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A-Level · Mechanics

M1 Mechanics: Quantities, Units and Kinematics

EDEXCEL 9MA0 · Calculator allowed · about 160 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question concerns quantities and units used in mechanics.

(a)State the SI unit used to measure (i) mass, (ii) time, (iii) displacement.(3)

(b)A runner completes a distance of 3.2 km in a time of 11 minutes 40 seconds, moving at a constant speed. Calculate the runner's average speed in m/s, giving your answer correct to 3 significant figures.(3)

(c)The mass of a small ball bearing is 4.5 grams. Express this mass in kilograms, giving your answer in standard form.(2)

(Total for Question 1 is 8 marks)

A car accelerates uniformly from rest to a speed of 24 m/s in 15 seconds. It then travels at this constant speed for a further 20 seconds before decelerating uniformly to rest in 8 seconds. The car is modelled as a particle moving in a straight line.

(a)Find the acceleration of the car during the first 15 seconds.(2)

(b)Find the distance travelled by the car during the first 15 seconds.(2)

(c)Find the deceleration of the car during the final 8 seconds.(2)

(d)Calculate the total distance travelled by the car during the whole 43 second journey.(3)

(e)Hence find the average speed of the car for the whole journey, giving your answer to 3 significant figures.(2)

(Total for Question 2 is 11 marks)

A stone is released from rest at the top of a vertical cliff and hits the sea 3.2 s later. The stone is modelled as a particle moving freely under gravity, with g = 9.8 m/s^2.

(a)Find the speed with which the stone hits the sea.(2)

(b)Find the height of the cliff above sea level.(3)

(c)State one modelling assumption, other than that the stone is a particle, that has been used in this question.(1)

(d)A second, identical stone is thrown vertically downwards from the same point on the cliff with initial speed 5 m/s. Using the height found in part (b), find how long this second stone takes to reach the sea, giving your answer to 3 significant figures.(3)

(Total for Question 3 is 9 marks)

The velocity-time graph for a cyclist's journey along a straight road consists of three straight line segments joining the points (0,0), (6,9), (18,9) and (24,0), where t is time in seconds and v is velocity in m/s.

(a)State what feature of a velocity-time graph represents the cyclist's acceleration.(1)

(b)Find the acceleration of the cyclist during the interval 0 ≤ t ≤ 6.(2)

(c)Find the acceleration of the cyclist during the interval 18 ≤ t ≤ 24.(2)

(d)Using the graph, calculate the total distance travelled by the cyclist.(3)

(e)Find the average speed of the cyclist for the whole journey.(2)

(Total for Question 4 is 10 marks)

At time t = 0, a particle P has position vector (-6i - 5j) m relative to a fixed origin O. The particle moves with constant velocity (3i + 4j) m/s. Unit vectors i and j are due east and due north respectively.

(a)Find the position vector of P at time t = 3 seconds.(2)

(b)Find the speed of P.(2)

(c)Find the value of t at which P is due north of O.(3)

(d)Find the distance of P from O when t = 5, giving your answer to 3 significant figures.(3)

(Total for Question 5 is 10 marks)

A particle moves in a straight line with constant acceleration. It passes through a point A with speed 4 m/s and, 6 seconds later, passes through a point B with speed 19 m/s. The distance AB is d metres.

(a)Find the acceleration of the particle.(2)

(b)Show that d = 69.(3)

(c)The particle continues to move with the same constant acceleration. Find the speed of the particle 2 seconds after it passes through B.(2)

(d)Find the total distance travelled by the particle from A until 2 seconds after it passes through B.(3)

(Total for Question 6 is 10 marks)

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

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

(b)Find an expression for the acceleration a m/s² of P at time t.(1)

(c)Find the values of t for which P is instantaneously at rest.(3)

(d)Find the acceleration of P at each of the times found in part (c), and hence determine which of these times corresponds to a local maximum displacement.(3)

(e)Find the total distance travelled by P in the interval 0 ≤ t ≤ 3.(4)

(Total for Question 7 is 13 marks)

A particle moves in a straight line. At time t seconds, the acceleration of the particle is a m/s², where a = 6 - 2t. When t = 0, the particle is at a fixed origin O and is moving with velocity 1 m/s in the positive direction.

(a)Find an expression for v, the velocity of the particle, in terms of t.(3)

(b)Find an expression for x, the displacement of the particle from O, in terms of t.(3)

(c)Find the maximum velocity of the particle and the value of t at which it occurs.(3)

(d)Find, to 3 significant figures, the value of t (other than t = 0) at which the particle returns to O.(4)

(Total for Question 8 is 13 marks)

At time t = 0, car A passes a fixed point O travelling in a straight line with speed 8 m/s and constant acceleration 1.5 m/s^2. At the same instant, car B is 10 m ahead of O on the same straight line, travelling in the same direction as A with constant speed 12 m/s (zero acceleration). Both cars are modelled as particles.

(a)Write down an expression, in terms of t, for the distance of A from O at time t.(1)

(b)Write down an expression, in terms of t, for the distance of B from O at time t.(1)

(c)Show that, at the time when A catches up with B, 3t² - 16t - 40 = 0.(3)

(d)Hence find the time at which A catches up with B, giving your answer to 3 significant figures.(3)

(e)Find the distance of A from O at the time found in part (d).(2)

(Total for Question 9 is 10 marks)

A particle P moves in a plane so that at time t seconds its velocity v m/s is given by v = (2t - 3)i + (4 - t)j. When t = 0, the position vector of P relative to a fixed origin O is (5i + 2j) m.

(a)Find the position vector of P at time t.(4)

(b)Show that the acceleration of P is constant, and find the magnitude of this acceleration.(4)

(c)Find the value of t at which P is moving parallel to the vector i.(2)

(d)Find the speed of P at the instant found in part (c).(3)

(Total for Question 10 is 13 marks)

A particle travels in a straight line from a point A to a point B, taking a total time of 40 seconds. The particle's velocity-time graph for the journey consists of three straight stages: Stage 1 (0 ≤ t ≤ 15): the particle accelerates uniformly from speed u m/s to speed 10 m/s, with constant acceleration 0.4 m/s^2. Stage 2 (15 ≤ t ≤ 15+T): the particle travels at a constant speed of 10 m/s for T seconds. Stage 3: the particle decelerates uniformly from 10 m/s to rest, arriving at B when t = 40. The total distance from A to B is 275 m.

(a)Find the value of u.(2)

(b)Find the distance travelled during Stage 1.(2)

(c)Show that T = 9.(4)

(d)Sketch the velocity-time graph for the whole journey, marking the values of t and v at each point where the gradient changes.(3)

(e)Find the deceleration during Stage 3.(3)

(Total for Question 11 is 14 marks)

A particle P moves in a straight line with constant acceleration. It passes through three points A, B and C, in that order, where B is the midpoint of AC. The speeds of P at A, B and C are p m/s, q m/s and r m/s respectively, where p, q, r > 0.

(a)Show that 2q² = p² + r^2.(4)

(b)Given that p = 3 and q = 5, find the value of r.(3)

(Total for Question 12 is 7 marks)

Mark scheme · M1 Mechanics: Quantities, Units and Kinematics

Question 1

(a) B1 kilogram (kg) for mass
(a) B1 second (s) for time
(a) B1 metre (m) for displacement
(a) Answer: (i) kilogram, kg (ii) second, s (iii) metre, m
(b) M1 convert 3.2 km to 3200 m and 11 min 40 s to 700 s
(b) M1 speed = distance / time = 3200 / 700 oe
(b) A1 awrt 4.57 m/s
(b) Answer: 4.57 m/s (3 s.f.)
(c) M1 divide by 1000 (1 g = 10^-3 kg)
(c) A1 4.5 x 10^-3 kg cao
(c) Answer: 4.5 x 10^-3 kg

Question 2

(a) M1 use a = (v - u)/t with v=24, u=0, t=15
(a) A1 1.6 m/s² cao
(a) Answer: 1.6 m/s²
(b) M1 s = 1/2(u+v)t with u=0, v=24, t=15, or ft from (a)
(b) A1 180 m cao
(b) Answer: 180 m
(c) M1 a = (0-24)/8
(c) A1 3 m/s² (deceleration) cao
(c) Answer: 3 m/s²
(d) M1 distance at constant speed = 24 x 20 = 480 m
(d) M1 distance decelerating = 1/2(24+0)(8) = 96 m
(d) A1 756 m cao (180+480+96)
(d) Answer: 756 m
(e) M1 average speed = total distance / total time = 756/43 ft from (d)
(e) A1 awrt 17.6 m/s
(e) Answer: 17.6 m/s (3 s.f.)

Question 3

(a) M1 v = u + at with u=0, a=9.8, t=3.2
(a) A1 awrt 31.4 m/s
(a) Answer: 31.4 m/s (3 s.f.)
(b) M1 s = ut + 1/2 a t² with u=0, a=9.8, t=3.2
(b) M1 s = 0.5*9.8*3.2² correctly evaluated
(b) A1 awrt 50.2 m
(b) Answer: 50.2 m (3 s.f.)
(c) B1 any valid assumption, e.g. air resistance is negligible / g is constant near the Earth's surface / the motion is vertical
(c) Answer: Air resistance on the stone is negligible (oe).
(d) M1 form 4.9t² + 5t - 50.176 = 0 (or equivalent using s=ut+1/2at²)
(d) M1 solve using the quadratic formula and reject the negative root
(d) A1 awrt 2.73 s
(d) Answer: 2.73 s (3 s.f.)

Question 4

(a) B1 the gradient (slope) of the graph
(a) Answer: The gradient of the velocity-time graph.
(b) M1 gradient = (9-0)/(6-0)
(b) A1 1.5 m/s² cao
(b) Answer: 1.5 m/s²
(c) M1 gradient = (0-9)/(24-18)
(c) A1 -1.5 m/s², i.e. a deceleration of 1.5 m/s² cao
(c) Answer: -1.5 m/s² (deceleration of 1.5 m/s²)
(d) M1 area of triangle (0 to 6) + rectangle (6 to 18) + triangle (18 to 24)
(d) M1 0.5*6*9 + 12*9 + 0.5*6*9 correctly evaluated
(d) A1 162 m cao
(d) Answer: 162 m
(e) M1 average speed = total distance/total time, ft from (d)
(e) A1 6.75 m/s cao
(e) Answer: 6.75 m/s

Question 5

(a) M1 r = r0 + vt applied with t=3
(a) A1 (3i + 7j) m cao
(a) Answer: (3i + 7j) m
(b) M1 speed = √3² + 4²
(b) A1 5 m/s cao
(b) Answer: 5 m/s
(c) M1 set the i-component of position vector equal to 0: -6+3t=0
(c) A1 t = 2 (s)
(c) A1 cao, with j-component positive confirming due north (not south)
(c) Answer: t = 2 s
(d) M1 find position vector at t=5: (9i+15j)
(d) M1 distance = √9²+15²
(d) A1 awrt 17.5 m
(d) Answer: 17.5 m (3 s.f.)

Question 6

(a) M1 a = (19-4)/6
(a) A1 2.5 m/s² cao
(a) Answer: 2.5 m/s²
(b) M1 use s = ut + 1/2 a t² with u=4, a=2.5 (ft from (a)), t=6
(b) M1 s = 4*6 + 0.5*2.5*36 evaluated
(b) A1 cso: d = 24+45 = 69
(b) Answer: d = 69 m (shown)
(c) M1 v = 19 + 2.5*2
(c) A1 24 m/s cao
(c) Answer: 24 m/s
(d) M1 total time from A = 6+2 = 8 s
(d) M1 s = 4*8 + 0.5*2.5*8² (or 69 + distance B to final point)
(d) A1 112 m cao
(d) Answer: 112 m

Question 7

(a) M1 differentiate x with respect to t
(a) A1 v = 3t² - 12t + 9 cao
(a) Answer: v = 3t² - 12t + 9
(b) B1 a = 6t - 12 (ft from (a))
(b) Answer: a = 6t - 12
(c) M1 set v=0: 3t²-12t+9=0, divide by 3
(c) M1 factorise (t-1)(t-3)=0
(c) A1 t=1 and t=3 cao
(c) Answer: t = 1 and t = 3
(d) B1 a(1) = -6 m/s²
(d) B1 a(3) = 6 m/s²
(d) A1 t=1 gives a local maximum (a<0 means v changes from + to -, so x has a maximum)
(d) Answer: a(1)=-6, a(3)=6; t=1 gives the local maximum displacement.
(e) M1 find x(0), x(1), x(3)
(e) A1 x(0)=0, x(1)=4, x(3)=0
(e) M1 distance = |x(1)-x(0)| + |x(3)-x(1)|
(e) A1 8 m cao
(e) Answer: 8 m

Question 8

(a) M1 integrate a with respect to t: v = 6t - t² + C
(a) A1 correct unsimplified integral
(a) A1 use v(0)=1 to find C=1, so v=6t-t²+1 cao
(a) Answer: v = 6t - t² + 1
(b) M1 integrate v with respect to t: x = 3t² - t³/3 + t + C2
(b) A1 correct unsimplified integral, ft from (a)
(b) A1 use x(0)=0 to find C2=0, so x=3t²-(t³)/3+t cao
(b) Answer: x = 3t² - (t³)/3 + t
(c) M1 set dv/dt = a = 0: 6-2t=0
(c) A1 t=3
(c) A1 v(3)=10 m/s cao
(c) Answer: Maximum velocity is 10 m/s, occurring at t = 3 s.
(d) M1 set x=0 and factor out t: t(-t²/3+3t+1)=0
(d) M1 form the quadratic t²-9t-3=0 (multiplying through by -3)
(d) M1 solve using the quadratic formula and reject the negative root
(d) A1 awrt 9.32 s
(d) Answer: t = 9.32 s (3 s.f.)

Question 9

(a) B1 8t + 0.75t² oe
(a) Answer: 8t + 0.75t²
(b) B1 10 + 12t
(b) Answer: 10 + 12t
(c) M1 set 8t+0.75t² = 10+12t, ft from (a) and (b)
(c) M1 rearrange to 0.75t²-4t-10=0
(c) A1 cso: multiply through by 4 to give 3t²-16t-40=0
(c) Answer: 3t² - 16t - 40 = 0 (shown)
(d) M1 apply the quadratic formula to 3t²-16t-40=0
(d) A1 reject the negative root
(d) A1 awrt 7.19 s
(d) Answer: t = 7.19 s (3 s.f.)
(e) M1 substitute t=7.19 (ft) into 8t+0.75t² or 10+12t
(e) A1 awrt 96.3 m
(e) Answer: 96.3 m (3 s.f.)

Question 10

(a) M1 integrate the i-component: t²-3t+C1
(a) A1 use r_i(0)=5 to give i-component t²-3t+5
(a) M1 integrate the j-component: 4t-0.5t²+C2
(a) A1 use r_j(0)=2 to give j-component 4t-0.5t²+2 cao
(a) Answer: r = (t² - 3t + 5)i + (4t - 0.5t² + 2)j
(b) B1 a = dv/dt = 2i - j
(b) B1 state a does not depend on t, so acceleration is constant
(b) M1 magnitude = √2²+(-1)²
(b) A1 awrt 2.24 m/s²
(b) Answer: a = (2i - j) m/s², constant; magnitude 2.24 m/s² (3 s.f.)
(c) M1 set the j-component of v equal to 0: 4-t=0
(c) A1 t=4 cao
(c) Answer: t = 4 s
(d) M1 substitute t=4 (ft) into v to get 5i+0j
(d) A1 correct components
(d) A1 speed = 5 m/s cao
(d) Answer: 5 m/s

Question 11

(a) M1 use v=u+at with v=10, a=0.4, t=15, rearranged for u
(a) A1 u=4 m/s cao
(a) Answer: u = 4 m/s
(b) M1 s = 1/2(u+v)t = 0.5*(4+10)*15, ft from (a)
(b) A1 105 m cao
(b) Answer: 105 m
(c) M1 distance in Stage 2 = 10T
(c) M1 Stage 3 duration = 40-15-T = 25-T, distance = 1/2*10*(25-T) = 125-5T
(c) dM1 form the total distance equation 105+10T+125-5T=275, dependent on both previous M marks
(c) A1 cso: solve to give T=9
(c) Answer: T = 9 (shown)
(d) B1 correct overall shape: rising line, horizontal line, falling line, all with positive velocity
(d) B1 t-values 15, 24 (15+T with T=9, ft) and 40 marked on the horizontal axis
(d) B1 v-values 4, 10 and 0 marked on the vertical axis at the correct points
(d) Answer: Straight lines from (0,4) to (15,10), (15,10) to (24,10), and (24,10) to (40,0).
(e) M1 Stage 3 duration = 25-T = 25-9=16 s, ft from (c)
(e) M1 a = (0-10)/16
(e) A1 0.625 m/s² cao
(e) Answer: 0.625 m/s²

Question 12

(a) M1 let AB=BC=d and constant acceleration a; use v²=u²+2as for A to B: q²=p²+2ad
(a) M1 use v²=u²+2as for B to C: r²=q²+2ad
(a) M1 eliminate 2ad by equating q²-p² = r²-q²
(a) A1 cso: rearrange to 2q² = p²+r²
(a) Answer: 2q² = p² + r² (shown)
(b) M1 substitute p=3, q=5 into 2q²=p²+r², ft from (a)
(b) M1 r² = 50-9 = 41
(b) A1 awrt 6.40 m/s
(b) Answer: r = 6.40 m/s (3 s.f.)

Mechanics: Quantities, Units and Kinematics - Worksheets, Questions and Revision