.8 m/s^2.

(a)Show that the time taken for the ball to reach the sea is exactly 5 seconds.(3)

(b)Find the horizontal distance from the base of the cliff to the point where the ball lands.(2)

(c)Find the speed of the ball as it hits the sea, giving your answer to 3 significant figures.(3)

(d)Find the angle the ball's velocity makes with the horizontal as it hits the sea, giving your answer to 3 significant figures.(2)

(Total for Question 4 is 10 marks)

A particle is projected from a point on horizontal ground with speed U at an angle α above the horizontal, and moves freely under gravity. Take g = 9.8 m/s^2.

(a)By considering the vertical component of the initial velocity, show that the greatest height reached by the particle is H = U² sin²(α) / (2g).(3)

(b)Given that U = 26 m/s and α = 38 degrees, find the greatest height reached, giving your answer to 3 significant figures.(2)

(c)By first finding the time taken to reach the greatest height, use a different suvat equation to verify your value of H from part (b).(3)

(Total for Question 5 is 8 marks)

A particle is projected from a point on horizontal ground with speed U m/s at an angle α above the horizontal, where 0 < α < 90, and lands on the same horizontal ground. You may quote the result that the horizontal range is R = U² sin(2alpha)/g without further proof.

(a)Explain why, for a fixed value of U, the range R is greatest when α = 45 degrees.(2)

(b)Given that U = 19.6 m/s, find the maximum possible range, giving your answer exactly.(2)

(c)In practice, sports scientists modelling the flight of a thrown object such as a javelin or a shot do not always find that the optimal launch angle for maximum distance is 45 degrees. Suggest two features of a real thrown object that this projectile model ignores, and for each one, briefly explain the effect it would have on the actual range achieved compared with the range predicted by the model.(4)

(Total for Question 6 is 8 marks)

A particle P is projected from the origin O with velocity (8i + 14.7j) m/s, and moves freely under gravity, taking g = 9.8 m/s² acting in the direction -j. At time t seconds after projection, the position vector of P relative to O is r metres.

(a)Find r in terms of t, giving your answer using the unit vectors i and j.(2)

(b)Find the value of t at which P is moving parallel to the vector (2i - j).(4)

(c)Find the speed of P at this instant.(2)

(Total for Question 7 is 8 marks)

A particle P is projected from the origin O with velocity (12i + 29.4j) m/s, and moves freely under gravity, taking g = 9.8 m/s² acting in the direction -j. At time t seconds after projection, the position vector of P relative to O is r metres.

(a)Find r in terms of t, giving your answer using the unit vectors i and j.(2)

(b)Find the value of t at which P is at its greatest height above the level of O.(2)

(c)Find, in the form ai + bj, the position vector of P at this instant.(3)

(d)Find the speed of P at this instant.(2)

(Total for Question 8 is 9 marks)

A sports analyst models the flight of a rugby ball, kicked for a conversion, as the motion of a particle projected from ground level, moving freely under gravity, with no forces other than gravity acting on it once it leaves the kicker's boot.

Discuss the validity of this projectile model for a real rugby conversion kick. Refer to at least three specific features of the real situation that the model does not take into account, and explain how each feature would affect the accuracy of the model's predicted range and/or time of flight.

(Total for Question 9 is 5 marks)

A cricketer, Amara, strikes a ball from ground level so that it leaves the bat with speed 27 m/s at an angle of 30 degrees above the horizontal, directly towards a vertical sight screen of height 6 m, which stands at a horizontal distance of 55 m from the point where the ball is struck, in the vertical plane of the ball's motion. The ball is modelled as a particle moving freely under gravity. Take g = 9.8 m/s^2.

Figure (to be drawn): Diagram: ball struck from ground level at 30 degrees to the horizontal, travelling towards a vertical sight screen of height 6 m standing 55 m away, in the same vertical plane as the ball's motion.

(a)Find the horizontal component of the ball's initial velocity, giving your answer to 3 significant figures.(2)

(b)Find the time taken for the ball to travel the 55 m horizontal distance to the sight screen, giving your answer to 3 significant figures.(2)

(c)Find the height of the ball above the ground at this instant, giving your answer to 3 significant figures.(3)

(d)Hence determine, with justification, whether the ball passes over the sight screen or strikes it.(2)

(Total for Question 10 is 9 marks)

Two students, Ryan and Sofia, are experimenting in a field. At the instant Ryan throws a beanbag from ground level with speed 25 m/s at an angle of 40 degrees above the horizontal, aimed directly at a second beanbag balanced on top of a vertical post standing at a horizontal distance of 15 m from Ryan, in the vertical plane of the throw, Sofia releases that second beanbag from rest, so that it falls freely under gravity. Both beanbags are modelled as particles moving freely under gravity, and they collide in mid-air. Take g = 9.8 m/s^2.

Figure (to be drawn): Diagram: Ryan at ground level throwing a beanbag at 40 degrees towards the top of a vertical post 15 m away, where a second beanbag is released from rest at the same instant.

(a)By setting up expressions for the horizontal and vertical displacements of each beanbag, and using the fact that Ryan's beanbag is aimed directly at the second beanbag's initial position, find the height of the post, giving your answer to 3 significant figures.(5)

(b)Find the time after the beanbags are released at which they collide, giving your answer to 3 significant figures.(2)

(c)State two modelling assumptions, beyond treating each beanbag as a particle moving freely under gravity, that are needed for the collision to occur exactly as described.(2)

(Total for Question 11 is 9 marks)

A particle is projected from a point O with speed 21 m/s at an angle of 35 degrees above the horizontal. The point O lies at the top of a straight slope which descends at a constant angle of 15 degrees below the horizontal, in the same vertical plane as the particle's motion, so that the particle lands at a point P on the slope. Horizontal and vertically-upward displacements from O are denoted x and y respectively (both in metres, with y measured positive upward). Take g = 9.8 m/s^2.

Figure (to be drawn): Diagram: point O at the top of a downward slope inclined at 15 degrees to the horizontal; a particle is projected from O at 35 degrees above the horizontal, in the same vertical plane as the slope, and lands on the slope at P.

(a)Write down expressions, in terms of t, for x and y.(2)

(b)Given that the slope satisfies the equation y = -x tan(15), find the time of flight from O to P, giving your answer to 3 significant figures.(5)

(c)Hence find the distance OP, giving your answer to 3 significant figures.(3)

(Total for Question 12 is 10 marks)

A small block of mass 2 kg is released from rest at the top of a rough plane inclined at 30 degrees to the horizontal. The coefficient of friction between the block and the plane is 0.2. The plane has length 5 m, measured along a line of greatest slope, and the bottom edge of the plane is 1.2 m above horizontal ground. Beyond this edge, the block leaves the plane and moves freely under gravity as a projectile until it lands on the ground. Take g = 9.8 m/s^2.

Figure (to be drawn): Diagram: rough plane inclined at 30 degrees to the horizontal, length 5 m measured along the slope, with its lower edge 1.2 m above horizontal ground; a block slides down from rest at the top.

(a)By resolving forces perpendicular and parallel to the plane, show that the acceleration of the block down the plane is given by a = g(sin(30) - 0.2cos(30)).(3)

(b)Find the value of this acceleration, giving your answer to 3 significant figures.(2)

(c)Find the speed of the block at the instant it leaves the plane, giving your answer to 3 significant figures.(2)

(d)At the instant the block leaves the plane, its velocity is directed at 30 degrees below the horizontal, with the magnitude found in part (c). Find the horizontal distance from the point where the block leaves the plane to the point where it lands on the ground, giving your answer to 3 significant figures.(6)

(Total for Question 13 is 13 marks)

In a warehouse, a technician called Tom sets up a demonstration. Block A, of mass 2 kg, lies on a rough horizontal table, 3 m from a smooth light pulley fixed at the edge of the table. A light inextensible string connects A, over the pulley, to block B, of mass 6 kg, which hangs freely 1.5 m above the floor. The coefficient of friction between A and the table is 0.2. The table edge is 1.25 m above the floor. The system is released from rest with the string taut, and B descends, pulling A towards the pulley. Take g = 9.8 m/s^2.

Figure (to be drawn): Diagram: block A on a rough horizontal table, 3 m from a smooth pulley at the table edge, connected by a string over the pulley to block B hanging 1.5 m above the ground; the table edge is 1.25 m above the ground.

(a)Show that, while both blocks are moving and the string remains taut, the acceleration of the system is 6.86 m s^-2.(3)

(b)Find the tension in the string while both blocks are moving.(2)

(c)Given that B strikes the floor after falling 1.5 m, find the speed of the blocks at this instant, giving your answer to 3 significant figures.(3)

(d)Once B lands, it remains on the floor and the string becomes slack. Find the speed of A as it reaches the edge of the table, given that A has a further 1.5 m to travel and now decelerates under friction alone, giving your answer to 3 significant figures.(3)

(e)A leaves the table edge horizontally with this speed and falls to the floor, which is 1.25 m below the table edge. Find the horizontal distance travelled by A from the table edge to the point where it lands, giving your answer to 3 significant figures.(3)

(Total for Question 14 is 14 marks)

Mark scheme · M8 Mechanics: Projectiles and Applications Depth

Question 1

M1 substitutes U = 18, α = 55 into R = U² sin(2alpha)/g (oe)
A1 correct unsimplified expression, e.g. R = 324 sin(110)/9.8
A1 R = awrt 31.1 (m), cao
Answer: R = awrt 31.1 m

Question 2

M1 rearranges R = U² sin(2alpha)/g to U² = Rg/sin(2alpha) and substitutes R = 40, α = 50 (oe)
A1 U² = awrt 398
A1 U = awrt 20.0 (m/s), cao
Answer: U = awrt 20.0 m/s

Question 3

(a) M1 identifies the initial vertical component of velocity, u_y = 22sin(42) (oe)
(a) M1 uses v² = u² - 2gs with v = 0 and their u_y, rearranged for s
(a) A1 H = awrt 11.1 (m), cao
(a) Answer: H = awrt 11.1 m
(b) M1 uses t = u_y/g with their u_y (or v = u - gt with v = 0)
(b) A1 t = awrt 1.50 (s), cao
(b) Answer: t = awrt 1.50 s

Question 4

(a) M1 uses s = ut + 0.5at² vertically with u = 0, a = 9.8, s = 122.5 (oe)
(a) M1 rearranges to t² = 25
(a) A1 t = 5, cso
(a) Answer: t = 5 s (shown)
(b) M1 horizontal distance = 17 x 5 (oe)
(b) A1 85 (m), cao
(b) Answer: 85 m
(c) M1 vertical speed at impact v_y = gt = 9.8 x 5 = 49 (oe)
(c) M1 combines components: speed = √17² + 49²
(c) A1 awrt 51.9 (m/s)
(c) Answer: 51.9 m/s (3sf)
(d) M1 tan(θ) = 49/17 (oe)
(d) A1 θ = awrt 70.9 (degrees)
(d) Answer: 70.9 degrees below the horizontal (3sf)

Question 5

(a) M1 identifies the initial vertical component of velocity, u_y = Usin(α)
(a) M1 uses v² = u² - 2gs with v = 0 (at the greatest height) and their u_y
(a) A1 rearranges to H = U² sin²(α)/(2g), cso
(a) Answer: H = U² sin²(α) / (2g) (shown)
(b) M1 substitutes U = 26, α = 38 into H = U² sin²(α)/(2g)
(b) A1 H = awrt 13.1 (m), cao
(b) Answer: H = awrt 13.1 m
(c) M1 finds the time to greatest height, t = u_y/g = 26sin(38)/9.8, ft
(c) M1 uses a different suvat equation, e.g. s = ((u+v)/2)t with u = u_y, v = 0, and their t
(c) A1 H = awrt 13.1 (m), matching (b), ft
(c) Answer: H = awrt 13.1 m (verified)

Question 6

(a) B1 sin(2alpha) has its greatest possible value, 1, when 2alpha = 90 degrees, since the sine function has maximum value 1
(a) B1 hence α = 45 degrees gives the maximum range, for a fixed value of U
(a) Answer: R is greatest when α = 45 degrees, since this makes sin(2alpha) = sin(90) = 1, its maximum possible value
(b) M1 substitutes α = 45 into R = U² sin(90)/g = U²/g
(b) A1 R = 39.2 (m), exact, cao
(b) Answer: R = 39.2 m (exact)
(c) B1 feature 1, e.g. air resistance (drag) on the object is ignored by the model
(c) B1 effect of feature 1, e.g. including drag would reduce the object's horizontal velocity throughout the flight, so the actual range would be smaller than the model predicts, and the optimal angle would typically be less than 45 degrees
(c) B1 feature 2, e.g. the object is released above ground level (from the thrower's hand or the kicker's boot), not from ground level as the model assumes; or the object's shape or spin can generate aerodynamic lift
(c) B1 effect of feature 2, e.g. a release height above ground level means the object stays airborne longer than the model predicts, increasing the actual range for the same speed and angle; or aerodynamic lift would increase the range and change the optimal angle
(c) Answer: E.g. air resistance (reduces range, lowers optimal angle) and release height above ground (increases range for the same speed and angle)

Question 7

(a) M1 uses r = ut + 0.5at² componentwise, with a = -9.8j
(a) A1 r = 8ti + (14.7t - 4.9t²)j, oe
(a) Answer: r = 8ti + (14.7t - 4.9t²)j (m)
(b) M1 differentiates r, or uses v = u + at, to obtain v(t) = 8i + (14.7 - 9.8t)j
(b) M1 sets up the condition for v to be parallel to 2i - j, e.g. (14.7 - 9.8t)/8 = -1/2 (oe)
(b) M1 rearranges to 9.8t = 18.7
(b) A1 t = awrt 1.91 (s), oe
(b) Answer: t = awrt 1.91 s
(c) M1 v = 8i - 4j, speed = √8² + (-4)²
(c) A1 awrt 8.94 (m/s), oe (accept 4sqrt(5))
(c) Answer: awrt 8.94 m/s (= 4sqrt(5))

Question 8

(a) M1 uses r = ut + 0.5at² componentwise, with a = -9.8j
(a) A1 r = 12ti + (29.4t - 4.9t²)j, oe
(a) Answer: r = 12ti + (29.4t - 4.9t²)j (m)
(b) M1 sets the vertical component of velocity to zero: 29.4 - 9.8t = 0 (oe, e.g. differentiates the j-component of r)
(b) A1 t = 3 (s), cao
(b) Answer: t = 3 s
(c) M1 substitutes t = 3 into the i-component of r, ft
(c) M1 substitutes t = 3 into the j-component of r, ft
(c) A1 36i + 44.1j (m), cao
(c) Answer: 36i + 44.1j (m)
(d) M1 recognises the vertical component of velocity is zero at the greatest height, so the speed equals the horizontal component of velocity
(d) A1 12 (m/s), cao
(d) Answer: 12 m/s

Question 9

Level 1 (1-2): Identifies one or two relevant features of the real kick that the model ignores, with little or no discussion of their effect on the modelled flight.
Level 2 (3-4): Identifies at least three relevant features and gives a generally correct explanation of the effect of at least one on the accuracy of the model, with some imprecision.
Level 3 (5): Identifies at least three relevant features and explains clearly and correctly how each affects the accuracy of the model's predicted range and/or time of flight, using appropriate mechanics reasoning throughout.
Indicative content:
- the ball is modelled as a particle, so its size, shape and rotation (e.g. spin imparted by the kicker) are ignored
- air resistance (drag) is assumed negligible, but a real rugby ball experiences significant air resistance, which would reduce its range and time of flight compared with the model's prediction
- the ball is released above ground level, from the kicking tee or the kicker's boot, rather than from ground level as the model assumes, so the true range would be greater than predicted for the same speed and angle
- a spinning or asymmetric ball can experience aerodynamic forces (such as swerve or additional lift) that are not included in a simple projectile model
- wind is not accounted for in the model, but a real kick can be significantly affected by a headwind, tailwind or crosswind, changing both the range and the direction of the ball's flight
- g is assumed constant throughout the flight, which is a reasonable assumption given the heights and distances involved, so this assumption has negligible effect on the comparison

Question 10

(a) M1 u_x = 27cos(30) (oe)
(a) A1 awrt 23.4 (m/s)
(a) Answer: u_x = awrt 23.4 m/s
(b) M1 t = 55/u_x, ft (a)
(b) A1 awrt 2.35 (s)
(b) Answer: t = awrt 2.35 s
(c) M1 identifies the initial vertical component u_y = 27sin(30) = 13.5
(c) dM1 substitutes into y = u_y t - 4.9t², using their t from (b), dependent on the previous mark
(c) A1 awrt 4.64 (m)
(c) Answer: awrt 4.64 m
(d) M1 compares their height from (c) with the screen height of 6 m
(d) A1 correct conclusion with reason: since 4.64 m < 6 m, the ball is below the top of the screen at this horizontal distance, so it strikes the sight screen rather than passing over it, ft
(d) Answer: The ball strikes the sight screen (its height there, awrt 4.64 m, is less than the screen's height of 6 m)

Question 11

(a) M1 horizontal displacement of the thrown beanbag: x = 25cos(40) t
(a) M1 vertical displacements: y1 = 25sin(40) t - 4.9t² (thrown beanbag) and y2 = h - 4.9t² (falling beanbag, released from height h)
(a) M1 uses the fact that Ryan aims directly at the post-top, i.e. tan(40) = h/15, or equivalently finds t = 15/(25cos(40)) from x = 15
(a) M1 sets y1 = y2 at the collision instant; the -4.9t² terms cancel, giving h = 25sin(40) x t = 15tan(40)
(a) A1 h = awrt 12.6 (m)
(a) Answer: h = awrt 12.6 m
(b) M1 t = 15/(25cos(40)), ft
(b) A1 awrt 0.783 (s)
(b) Answer: t = awrt 0.783 s
(c) B1 the second beanbag is released from rest with no horizontal velocity, so it falls vertically downward only
(c) B1 nothing obstructs the path of either beanbag before the collision point, and the beanbags are released and thrown at precisely the same instant
(c) Answer: E.g. the released beanbag falls with no horizontal velocity; there is no obstruction between the two beanbags before they collide

Question 12

(a) B1 x = 21cos(35) t, oe
(a) B1 y = 21sin(35) t - 4.9t², oe
(a) Answer: x = 21cos(35) t; y = 21sin(35) t - 4.9t²
(b) M1 substitutes the expressions for x and y from (a) into y = -x tan(15)
(b) M1 collects terms and factors out t, noting t is not 0 at P
(b) M1 rearranges the resulting linear equation to make t the subject
(b) A1 unsimplified t = (21sin(35) + 21cos(35)tan(15)) / 4.9, oe
(b) A1 t = awrt 3.40 (s), cao
(b) Answer: t = awrt 3.40 s
(c) M1 x = awrt 58.5 (m), using their t from (b), ft
(c) M1 OP = x/cos(15), oe (or uses Pythagoras with x and y)
(c) A1 OP = awrt 60.5 (m), ft
(c) Answer: OP = awrt 60.5 m

Question 13

(a) M1 resolves perpendicular to the plane: R = 2g cos(30)
(a) M1 resolves along the plane (down positive), using F = 0.2R: 2g sin(30) - 0.2(2g cos(30)) = 2a
(a) A1 mass cancels throughout, giving a = g(sin(30) - 0.2cos(30)), cso
(a) Answer: a = g(sin(30) - 0.2cos(30)) (shown)
(b) M1 substitutes g = 9.8 and evaluates sin(30), cos(30)
(b) A1 a = awrt 3.20 (m/s²), cao
(b) Answer: a = awrt 3.20 m/s²
(c) M1 uses v² = u² + 2as with u = 0, s = 5, and their a from (a)/(b), ft
(c) A1 v = awrt 5.66 (m/s), cao
(c) Answer: v = awrt 5.66 m/s
(d) M1 resolves the velocity into components: u_x = v cos(30), u_y = v sin(30), ft their v from (c)
(d) A1 u_x = awrt 4.90 (m/s) and u_y = awrt 2.83 (m/s)
(d) M1 sets up the vertical equation of motion for the 1.2 m fall: 1.2 = u_y t + 4.9t², oe
(d) A1 t = awrt 0.284 (s)
(d) M1 horizontal distance = u_x x t, ft
(d) A1 awrt 1.39 (m)
(d) Answer: awrt 1.39 m

Question 14

(a) M1 equation of motion for B (down positive): 6g - T = 6a
(a) M1 equation of motion for A: T - 0.2(2)(9.8) = 2a, using R = 2g for the normal reaction on the horizontal table
(a) A1 adds the two equations to eliminate T; mass simplifies to give a = 6.86, cso
(a) Answer: a = 6.86 m s^-2 (shown)
(b) M1 substitutes a = 6.86 into either equation of motion
(b) A1 T = 17.64 (N), cao
(b) Answer: T = 17.64 N
(c) M1 uses v1² = u² + 2as with u = 0, s = 1.5, and a = 6.86
(c) A1 v1² = 20.58, unsimplified
(c) A1 v1 = awrt 4.54 (m/s)
(c) Answer: v1 = awrt 4.54 m/s
(d) B1 new deceleration a2 = 0.2(9.8) = 1.96 (m/s²), since the normal reaction and hence friction on A are unchanged once the string is slack
(d) dM1 uses v2² = v1² - 2(1.96)(1.5), ft their v1 from (c), dependent on the previous mark
(d) A1 v2 = awrt 3.83 (m/s)
(d) Answer: v2 = awrt 3.83 m/s
(e) M1 uses 1.25 = 4.9t² (or equivalent) to find the time of fall
(e) M1 horizontal distance = v2 x t, ft their v2 from (d)
(e) A1 awrt 1.94 (m)
(e) Answer: awrt 1.94 m

Mechanics: Projectiles and Applications Depth - Worksheets, Questions and Revision