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

19 original exam-style questions - 3 pages of questions with a full mark scheme - free printable PDF.

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A-Level · Mechanics

M1DB Mechanics: Quantities, Units and Kinematics: Fluency and Exam Drill (Part 2)

EDEXCEL 9MA0 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

State whether each of the following is a vector quantity or a scalar quantity: (a) speed, (b) displacement.

(Total for Question 1 is 1 mark)

State the SI unit of speed.

(Total for Question 2 is 1 mark)

Convert a speed of 90 km/h into m/s.

(Total for Question 3 is 1 mark)

Convert a time of 2 hours 15 minutes into seconds.

(Total for Question 4 is 1 mark)

A parcel moves at a constant speed of 0.8 m/s along a conveyor belt for 25 seconds. Find the distance it travels.

(Total for Question 5 is 1 mark)

A skateboarder has initial speed 1.5 m/s and accelerates uniformly at 0.6 m/s² for 12 seconds. Find her final speed.

(Total for Question 6 is 2 marks)

A delivery van slows down uniformly from 18 m/s to 6 m/s in 8 seconds. Find its deceleration.

(Total for Question 7 is 2 marks)

A sledge decelerates uniformly from 9 m/s to rest in 3 seconds. Find the distance it travels while decelerating.

(Total for Question 8 is 2 marks)

A shopping trolley starts with speed 1 m/s and accelerates uniformly at 0.5 m/s². Find the distance it travels in the first 6 seconds.

(Total for Question 9 is 2 marks)

A car is travelling in a straight line and brakes with constant deceleration 4 m/s², coming to rest after travelling a further 32 m. Find the speed of the car at the instant the brakes were applied.

(Total for Question 10 is 2 marks)

A stone falls freely from rest under gravity, where g = 9.8 m/s². Find the time taken for the stone to reach a speed of 24.5 m/s.

(Total for Question 11 is 2 marks)

A particle P has position vector (3i + j) m at time t = 0 and moves with constant velocity (-2i + 5j) m/s. Find the value of t at which the position vector of P has zero i-component (that is, the instant at which P crosses the y-axis).

(Total for Question 12 is 3 marks)

In the equation s = ut + 1/2 a t², s is measured in metres, u in m/s, t in seconds and a in m/s². Show that the term 1/2 a t² has units of metres.

(Total for Question 13 is 2 marks)

A ball is thrown vertically upwards from ground level with initial speed 14 m/s and is modelled as a particle moving freely under gravity. Using g = 9.8 m/s², find the maximum height reached by the ball above the point of projection, and find the total time for which the ball is in the air before it returns to the point of projection, giving your final answer to 3 significant figures.

(Total for Question 14 is 4 marks)

A cyclist has velocity (3i + 4j) m/s relative to the ground, where i and j are unit vectors due east and due north respectively. A steady wind gives the cyclist an additional velocity of (-i + 2j) m/s relative to the ground. Find the cyclist's resultant velocity vector, and find her speed, giving your answer to 3 significant figures.

(Total for Question 15 is 4 marks)

A particle moves so that its position vector at time t seconds is r = (2t² - 3t)i + 4tj metres, where i and j are perpendicular unit vectors. Find the velocity vector of the particle at time t = 2 seconds, and find its speed at this instant, giving your answer to 3 significant figures.

(Total for Question 16 is 4 marks)

A student is asked to find the distance travelled by a particle that decelerates uniformly from 20 m/s to 8 m/s over 6 seconds. Their working is shown below:
s = (u + v)t
s = (20 + 8) x 6
s = 168
Identify the error in the student's working, and find the correct value of s.

(Total for Question 17 is 3 marks)

A particle P moves along the x-axis. At time t seconds its acceleration is a = (6t - 4) m/s². When t = 0, the velocity of P is 3 m/s and the displacement of P from a fixed point O is 0. Find an expression for the velocity v m/s of P at time t, and find an expression for the displacement x metres of P from O at time t.

(Total for Question 18 is 4 marks)

A cyclist sets off from rest and accelerates uniformly to a speed of 8 m/s over the first 4 seconds of her journey. She then travels at this constant speed for a further T seconds, before decelerating uniformly to rest over the next 5 seconds. The velocity-time graph for her journey is shown below. Given that the total distance she travels is 100 m, find the value of T.

Figure (to be drawn): Velocity-time graph: velocity (m/s) on the vertical axis, time (s) on the horizontal axis, starting at the origin (0, 0). A straight line rises from (0, 0) to (4, 8). A horizontal line continues from (4, 8) to (4+T, 8). A straight line then falls from (4+T, 8) to (4+T+5, 0), meeting the time axis there.

(Total for Question 19 is 4 marks)

Mark scheme · M1DB Mechanics: Quantities, Units and Kinematics: Fluency and Exam Drill (Part 2)

Question 1

B1 speed correctly identified as a scalar and displacement correctly identified as a vector, both required
Answer: Speed is a scalar quantity; displacement is a vector quantity.

Question 2

B1 m/s (or m s^-1) oe
Answer: m/s

Question 3

B1 25 cao
Answer: 25 m/s

Question 4

B1 8100 cao
Answer: 8100 s

Question 5

B1 20 cao
Answer: 20 m

Question 6

M1 use v = u + at with u=1.5, a=0.6, t=12
A1 8.7 m/s cao
Answer: 8.7 m/s

Question 7

M1 a = (6-18)/8
A1 1.5 m/s² cao (deceleration)
Answer: 1.5 m/s²

Question 8

M1 s = 1/2(u+v)t with u=9, v=0, t=3
A1 13.5 m cao
Answer: 13.5 m

Question 9

M1 s = ut + 1/2 a t² with u=1, a=0.5, t=6
A1 15 m cao
Answer: 15 m

Question 10

M1 use v² = u² + 2as with v=0, a=-4, s=32 (or v²=u²-2as with a=4)
A1 16 m/s cao
Answer: 16 m/s

Question 11

M1 use v = u + at (or t = v/g) with u=0, v=24.5, a=9.8
A1 2.5 s cao
Answer: 2.5 s

Question 12

M1 form r(t) = (3-2t)i + (1+5t)j
M1 set the i-component 3-2t = 0 and solve for t
A1 t = 1.5 cao
Answer: t = 1.5 s

Question 13

B1 units of a t² = (m s^-2)(s²) = m, with the s² terms shown cancelling
B1 conclusion that 1/2 a t² therefore has units of metres, consistent with s cso
Answer: 1/2 a t² has units of metres, consistent with s.

Question 14

M1 use v² = u² - 2gs with v=0, u=14, g=9.8 to find the maximum height
A1 10 m cao
M1 use symmetry (T = 2u/g) or v = u - gt to find the total time of flight
A1 awrt 2.86 s
Answer: Maximum height = 10 m; total time in the air = 2.86 s (3 s.f.)

Question 15

M1 add the velocity components: (3-1)i + (4+2)j
A1 (2i + 6j) m/s cao
M1 speed = √2² + 6²
A1 awrt 6.32 m/s
Answer: Resultant velocity = (2i + 6j) m/s; speed = 6.32 m/s (3 s.f.)

Question 16

M1 differentiate: v = (4t-3)i + 4j
M1 substitute t=2
A1 (5i + 4j) m/s cao
A1 awrt 6.40 m/s
Answer: v(2) = (5i + 4j) m/s; speed = 6.40 m/s (3 s.f.)

Question 17

B1 identifies that the student has omitted the factor of 1/2 (used s=(u+v)t instead of s=1/2(u+v)t)
M1 correct method: s = 1/2(20+8)(6)
A1 84 m cao
Answer: The student omitted the factor of 1/2 in the formula; the correct distance is s = 84 m.

Question 18

M1 integrate a to get v = 3t² - 4t + C
A1 use v=3 at t=0 to find C=3, so v = 3t² - 4t + 3 cao
M1 integrate v to get x = t³ - 2t² + 3t + D
A1 use x=0 at t=0 to find D=0, so x = t³ - 2t² + 3t cao
Answer: v = 3t² - 4t + 3; x = t³ - 2t² + 3t

Question 19

M1 distance in the acceleration phase = 1/2 x 4 x 8 = 16
M1 distance in the deceleration phase = 1/2 x 5 x 8 = 20
M1 set up 16 + 8T + 20 = 100 (or equivalent)
A1 T = 8 cao
Answer: T = 8 s

Question 19

4 marks

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

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