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

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Type

Multiple choice Written

Difficulty

Easy Medium Hard

Question 1

Paper 2Easy9 marks

A cart moves along a straight track in the positive $x$ -direction. Figure 1 shows its velocity $v$ as a function of time $t$ At $t=0$ the cart is at $x=5.0\ \mathrm{m}$

(a)

State the velocity of the cart at $t=4.0\ \mathrm{s}$

[1]

(b)

Determine the acceleration of the cart between $t=2.0\ \mathrm{s}$ and $t=6.0\ \mathrm{s}$

[2]

(c)

Calculate the displacement of the cart between $t=0$ and $t=2.0\ \mathrm{s}$

[1]

(d)

Calculate the displacement of the cart between $t=2.0\ \mathrm{s}$ and $t=6.0\ \mathrm{s}$

[2]

(e)

State the acceleration of the cart between $t=6.0\ \mathrm{s}$ and $t=8.0\ \mathrm{s}$

[1]

(f)

Determine the position of the cart at $t=6.0\ \mathrm{s}$ using your answers to (c) and (d).

[2]

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

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 3

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 4

Paper 2Easy6 marks

An electric winch raises a crate vertically at constant speed. Figure 1 shows the power transfers while the winch is operating.

(a)

State the energy form supplied to the winch and the useful energy form gained by the crate.

[2]

(b)

Calculate the efficiency of the winch.

[2]

(c)

Calculate the useful energy transferred to the crate in 25 s at the operating conditions shown.

[2]

Question 5

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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