Mechanics: Quantities, Units and Kinematics (Part 2) - Worksheets, Questions and Revision

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A-Level · Mechanics

M1B Mechanics: Quantities, Units and Kinematics (Part 2)

EDEXCEL 9MA0 · Calculator allowed · about 160 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question concerns quantities, units and unit conversions used in mechanics.

(a)State the SI unit used to measure (i) force, (ii) velocity, (iii) acceleration.(3)

(b)A quantity has units kg m s^-2. State what physical quantity this represents.(1)

(c)A cricket ball is bowled at a speed of 89 mph. Given that 1 mile = 1609 m, find this speed in m/s, giving your answer to 3 significant figures.(3)

(d)A student is asked to convert 54 km/h into m/s and writes: '54 km/h = 54 / 1000 / 3600 = 1.5 x 10^-5 m/s'. Identify the error in the student's method, and hence find the correct value of 54 km/h in m/s.(2)

(Total for Question 1 is 9 marks)

A go-kart moves along a straight track with constant acceleration. The table below shows its velocity v m/s at time t seconds after it passes a marker post P.

t (s): 0, 4, 8, 12
v (m/s): 3, 11, 19, 27

(a)Show that the go-kart's acceleration is constant, and state its value.(3)

(b)Using the go-kart's speed at P (t = 0), calculate the distance it travels from P to the point where t = 12.(3)

(c)After t = 12, the go-kart decelerates uniformly, coming to rest 6 seconds later. Find the magnitude of this deceleration.(2)

(d)Find the total distance travelled by the go-kart from P until it comes to rest.(3)

(Total for Question 2 is 11 marks)

A ball is thrown vertically upwards from a point 1.5 m above horizontal ground with speed 14 m/s. The ball is modelled as a particle moving freely under gravity, with g = 9.8 m/s².

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

(b)Find the time taken for the ball to return to the point from which it was thrown.(2)

(c)Find the total time from the instant of release until the ball hits the ground, giving your answer to 3 significant figures.(4)

(d)State, with a reason, the likely effect on the ball's actual maximum height, compared with the value found in part (a), if air resistance is not in fact negligible.(1)

(Total for Question 3 is 10 marks)

A remote-controlled toy car moves along a straight track. Its displacement s metres from a fixed point O at time t seconds is shown in the displacement-time graph, which consists of three straight line segments joining the points (0, 0), (5, 15), (9, 15) and (14, -5).

(a)State what feature of a displacement-time graph represents the velocity of the car.(1)

(b)Find the velocity of the car during the interval 0 ≤ t ≤ 5.(2)

(c)Describe the motion of the car during the interval 5 ≤ t ≤ 9.(1)

(d)Find the velocity of the car during the interval 9 ≤ t ≤ 14, and state what the sign of your answer tells you about its direction of motion.(3)

(e)Find the value of t at which the car passes through O.(3)

(f)Find the total distance travelled by the car during the interval 0 ≤ t ≤ 14.(3)

(Total for Question 4 is 13 marks)

A ship S has position vector (2i + 5j) km relative to a lighthouse L at the instant t = 0, where time is measured in hours and i and j are unit vectors due east and due north respectively. The ship moves with constant velocity (6i - 3j) km/h.

(a)Find the position vector of S, relative to L, at t = 2.(2)

(b)Find the speed of the ship, giving your answer to 3 significant figures.(2)

(c)Find the time at which the ship is due east of the lighthouse.(2)

(d)At t = 3, a rescue boat sets off from the lighthouse L and travels directly towards the ship's position at that instant, at a constant speed of 25 km/h. Find, to the nearest minute, how long the rescue boat takes to reach that position.(4)

(Total for Question 5 is 10 marks)

A cyclist rides from a depot D to a shop S, a distance of 4.5 km, at a constant average speed of 15 km/h. She stays at the shop for 6 minutes, then cycles back to D along the same route at a constant average speed of 12 km/h.

(a)Find the total time for which the cyclist is actually moving (that is, excluding the 6-minute stop), giving your answer in minutes.(2)

(b)Find her average speed for the whole 9 km round trip, using only the time she is actually moving, giving your answer in km/h to 3 significant figures.(2)

(c)State the value of her average velocity for the whole round trip (from leaving D to returning to D), including the time spent at the shop, and explain your answer.(2)

(d)Explain why the average speed found in part (b) is not equal to the mean of her two average speeds, 15 km/h and 12 km/h.(2)

(Total for Question 6 is 8 marks)

Two cyclists, A and B, start 250 m apart on a straight, flat cycle path and move directly towards each other, setting off at the same instant. Cyclist A moves at a constant speed of 6 m/s. Cyclist B starts from rest and moves with constant acceleration 0.4 m/s². Let t seconds be the time after they set off.

(a)Write down, in terms of t, an expression for the distance travelled by cyclist A, and an expression for the distance travelled by cyclist B.(2)

(b)Show that, at the instant the cyclists meet, t² + 30t - 1250 = 0.(2)

(c)Hence find the time at which the cyclists meet, giving your answer to 3 significant figures.(3)

(d)Find the distance of the meeting point from cyclist A's starting point.(2)

(e)State one limitation of modelling the two cyclists as particles in this context.(1)

(Total for Question 7 is 10 marks)

A delivery robot starts from rest at one end of a straight corridor and accelerates uniformly at 0.5 m/s² for T seconds, reaching a maximum speed of V m/s. It then decelerates uniformly at 1 m/s², coming to rest exactly at the far end of the corridor, a distance of 54 m from its starting point.

(a)Show that T = 12.(4)

(b)Find the robot's maximum speed, V.(2)

(c)Find the total time taken by the robot to travel the length of the corridor.(2)

(d)Sketch the velocity-time graph for the whole journey, marking the coordinates of the point at which the velocity is greatest.(3)

(e)Find the robot's average speed for the whole 54 m journey.(2)

(Total for Question 8 is 13 marks)

A particle P moves along the x-axis. At time t seconds (0 ≤ t ≤ 6), the velocity of P is v m/s, where v = 3t² - 16t + 20.

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

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

(c)Given that P is at the point where x = 5 when t = 0, find an expression for the displacement x of P from the origin, in terms of t.(3)

(d)Find the total distance travelled by P during the interval 0 ≤ t ≤ 4.(4)

(Total for Question 9 is 12 marks)

A particle P moves in a plane. At time t seconds (t ≥ 0), the position vector of P relative to a fixed origin O is r = (2t² - 8t)i + ((1/3)t³ - 2t)j metres, where i and j are unit vectors due east and due north respectively.

(a)Find v, the velocity of P at time t.(2)

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

(c)Find the exact value of t, where t > 0, at which P is moving parallel to i.(2)

(d)Find, in the form π + qj, the acceleration of P at the instant found in part (b).(2)

(e)Find, to the nearest degree, the bearing on which P is travelling at the instant t = 0.(3)

(Total for Question 10 is 11 marks)

A particle moves in a straight line with constant acceleration. It passes through three points D, E and F, in that order, at times t=0, t=T and t=2T respectively, where T is a positive constant. The speeds of the particle at D, E and F are d m/s, e m/s and f m/s respectively, where d, e, f > 0.

(a)Show that e = (d + f)/2.(4)

(b)Given that d=5 and f=17, find the value of e.(1)

(c)Given further that T=4, find the acceleration of the particle.(2)

(d)Find the distance DF.(3)

(Total for Question 11 is 10 marks)

Two particles, A and B, are 200 m apart on a straight horizontal road and move directly towards each other, starting at the same instant. Particle A starts with speed 4 m/s and accelerates at a constant 1.2 m/s² towards B. Particle B starts with speed 10 m/s and decelerates at a constant 0.6 m/s² towards A. Both particles are modelled as moving in a straight line until either they collide or B comes to rest, whichever happens first.

(a)Show that, provided the particles collide before B comes to rest, the time t seconds at which they meet satisfies 3t² + 140t - 2000 = 0.(4)

(b)Solve the equation in part (a) to find the time at which the particles meet, giving your answer to 3 significant figures.(3)

(c)Find the time at which B would come to rest if it had not already collided with A, and use this to verify that the collision found in part (b) occurs while B is still moving.(3)

(d)Find the distance from A's starting point at which the collision occurs, giving your answer to 3 significant figures.(2)

(Total for Question 12 is 12 marks)

Mark scheme · M1B Mechanics: Quantities, Units and Kinematics (Part 2)

Question 1

(a) B1 newton (N) for force
(a) B1 metre per second (m/s) for velocity
(a) B1 metre per second squared (m/s²) for acceleration
(a) Answer: (i) newton, N (ii) metre per second, m/s (iii) metre per second squared, m/s²
(b) B1 force (since F = ma has units kg x m/s²)
(b) Answer: Force.
(c) M1 convert 89 miles to metres: 89 x 1609 = 143201
(c) M1 divide by 3600 to convert per hour to per second
(c) A1 awrt 39.8 m/s
(c) Answer: 39.8 m/s (3 s.f.)
(d) M1 identify the error: to convert km/h to m/s you multiply by 1000 (km to m) and divide by 3600 (per hour to per second); the student divided by 1000 instead of multiplying
(d) A1 15 m/s cao
(d) Answer: The student divided by 1000 instead of multiplying; the correct value is 15 m/s.

Question 2

(a) M1 calculate at least two of the gradients between consecutive readings, e.g. (11-3)/4 and (19-11)/4
(a) A1 cso: all three gradients equal 2, confirming constant acceleration
(a) B1 a = 2 m/s²
(a) Answer: Constant acceleration confirmed; a = 2 m/s².
(b) M1 use s = ut + 1/2 a t² with u = 3, a = 2 (ft from (a)), t = 12
(b) M1 s = 3(12) + 0.5(2)(12²) correctly evaluated
(b) A1 180 m cao
(b) Answer: 180 m
(c) M1 a = (0 - 27)/6
(c) A1 4.5 m/s² (deceleration) cao
(c) Answer: 4.5 m/s²
(d) M1 distance during deceleration phase = 1/2(27+0)(6), ft from (c)
(d) M1 add to the distance found in part (b), ft
(d) A1 261 m cao
(d) Answer: 261 m

Question 3

(a) M1 use v² = u² - 2gs with v=0, u=14, g=9.8 to find the height above the point of release
(a) M1 s = 14²/(2 x 9.8) correctly evaluated
(a) A1 11.5 m cao (height above release point of 10 m, plus the initial 1.5 m)
(a) Answer: 11.5 m
(b) M1 t = 2u/g
(b) A1 awrt 2.86 s
(b) Answer: 2.86 s (3 s.f.)
(c) M1 form 4.9t² - 14t - 1.5 = 0 (or equivalent using s = ut - 1/2 g t² with displacement -1.5 from the release point)
(c) M1 rearrange correctly into a solvable quadratic
(c) M1 apply the quadratic formula
(c) A1 awrt 2.96 s, rejecting the negative root
(c) Answer: 2.96 s (3 s.f.)
(d) B1 the actual maximum height would be less than 11.5 m, because air resistance would act against the ball's motion as it rises, decelerating it further (oe)
(d) Answer: The actual maximum height would be less than 11.5 m, because air resistance decelerates the ball further as it rises (oe).

Question 4

(a) B1 the gradient (slope) of the graph
(a) Answer: The gradient of the displacement-time graph.
(b) M1 gradient = (15-0)/(5-0)
(b) A1 3 m/s cao
(b) Answer: 3 m/s
(c) B1 the car is stationary (at rest), remaining 15 m from O
(c) Answer: The car is stationary, 15 m from O.
(d) M1 gradient = (-5-15)/(14-9)
(d) A1 -4 m/s cao
(d) B1 the negative sign shows the car is moving in the negative direction (back towards and past O)
(d) Answer: -4 m/s; the car is moving in the negative direction, back towards and through O.
(e) M1 set up a linear equation for the third segment, e.g. s = 15 - 4(t-9), ft from (d)
(e) M1 solve 15 - 4(t-9) = 0
(e) A1 t = 12.75 cao
(e) Answer: t = 12.75 s
(f) M1 distance in the first segment = 15 (0 to 5), distance in the second segment = 0 (5 to 9)
(f) M1 distance in the third segment = |(-5)-15| = 20 (9 to 14)
(f) A1 35 m cao, not the net displacement of -5 m
(f) Answer: 35 m

Question 5

(a) M1 r = r0 + vt applied with t=2
(a) A1 (14i - j) km cao
(a) Answer: (14i - j) km
(b) M1 speed = √6² + 3²
(b) A1 awrt 6.71 km/h
(b) Answer: 6.71 km/h (3 s.f.)
(c) M1 set the j-component of the position vector equal to 0: 5-3t=0
(c) A1 t = 5/3 hours oe (1 hour 40 minutes), with the i-component positive confirming due east
(c) Answer: t = 5/3 hours (1 hour 40 minutes)
(d) M1 find the position vector at t=3: 20i - 4j
(d) M1 find the magnitude of this vector: √20² + 4²
(d) M1 divide by 25 and convert the time from hours to minutes
(d) A1 awrt 49 minutes
(d) Answer: 49 minutes (nearest minute)

Question 6

(a) M1 time out = 4.5/15, time back = 4.5/12 (in hours or minutes)
(a) A1 40.5 minutes cao
(a) Answer: 40.5 minutes
(b) M1 average speed = total distance/moving time, ft from (a)
(b) A1 awrt 13.3 km/h
(b) Answer: 13.3 km/h (3 s.f.)
(c) B1 0 km/h
(c) B1 because her overall displacement is zero, as she returns to her starting point D (oe)
(c) Answer: 0 km/h, because her overall displacement is zero (she finishes back at D).
(d) B1 the mean of 15 and 12 is 13.5 km/h, which does not equal the 13.3 km/h found in (b)
(d) B1 because average speed (total distance/total time) weights each leg by the time spent travelling it, and she spends unequal times on the two equal-length legs since her speeds differ (oe)
(d) Answer: The simple mean (13.5 km/h) ignores that she spends different amounts of time on each leg; average speed is weighted by time, not by simply averaging the two speeds.

Question 7

(a) B1 A: 6t
(a) B1 B: 0.2t²
(a) Answer: A: 6t; B: 0.2t² (metres)
(b) M1 6t + 0.2t² = 250, ft from (a)
(b) A1 cso: multiply by 5 to obtain t² + 30t - 1250 = 0
(b) Answer: t² + 30t - 1250 = 0 (shown)
(c) M1 apply the quadratic formula to t²+30t-1250=0
(c) A1 reject the negative root
(c) A1 awrt 23.4 s
(c) Answer: t = 23.4 s (3 s.f.)
(d) M1 substitute t=23.406 (ft) into 6t
(d) A1 awrt 140 m
(d) Answer: 140 m (3 s.f.)
(e) B1 any valid limitation, e.g. it ignores the length/size of the bicycles and riders, so in reality they could meet (or need to brake) slightly before the calculated point; or it ignores reaction time
(e) Answer: Modelling the cyclists as particles ignores the size of the bicycles and riders, so in practice they would meet slightly before the calculated point (oe).

Question 8

(a) M1 V = 0.5T
(a) M1 distance in the acceleration phase = 1/2(0.5)(T²) = 0.25T² oe
(a) M1 deceleration phase lasts V/1 = 0.5T seconds and covers 1/2(V)(0.5T) = 0.125T² oe; form the total-distance equation 0.25T²+0.125T²=54, dependent on both previous M marks
(a) A1 cso: 0.375T²=54 gives T²=144, T=12
(a) Answer: T = 12 (shown)
(b) M1 V = 0.5(12), ft from (a)
(b) A1 6 m/s cao
(b) Answer: 6 m/s
(c) M1 deceleration time = V/1 = 6, ft; total time = 12+6
(c) A1 18 s cao
(c) Answer: 18 s
(d) B1 correct overall shape: straight line rising from (0,0), straight line falling to the t-axis, both with positive velocity
(d) B1 t-values 12 and 18 marked on the horizontal axis, ft
(d) B1 v-value 6 marked on the vertical axis at t=12, ft
(d) Answer: Straight lines from (0,0) to (12,6), then (12,6) to (18,0).
(e) M1 average speed = 54/18, ft from (c)
(e) A1 3 m/s cao
(e) Answer: 3 m/s

Question 9

(a) M1 differentiate v with respect to t
(a) A1 a = 6t - 16 cao
(a) Answer: a = 6t - 16
(b) M1 set v=0: 3t²-16t+20=0
(b) M1 apply the quadratic formula (or factorise)
(b) A1 t=2 and t=10/3 (awrt 3.33), both cao
(b) Answer: t = 2 and t = 10/3 (awrt 3.33)
(c) M1 integrate v with respect to t: x = t³-8t²+20t+C
(c) A1 correct unsimplified integral, ft
(c) A1 use x(0)=5 to find C=5, giving x=t³-8t²+20t+5 cao
(c) Answer: x = t³ - 8t² + 20t + 5
(d) M1 evaluate x at the turning points found in (b) and at the interval endpoints: t=0, 2, 10/3, 4, ft
(d) A1 x(0)=5, x(2)=21, x(10/3)=535/27 (awrt 19.8), x(4)=21
(d) M1 sum the magnitudes of the changes between consecutive values, since P changes direction at each turning point
(d) A1 awrt 18.4 m
(d) Answer: 18.4 m (3 s.f.)

Question 10

(a) M1 differentiate the i-component: 4t-8
(a) A1 differentiate the j-component: t²-2, both correct
(a) Answer: v = (4t - 8)i + (t² - 2)j
(b) M1 set the i-component of v to 0: 4t-8=0
(b) A1 t=2 cao
(b) Answer: t = 2 s
(c) M1 set the j-component of v to 0: t²-2=0
(c) A1 t=√2, rejecting the negative root
(c) Answer: t = √2 s
(d) M1 differentiate v to find a(t)=4i+2tj
(d) A1 substitute t=2, ft, to give 4i+4j
(d) Answer: a = 4i + 4j (m/s²)
(e) M1 find v(0) = -8i - 2j
(e) M1 use right-angled trigonometry, e.g. the angle south of due west = arctan(2/8)
(e) A1 awrt 256 degrees
(e) Answer: 256 degrees (nearest degree)

Question 11

(a) M1 use v=u+at at E: e=d+aT
(a) M1 use v=u+at at F: f=d+2aT
(a) M1 eliminate a, e.g. substitute aT=e-d into the equation for f
(a) A1 cso: rearrange to e=(d+f)/2
(a) Answer: e = (d + f)/2 (shown)
(b) B1 e=11, ft from (a)
(b) Answer: e = 11 m/s
(c) M1 a=(e-d)/T, ft
(c) A1 1.5 m/s² cao
(c) Answer: 1.5 m/s²
(d) M1 use the average velocity over D to F, (d+f)/2, or an equivalent SUVAT method, ft
(d) M1 multiply by the total time 2T, ft
(d) A1 88 m cao
(d) Answer: 88 m