M107 assessment Kinematics 1 SOLUTIONS

Do the questions as a test – circle questions you cannot answer

Red

1) Frederica drops a stone down a well. She measures the time it takes to hit the water at the bottom as 4 seconds. Estimate the depth of the well and state two assumptions you have made in modelling your solution s=¿ s u=0 v=¿ a=+9.8 t=4

s=ut+ 12

a t 2 ¿0+ 12

(9.8 ) (16 )=¿

2) A Ferrari accelerates constantly for 3 seconds to reach the speed limit of 30 mphIf the distance covered in the 3 seconds is 20 m, calculate the initial speed of the car. Use 1 mile = 1.6 km s=20 u=u

v=30× 10001.6 × 3600

=403 m/s

a=¿ t=3

s=( u+v2 )t=(u+40 /3

2 )3=20 gives u=0m/s

3)

A car moves along a straight horizontal road. At time t=0, the velocity of the car is U ms-1. The car then accelerates with constant acceleration a ms-2 for T seconds. The car travels a distance D metres during these T seconds. The diagram shows the velocity-time graph for the motion of the car for 0≤ t ≤T . Using the graph, show that D=U T +½a T 2 . (No credit will be given for answers which use any of the kinematics (suvat) formulae listed under Mechanics in the AS Mathematics section of the formulae booklet.) [4]

e.g. dissection to triangle + rectangle

using gradient of graph is (acceleration) a= hT gives h=aT

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M107 assessment Kinematics 1 SOLUTIONS

Area under graph (distance travelled ) D=UT + 12(h)× T

Combine equations SO D=UT + 12

hT 2 A1

Question from 2017 assessment materials for AS

4) A man throws a stone on a level playing field. Assuming the stone is thrown at

initial speed 18 m/s, from a height of 2 m and at an angle of 40, how far will it have travelled horizontally from the man by the time it hits the ground? [6]Vertically u=18 sin 40

s=ut+ 12

a t 2 gives −2=18 sin 40 t+ 12(−9.8)t 2 solving using quadratic formulas

gives t=18sin 40 ±√18 sin 402−4 (4.9)(−2)2(4.9)

=2.52 secs

Horizontal velocity u=18 cos40s=ut gives s=¿

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5) A girl throws a stone which breaks a window 2 seconds later. The speed of projection is 20 ms-1 and the angle of projection is 60. Assuming that the stone can be modelled as a particle moving with constant acceleration, find the horizontal and vertical components of the velocity of the stone just before impact [4]Horizontally u=v=20cos 60=10

Vertically u=20sin 60 ,a=−9.8 , t=2Using v=u+at gives v=20 sin 60+(−9.8 ) (2 )=−2.28

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Amber

6) Figure 1

Figure 1 shows the speed-time graph of a cyclist moving on a straight road over a 7 s period. The sections of the graph from t = 0 to t = 3, and from t = 3 to t = 7, are straight lines. The section from t = 3 to t = 7 is parallel to the t-axis.State what can be deduced about the motion of the cyclist from the fact that:

a) the graph from t = 0 to t = 3 is a straight line, [1]

b) the graph from t = 3 to t = 7 is parallel to the t-axis. [1]

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5

v (m s–1)

t (s)O 73

2

v

25

010 18 30

t

M107 assessment Kinematics 1 SOLUTIONS

c) Find the distance travelled by the cyclist during this 7 s period. [4]

a) Constant acceleration B1

b) Constant speed/velocity B1

c) Distance = ½ (2 + 5) x 3, + (4 x 5) = 30.5 m

M1 A1 B1

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7) A car is moving along a straight horizontal road. At time t = 0, the car passes a point A with speed 25 m s–1. The car moves with constant speed 25 m s–1 until t = 10 s. The car then decelerates uniformly for 8 s. At time t = 18 s, the speed of the car is V m s–1 and this speed is maintained until the car reaches the point B at time t = 30 s.

a) Sketch a speed–time graph to show the motion of the car from A  to B [3]

Given that AB = 526 m, find

b) the value of V, [5]

c) the deceleration of the car between t = 10 s and t = 18 s. [3]

a)

2 horizontal lines B1Joined by straight line sloping down B125, 10, 18, 30 oe B1

b) 25×10+ 12 (25+V )×8+12×V=526

Solving to: V = 11 M1A1A1

DM1 A1

c) “v = u + at” so 11 = 25 – 8a ft their Va = 1.75 (ms-2)

M1 A1ft A1

8) A firework rocket starts from rest at ground level and moves vertically. In the first 6s of its motion, the rocket rises 29m. The rocket is modelled as a particle moving with constant acceleration ams–2. Find

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M107 assessment Kinematics 1 SOLUTIONS

a) the value of a, [2]

b) the speed of the rocket 6s after it has left the ground [2]

After 6s, the rocket burns out. The motion of the rocket is now modelled as that of a particlemoving freely under gravity

c) Find the height of the rocket above the ground 9s after it has left the ground [4]

9) A racing car moves along a straight horizontal road with constant acceleration. It passes the point O with speed 12 m/s. It passes the point A 5s later with speed 58 m/s

a) Find the acceleration of thecar [2]

b) Find the distance OA [3]

The point B is the midpoint of AO

c) Find the speed of the car when it passes point B [3]

10)

A firework rocket starts from rest at ground level and moves vertically. In the first 3 s of its motion, the rocket rises 27 m. The rocket is modelled as a particle moving with constant acceleration a m s–2. Find

a) the value of a, [2]

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M107 assessment Kinematics 1 SOLUTIONS

b) the speed of the rocket 3 s after it has left the ground. [2]

After 3 s, the rocket burns out. The motion of the rocket is now modelled as that of a particle moving freely under gravity.

c) Find the height of the rocket above the ground 5 s after it has left the ground. [4]

a) 27 = 0 + ½ × a × 32, so a = 6 M1A1

b) v = 6 × 3 = 18 ms -1 M1

A1ft

c) From t = 3 to t = 5, s = 18 × 2 - ½ × 9.8 × 22 Total height = s + 27 = 43.4 m

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A1ft

M1 A1

Green11)

A car moves along a straight horizontal road. At time t=0, the velocity of the car is U ms-1. The car then accelerates with constant acceleration a ms-2 for T seconds. The car travels a distance D metres during these T seconds. The diagram shows the velocity-time graph for the motion of the car for 0 ≤ t ≤T . Using the graph, show that D=U T +½a T 2 . (No credit will be given for answers which use any of the kinematics (suvat) formulae listed under Mechanics in the AS Mathematics section of the formulae booklet.) [4]

e.g. dissection to triangle + rectangle

using gradient of graph is (acceleration) a= hT gives h=aT

Area under graph (distance travelled ) D=UT + 12(h)× T

Combine equations SO D=UT + 12

hT 2

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M1A1

A1

Question from 2017 assessment materials for AS

12)

A car moves along a horizontal straight road, passing two points A and B. At A the speed of the car is 15 m s–1. When the driver passes A, he sees a warning sign W ahead of him, 120 m away. He immediately applies the brakes and the car decelerates

5

5

15

16 22 t

V

M107 assessment Kinematics 1 SOLUTIONS

with uniform deceleration, reaching W with speed 5 m s–1. At W, the driver sees that the road is clear. He then immediately accelerates the car with uniform acceleration for 16 s to reach a speed of V m s–1 (V > 15). He then maintains the car at a constant speed of V m s–1. Moving at this constant speed, the car passes B after a further 22 s.

a) Sketch a speed-time graph to illustrate the motion of the car as it moves from A to B. [3]

b) Find the time taken for the car to move from A to B. [3]

The distance from A to B is 1 km.

c) Find the value of V. [5]

a)

Shape B1, Shape for last 22 secs (with V > 15) B1 Figures B1

b) 12 (15+5)×t=120 So: t = 12,so: T = 12 + 16 + 22 = 50 secs

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c) 120+ 12 (V +5)×16+22 V=1000

Solve: 30V = 840, so: V = 28 M1 B1 A1

DM1 A1

13)

A ball is projected vertically upwards with speed 21 m s–1 from a point A, which is 1.5 m above the ground. After projection, the ball moves freely under gravity until it reaches the ground. Modelling the ball as a particle, find

a) the greatest height above A reached by the ball, [3]

b) the speed of the ball as it reaches the ground, [3]

c) the time between the instant when the ball is projected from A and the instant when the ball reaches the ground

[4]

a) v2 = u2 + 2as so 02 = 212 − 2×9.8× h h = 22.5+1.5 = 24 m

M1 A1

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b) v2 = u2 + 2as so v2 = 02 + 2×9.8× (22.5+1.5) (= 470.4) or equivalent v ≈ 22 (ms−1 ) accept 21.7

M1 A1

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M107 assessment Kinematics 1 SOLUTIONS

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c) v = u + at so −( √ 470.4) = 21− 9.8t or equivalent − 1 each errort ≈ 4.4 (s) accept 4.36

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14)

A and B are two bus stops on a straight horizontal road. A bus passes A travelling towards B at a constant velocity of 16ms−1. The bus continues at this velocity for T seconds. It then decelerates at a constant rate for the next 7s until it comes to rest at B

a) Sketch a velocity–time graph for the motion of the bus [2]

b) Find the deceleration of the bus [2]

c) Find, in terms of T, the distance travelled by the bus [3]

6s after the bus passes A, a car leaves A and travels towards B. The car  moves from rest with a constant acceleration of 36ms−2 The car and bus reach B at the same time

d) Find the distance between A and B [7]

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M107 assessment Kinematics 1 SOLUTIONS

15)

A boy throws a stone with speed U m s−1 from a point O at the top of a vertical cliff. The point O is 18 m above sea level. The stone is thrown at an angle α

above the horizontal, where tan∝=34 .

The stone hits the sea at the point S which is at a horizontal distance of 36 m from the foot of the cliff, as shown The stone is modelled as a particle moving freely under gravity with g = 10 m s−2

a) Find the value of U, [6]

b) Find the speed of the stone when it is 10.8 m above sea level, giving your answer to 2 significant figures [5]

c) Suggest two improvements that could be made to the model. [2]

a) Horizontally s=ut gives 36=Ut cos∝Vertically s=ut+ 1

2a t 2 gives −18=Utsin∝−1

2(−10)t 2

Solving simultaneously gives U = 15

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b) Horizontally U cos∝=12 Vertically v2=u2+2 as gives v2=(U sin∝ )2+2(−10)(10.8−18) Gives v = 15Combine horizontal and vertical components √122+152=√369= 19 (2sf)

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c) Two of Include air resistance in model Include wind effects in model Include dimensions of the stone in the model Use a more accurate value of g B1B1

Question comes from 2017 specimen assessment materials

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