IB Physics SL Theme A: Space, time and motion questions

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Question 1

Multiple choiceMedium1 mark

A particle moves along the directed route shown. The horizontal and vertical coordinates give its east and north positions, respectively, in metres.

Which row gives the displacement vector components (east, north) and the total distance travelled?

	Displacement components (east, north) / m	Distance travelled / m
A.	(3, 4)	15
B.	(4, 3)	15
C.	(−3, −4)	15
D.	(3, 4)	11

Show answer and explanation

Answer: A. Displacement is final position minus initial position; distance is the length of the route. From S(0,0) to F(3,4), the displacement components are (3−0, 4−0) = (3, 4) m. The route length is 6 + 4 + 4 + 1 = 15 m.

B: Interchanges the east and north components.
C: Reverses the change in position, using start minus finish.
D: Omits the 4 m segment from Q to R when adding the route lengths.
Remember: displacement depends on the initial and final positions; distance includes the whole path.

Question 2

Paper 2Medium6 marks

A trolley of mass $24\ \mathrm{kg}$ is pulled horizontally along a level track by a rope. The rope exerts a forward force of $48\ \mathrm{N}$ while the track exerts a frictional force of $12\ \mathrm{N}$ backwards. These forces remain constant for $4.0\ \mathrm{s}$ At the start of this interval, the trolley is moving forwards at $1.5\ \mathrm{m\,s^{-1}}$ Ignore any other horizontal forces.

The rope force is then reduced, and the trolley continues at constant velocity. The frictional force remains $12\ \mathrm{N}$

(a)

Calculate the trolley’s acceleration during the $4.0\ \mathrm{s}$ interval.

[2]

(b)

Determine the trolley’s speed at the end of the $4.0\ \mathrm{s}$ interval.

[1]

(c)

Using Newton’s first law, explain why the horizontal resultant force on the trolley is zero when it moves at constant velocity.

[1]

(d)

State the Newton’s third-law force pair corresponding to the frictional force exerted by the track on the trolley.

[2]

Question 3

Multiple choiceMedium1 mark

A constant force of magnitude $10.0\ \text{N}$ acts on a box, directed $30.0^\circ$ north of east. The box moves $6.0\ \text{m}$ east and then $8.0\ \text{m}$ north. What is the total work done by the force on the box?

Show answer and explanation

Answer: D. Work on each segment is the force component along that segment multiplied by its length. The east and north components are $10.0\cos30.0^\circ=8.66\ \text{N}$ and $10.0\sin30.0^\circ=5.00\ \text{N}$ Thus $W=(8.66)(6.0)+(5.00)(8.0)=91.96\ \text{J}$ which is $92\ \text{J}$ to two significant figures. A: subtracts the northward contribution instead of adding it, giving $52-40=12\ \text{J}$ B: includes only the work during the northward displacement, $40\ \text{J}$ C: includes only the work during the eastward displacement, $52\ \text{J}$ Remember: add the work contributions from successive displacements, including their signs.

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