Exam-style Questions: Vectors and Scalars and Linear Motion
-
Fig. 1.1 shows a table of vector and scalar quantities.
speed, acceleration energy, power force, pressure velocity, displacement Fig. 1.1
In the blank spaces provided in Fig. 1.1, label the pair of quantities that are both vectors with a V and the pair that are both scalars with an S
(Marks available: 2)
Answer outline and marking scheme for question: 1
speed, acceleration energy, power S force, pressure velocity, displacement V (Marks available: 2)
-
a) Fig. 4.1 shows a car of mass 1200 kg, pulling a caravan of mass 400 kg along a horizontal road.
Fig. 4.2 shows the variation with time t of the velocity v of the car as it accelerates from rest.
Calculate, for the first 20s of the journey:
i) the acceleration of the car
acceleration = ................................ m s-1
(2 Marks)
ii) the resultant force acting on the car
resultant force on the car = ................................ N
(2 Marks)
iii) the resultant force acting on the caravan
resultant force on the caravan = ................................ N
(2 Marks)
b) i) Use the graph in Fig. 4.2 to describe how the acceleration changes after the first 20 s of the journey.
(1 Mark)
ii) Suggest a possible reason for this change.
(1 Mark)
c) Fig. 2.1 shows a boy on a sledge travelling down a slope. The boy and sledge have a total mass of 60 kg and are travelling at a constant speed. The angle of the slope to the horizontal is 35°. All the forces acting on the boy and sledge are shown on Fig. 2.1 and in a force diagram in Fig. 2.2.
i) Calculate the magnitude of W, the total weight of the boy and sledge.
weight W = .............................................. N
(1 Mark)
ii) Determine the magnitude of the resistive force R. You may find it helpful to draw a vector triangle.
resistive force R = ....................................... N
(4 Marks)
iii) Determine the component of the weight W that acts perpendicular to the slope.
component of W = ..................................... N
(2 Marks)
iv) State and explain why the boy travelling at constant speed even though he is moving down a slope.
(2 Marks)
(Marks available: 17)
Answer outline and marking scheme for question: 2
a) i) Acceleration = 13 / 20 or gradient attempted
= 0.65 (m s-2) ± 0.01
(2 Marks)
ii) force = ma / 1200 x 0.65 ecf (b)(i)
= 780 (N)
(2 Marks)
iii) force = 400 x 0.65 ecf (b)(i)
= 260 N
(2 Marks)
b) i) (gradient is less hence) acceleration is less / reaches terminal velocity
(1 Mark)
ii) resultant force is less / resistive forces are increasing / driver eases off the accelerator / climbing a hill
(1 Mark)
c) i) weight = 60 x 9.81 (9.8 allowed, 10 not allowed)
= 589 (N) (590 allowed)
(1 Mark)
ii) Point for correct shape triangle/link given between R and the component of the weight down the slope e.g. R = Wcos55
Point for scale given / working shown
Point for resistive force R = 340 (N) (± 20 (N) for scale diagram)
(4 Marks)
iii) Component of W = 589 cos 35
= 482 (N) (480 allowed)
(or from triangle and = -P)
If the answers are reversed in b) and c) then -1 points
(2 Marks)
iv) Resultant force is zero (so no acceleration) / in equilibrium force up the slope equals force down the slope
(restrictive force up slope = component of W down slope scores two)
(2 Marks)
(Marks available: 17)