Mechanics: Kinematics Depth - Worksheets, Questions and Revision

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A-Level · Mechanics

M5 Mechanics: Kinematics Depth

EDEXCEL 9MA0 · Calculator allowed · about 155 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Priya cycles from rest along a straight, horizontal cycle path with constant acceleration 1.5 m/s² for 8 seconds. Model Priya as a particle.

(a)Find Priya's speed at the end of the 8 seconds.(2)

(b)Find the distance Priya travels in these 8 seconds.(2)

(Total for Question 1 is 4 marks)

A parcel trolley moves in a straight line across a warehouse floor. It passes point A with speed 3 m/s and constant acceleration 0.7 m/s². It passes point B, 9 m from A in the direction of motion, T seconds after passing A. Model the trolley as a particle.

(a)Show that 7T² + 60T - 180 = 0.(2)

(b)Hence find the value of T, giving your answer to 3 significant figures.(3)

(Total for Question 2 is 5 marks)

Amara throws a ball vertically upwards from a point 1.2 m above horizontal ground with speed 14 m/s. Model the ball as a particle moving freely under gravity, and use g = 9.8 m/s².

(a)Find the greatest height above the ground reached by the ball.(3)

(b)Find the time taken for the ball to hit the ground, giving your answer to 3 significant figures.(3)

(Total for Question 3 is 6 marks)

A train travels along a straight section of track between station P and station Q. It starts from rest at P and accelerates uniformly to a speed of 30 m/s in 40 seconds. It then travels at this constant speed for 90 seconds, before decelerating uniformly to rest at Q, coming to rest 25 seconds after the deceleration begins. Model the train as a particle.

(a)Find the total distance from P to Q.(4)

(b)Find the average speed for the whole journey from P to Q, giving your answer to 3 significant figures.(2)

(Total for Question 4 is 6 marks)

Two cars, A and B, pass the same road sign at the same instant, travelling in the same direction along a straight road. Car A passes the sign with speed 22 m/s and then travels at this constant speed. Car B passes the sign from rest at this same instant, with constant acceleration 2 m/s². Let t seconds be the time after the cars pass the sign.

(a)Write down expressions, in terms of t, for the displacement from the sign of car A and of car B.(2)

(b)Find the value of t at which car B catches up with car A, and find the distance from the sign at which this happens.(3)

(c)Find the speed of car B at the instant it catches up with car A.(2)

(Total for Question 5 is 7 marks)

A particle P moves along the x-axis. At time t seconds (t ≥ 0), the displacement of P from a fixed origin O is x metres, where x = t³ - 6t² + 9t + 2.

(a)Find an expression for the velocity, v m/s, of P at time t seconds.(2)

(b)Find an expression for the acceleration, a m/s², of P at time t seconds.(2)

(c)Find the values of t for which P is instantaneously at rest.(2)

(d)Find the acceleration of P at the instant when t = 3.(2)

(Total for Question 6 is 8 marks)

A particle Q moves in a straight line. At time t seconds, the acceleration of Q is a m/s², where a = 4 - 6t. When t = 0, the velocity of Q is 3 m/s and the displacement of Q from a fixed point O is 0.

(a)Show that the velocity of Q at time t seconds is given by v = 3 + 4t - 3t².(3)

(b)Find the displacement of Q from O when t = 2 seconds.(3)

(c)Find the value of t, other than t = 0, at which Q returns to O.(2)

(Total for Question 7 is 8 marks)

A particle R moves in a straight line so that its velocity, v m/s, at time t seconds (0 ≤ t ≤ 6) is given by v = t² - 6t + 8.

(a)Find the values of t at which R is instantaneously at rest.(2)

(b)Find the acceleration of R at time t = 5 seconds, and state, with a reason, whether R is speeding up or slowing down at this instant.(3)

(c)Find the total distance travelled by R in the interval 0 ≤ t ≤ 6.(4)

(Total for Question 8 is 9 marks)

A particle P moves along the x-axis. At time t seconds, the velocity of P is v m/s, where v = 6sin(2t), for 0 ≤ t ≤ π. When t = 0, P is at the origin O.

(a)Find an expression for the acceleration, a m/s², of P at time t seconds.(2)

(b)Find the values of t, for 0 ≤ t ≤ π, at which P is instantaneously at rest.(3)

(c)Find an expression, in terms of t, for the displacement x metres of P from O at time t seconds.(3)

(d)Find the greatest distance of P from O in the interval 0 ≤ t ≤ π, justifying that this value is a maximum.(3)

(Total for Question 9 is 11 marks)

A stone is projected horizontally with speed 15 m/s from the top of a vertical cliff of height 45 m above the sea. The stone lands in the sea at point L. Model the stone as a particle moving freely under gravity, and use g = 9.8 m/s².

(a)Find the time taken for the stone to reach the sea, giving your answer to 3 significant figures.(3)

(b)Find the horizontal distance from the base of the cliff to L, giving your answer to 3 significant figures.(2)

(c)Find the speed of the stone as it hits the sea, giving your answer to 3 significant figures.(4)

(Total for Question 10 is 9 marks)

A ball is projected from a point O on horizontal ground with speed U m/s at an angle θ above the horizontal, where tan(θ) = 3/4. The ball moves freely under gravity and lands on the ground at the same level as O. Take U = 21 and use g = 9.8 m/s².

(a)Show that the horizontal and vertical components of the initial velocity of the ball are 16.8 m/s and 12.6 m/s respectively.(2)

(b)Find the greatest height above the ground reached by the ball.(2)

(c)Show that the time of flight of the ball is 18/7 seconds.(3)

(d)Hence show that the horizontal range of the ball is 43.2 m.(3)

(Total for Question 11 is 10 marks)

A ball is projected from a point A, which is 6 m above horizontal ground, with speed 25 m/s at an angle of elevation α, where sin(α) = 7/25. The ball moves freely under gravity until it hits the ground at point B. Use g = 9.8 m/s².

(a)Show that the initial horizontal and vertical components of velocity are 24 m/s and 7 m/s respectively.(2)

(b)Find the greatest height above the ground reached by the ball.(3)

(c)Show that the time taken for the ball to reach the ground satisfies 4.9t² - 7t - 6 = 0, and hence find this time, giving your answer to 3 significant figures.(5)

(d)Find the horizontal distance from A to B, giving your answer to 3 significant figures.(2)

(Total for Question 12 is 12 marks)

A particle is projected vertically upwards from the top of a tower 25 m high, with initial speed 18 m/s. Take the ground at the foot of the tower as the origin, with distances measured vertically upwards, and use g = 9.8 m/s².

(a)Find the greatest height above the ground reached by the particle.(3)

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

(c)Show that the times at which the particle is at a height of 40 m above the ground satisfy 4.9t² - 18t + 15 = 0, and hence find both times, giving your answers to 3 significant figures.(4)

(d)Find the time taken for the particle to hit the ground, giving your answer to 3 significant figures.(3)

(Total for Question 13 is 12 marks)

A particle P moves along the x-axis. At time t seconds (t ≥ 0), the acceleration of P is a m/s², where a = 6t - 18. When t = 0, P is at the origin O and moving with velocity 15 m/s in the positive x-direction.

(a)Show that the velocity of P at time t seconds is given by v = 3t² - 18t + 15.(3)

(b)Find the values of t at which P is instantaneously at rest.(2)

(c)Find an expression for the displacement, x metres, of P from O at time t seconds.(3)

(d)Find the total distance travelled by P in the interval 0 ≤ t ≤ 5.(5)

(Total for Question 14 is 13 marks)

Mark scheme · M5 Mechanics: Kinematics Depth

Question 1

(a) M1 uses v = u + at with u = 0, a = 1.5, t = 8
(a) A1 v = 12 (m/s) cao
(a) Answer: 12 m/s
(b) M1 uses s = ut + 0.5at² (or equivalent) with correct values
(b) A1 s = 48 (m) cao
(b) Answer: 48 m

Question 2

(a) M1 substitutes into s = ut + 0.5at² with s = 9, u = 3, a = 0.7: 9 = 3T + 0.35T²
(a) A1 multiplies through by 20 and rearranges to 7T² + 60T - 180 = 0 cso
(a) Answer: 7T² + 60T - 180 = 0 (shown)
(b) M1 uses the quadratic formula on 7T² + 60T - 180 = 0
(b) A1 T = awrt 2.35 (s)
(b) B1 rejects the negative root since T must be positive (time)
(b) Answer: T = 2.35 s (3sf)

Question 3

(a) M1 uses v² = u² - 2gh with v = 0, u = 14, g = 9.8
(a) A1 height above point of projection = 10 (m)
(a) A1 adds 1.2 m: greatest height above ground = 11.2 (m) cao
(a) Answer: 11.2 m
(b) M1 uses s = ut - 0.5gt² with displacement -1.2 m from the point of projection (or equivalent, measuring from the ground)
(b) A1 correct quadratic 4.9t² - 14t - 1.2 = 0 (oe)
(b) A1 t = awrt 2.94 (s), rejecting the negative root
(b) Answer: t = 2.94 s (3sf)

Question 4

(a) M1 finds the distance in the acceleration phase: 0.5(0+30)(40) = 600 (m)
(a) B1 finds the distance in the constant-speed phase: 30 * 90 = 2700 (m)
(a) M1 finds the distance in the deceleration phase: 0.5(30+0)(25) = 375 (m)
(a) A1 total distance = 3675 (m) cao
(a) Answer: 3675 m
(b) M1 uses total time = 40 + 90 + 25 = 155 (s), with their total distance
(b) A1 average speed = awrt 23.7 (m/s) cao
(b) Answer: 23.7 m/s (3sf)

Question 5

(a) B1 displacement of A: s_A = 22t
(a) B1 displacement of B: s_B = t² (from s = 0 + 0.5(2)t²)
(a) Answer: s_A = 22t; s_B = t²
(b) M1 sets s_A = s_B: 22t = t²
(b) A1 t = 22 (s), rejecting t = 0
(b) A1 distance from sign = 484 (m) cao
(b) Answer: t = 22 s; distance = 484 m
(c) M1 uses v = u + at with u = 0, a = 2, their t = 22
(c) A1 v = 44 (m/s) cao
(c) Answer: 44 m/s

Question 6

(a) M1 differentiates x with respect to t
(a) A1 v = 3t² - 12t + 9 cao
(a) Answer: v = 3t² - 12t + 9
(b) M1 differentiates their v with respect to t
(b) A1 a = 6t - 12 cao
(b) Answer: a = 6t - 12
(c) M1 sets v = 0 and divides by 3: t² - 4t + 3 = 0, then factorises
(c) A1 t = 1 and t = 3 cao
(c) Answer: t = 1 s and t = 3 s
(d) M1 substitutes t = 3 into their expression for a
(d) A1 a = 6 (m/s²) cao
(d) Answer: 6 m/s²

Question 7

(a) M1 integrates a with respect to t: v = 4t - 3t² + c
(a) B1 uses t = 0, v = 3 to find c = 3
(a) A1 v = 3 + 4t - 3t² cso
(a) Answer: v = 3 + 4t - 3t² (shown)
(b) M1 integrates their v with respect to t: s = 3t + 2t² - t³ + c, using s = 0 at t = 0 to give c = 0
(b) M1 substitutes t = 2 into their expression for s
(b) A1 s = 6 (m) cao
(b) Answer: 6 m
(c) M1 sets s = 0 and factorises out t: t(3 + 2t - t²) = 0, then solves the quadratic factor
(c) A1 t = 3 (s), rejecting t = -1 since t ≥ 0
(c) Answer: t = 3 s

Question 8

(a) M1 sets v = 0 and factorises: (t-2)(t-4) = 0
(a) A1 t = 2 and t = 4 cao
(a) Answer: t = 2 s and t = 4 s
(b) M1 differentiates v to find a = 2t - 6, and substitutes t = 5
(b) A1 a = 4 (m/s²) cao
(b) B1 speeding up, since v(5) = 3 (> 0) and a = 4 (> 0) act in the same direction (dependent on correct a and correct v(5))
(b) Answer: a = 4 m/s²; R is speeding up
(c) M1 integrates v to find s(t) = t³/3 - 3t² + 8t, with s(0) = 0
(c) dM1 evaluates s at t = 2, t = 4 and t = 6, recognising R changes direction at t = 2 and t = 4
(c) A1 distances 20/3 m, 4/3 m and 20/3 m for the three stages (oe, e.g. decimals)
(c) A1 total distance = 44/3 (m) (oe, awrt 14.7 m) cao
(c) Answer: 44/3 m (awrt 14.7 m)

Question 9

(a) M1 differentiates v with respect to t, using the chain rule
(a) A1 a = 12cos(2t) cao
(a) Answer: a = 12cos(2t)
(b) M1 sets v = 0: 6sin(2t) = 0, so sin(2t) = 0
(b) A1 2t = 0, π, 2pi within the range 0 ≤ 2t ≤ 2pi
(b) A1 t = 0, π/2, π (all three values, oe exact form) cao
(b) Answer: t = 0, π/2, π seconds
(c) M1 integrates v with respect to t: x = -3cos(2t) + c
(c) A1 uses x = 0 at t = 0 to find c = 3
(c) A1 x = 3 - 3cos(2t) (oe, e.g. 3(1 - cos(2t))) cao
(c) Answer: x = 3 - 3cos(2t)
(d) M1 recognises x is greatest when cos(2t) is least, i.e. cos(2t) = -1
(d) A1 this occurs at t = π/2 (matches a value found in part (b) where v = 0, confirming a turning point)
(d) A1 greatest distance = 6 (m) cao
(d) Answer: 6 m, at t = π/2 s

Question 10

(a) M1 uses s = ut + 0.5at² vertically with u = 0, a = 9.8, s = 45
(a) A1 t² = 45/4.9 = awrt 9.18
(a) A1 t = awrt 3.03 (s) cao
(a) Answer: t = 3.03 s (3sf)
(b) M1 horizontal distance = 15 * their t
(b) A1 awrt 45.5 (m) cao
(b) Answer: 45.5 m (3sf)
(c) M1 finds the vertical component of velocity at impact: v_y = gt
(c) A1 v_y = awrt 29.7 (m/s)
(c) M1 combines horizontal and vertical components using Pythagoras: speed = √15² + v_y²
(c) A1 speed = awrt 33.3 (m/s) cao
(c) Answer: 33.3 m/s (3sf)

Question 11

(a) B1 uses tan(θ) = 3/4 to identify sin(θ) = 3/5 and cos(θ) = 4/5 (3-4-5 triangle), and finds horizontal component 21 * 4/5 = 16.8 cso
(a) B1 finds vertical component 21 * 3/5 = 12.6 cso
(a) Answer: 16.8 m/s and 12.6 m/s (shown)
(b) M1 uses v² = u² - 2gh vertically with v = 0, u = 12.6
(b) A1 h = 8.1 (m) cao
(b) Answer: 8.1 m
(c) M1 sets vertical displacement to zero: 0 = 12.6t - 4.9t², factorises to t(12.6 - 4.9t) = 0
(c) A1 rejects t = 0 and solves 4.9t = 12.6
(c) A1 t = 18/7 (s) (oe, awrt 2.57 s) cso
(c) Answer: T = 18/7 s (= 2.57 s to 3sf)
(d) M1 uses range = horizontal component of velocity * time of flight
(d) A1 R = 16.8 * 18/7
(d) A1 R = 43.2 (m) cso
(d) Answer: 43.2 m (shown)

Question 12

(a) B1 uses sin(α) = 7/25 to identify cos(α) = 24/25 (7-24-25 triangle), and finds horizontal component 25 * 24/25 = 24 cso
(a) B1 finds vertical component 25 * 7/25 = 7 cso
(a) Answer: 24 m/s and 7 m/s (shown)
(b) M1 uses v² = u² - 2gh vertically with v = 0, u = 7, to find height above A
(b) A1 height above A = 2.5 (m)
(b) A1 adds 6 m: greatest height above ground = 8.5 (m) cao
(b) Answer: 8.5 m
(c) M1 sets up the vertical displacement from A to the ground (6 m below A) as 7t - 4.9t² = -6
(c) A1 rearranges to 4.9t² - 7t - 6 = 0 cso
(c) dM1 uses the quadratic formula on 4.9t² - 7t - 6 = 0
(c) A1 t = (7 + √166.6)/9.8 evaluated correctly
(c) A1 t = awrt 2.03 (s), rejecting the negative root
(c) Answer: 4.9t² - 7t - 6 = 0 (shown); t = 2.03 s (3sf)
(d) M1 horizontal distance = 24 * their t
(d) A1 awrt 48.8 (m) cao
(d) Answer: 48.8 m (3sf)

Question 13

(a) M1 uses v² = u² - 2gh with v = 0, u = 18, to find height above the tower
(a) A1 height above tower = awrt 16.5 (m)
(a) A1 adds 25 m: greatest height above ground = awrt 41.5 (m) cao
(a) Answer: 41.5 m (3sf)
(b) M1 uses v = u - gt with v = 0, u = 18
(b) A1 t = awrt 1.84 (s) cao
(b) Answer: 1.84 s (3sf)
(c) M1 sets height 25 + 18t - 4.9t² = 40 and rearranges
(c) A1 4.9t² - 18t + 15 = 0 cso
(c) M1 uses the quadratic formula, with discriminant 18² - 4(4.9)(15) = 30
(c) A1 t = awrt 1.28 (s) and t = awrt 2.40 (s), both values given
(c) Answer: 4.9t² - 18t + 15 = 0 (shown); t = 1.28 s and t = 2.40 s (3sf)
(d) M1 sets 25 + 18t - 4.9t² = 0, i.e. 4.9t² - 18t - 25 = 0
(d) A1 uses the quadratic formula with discriminant 18² + 4(4.9)(25) = 814
(d) A1 t = awrt 4.75 (s), rejecting the negative root
(d) Answer: 4.75 s (3sf)