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

Question 1

Multiple choiceMedium1 mark

A small ball is launched from a platform. Air resistance is negligible and $g=10\ \text{m s}^{-2}$ The figure shows the launch conditions. What is the horizontal distance from the point directly below the launch point to where the ball first reaches the ground?

Show answer and explanation

Answer: B. Taking upward as positive, vertical motion gives $0=10+5t-5t^2$ so the positive time to reach the ground is $t=2.0\ \text{s}$ The horizontal speed is $10\cos30^\circ\ \text{m s}^{-1}$ giving a horizontal distance of $(10\cos30^\circ)(2.0)=17.3\ \text{m}$

A: uses the vertical component, $10\sin30^\circ$ as the horizontal speed.
C: uses the launch speed as the horizontal speed, omitting the cosine component.
D: doubles the flight time after finding it, adding an extra descent interval.

Remember: horizontal and vertical components determine different parts of projectile motion.

Question 2

Paper 2Medium8 marks

A pedestrian follows the three sections of the route shown in Figure 1, starting at O and finishing at P.

(a)

Distinguish distance travelled from displacement for the walk shown.

[2]

(b)

Calculate the distance travelled by the pedestrian.

[2]

(c)

On Figure 1, draw an arrow to represent the pedestrian’s displacement from O to P.

[1]

(d)

Determine the magnitude and direction of the displacement from O to P. Give the direction as an angle measured north of east.

[3]

Question 3

Multiple choiceMedium1 mark

A 3.0 kg box is stationary on a horizontal platform. A person exerts a 10 N downward force on the box. Take $g=10\ \text{N kg}^{-1}$ What is the magnitude of the vertical force that the box exerts on the platform?

Show answer and explanation

Answer: C. Translational equilibrium of the box gives $N - mg - 10 = 0$ so the platform’s force on the box is $N = (3.0)(10)+10 = 40\ \text{N}$ By Newton’s third law, the box exerts an equal-magnitude force on the platform.

A: Treats the third-law forces as cancelling, although they act on different objects, giving $0\ \text{N}$
B: Balances only the box’s weight, omitting the person’s downward force, giving $30\ \text{N}$
D: Counts the person’s downward force twice, giving $30+2(10)=50\ \text{N}$
Remember: sum forces on one object for equilibrium; third-law pairs act on different objects.

Question 4

Paper 2Medium4 marks

A block is attached to a fixed support by a stretched horizontal spring and rests on a rough horizontal table. It remains at rest, and no other horizontal forces act on it. Figure 1 gives the block’s mass and the spring’s constant and extension. Use $g=9.81\ \mathrm{m\,s^{-2}}$

(a)

Calculate the magnitude of the force exerted by the spring on the block.

[1]

(b)

Determine the minimum coefficient of static friction between the block and the table for the block to remain at rest.

[3]

Question 5

Multiple choiceHard1 mark

A body of mass $M$ moves at speed $u$ along the $+x$ -direction and explodes into two pieces of masses $\alpha M$ and $(1-\alpha)M$ where $0<\alpha<1$ The piece of mass $\alpha M$ moves at speed $v$ perpendicular to the original motion. The external impulse during the explosion is negligible. What is the speed of the other piece divided by $u$

Show answer and explanation

Answer: C. Total momentum is conserved as a vector. The other piece must have momentum components $Mu$ and $-\alpha Mv$ so its speed is

\frac{\sqrt{(Mu)^2+(\alpha Mv)^2}}{(1-\alpha)M}.

Dividing by $u$ gives $\dfrac{\sqrt{1+\alpha^2(v/u)^2}}{1-\alpha}$

A: assigns mass $\alpha M$ rather than $(1-\alpha)M$ to the other piece.
B: adds the perpendicular momentum components as scalar magnitudes.
D: uses only the $x$ -component and omits the momentum needed to cancel the first piece’s perpendicular component.
Remember: conserve momentum in each perpendicular direction before finding a resultant speed.

Question 6

Paper 2Medium13 marks

A $16\ \mathrm{kg}$ sled moves horizontally along a level track. A rope pulls the sled through a displacement of $6.0\ \mathrm{m}$ with a constant tension of $50\ \mathrm{N}$ at $35^\circ$ above the horizontal. A constant friction force of $18\ \mathrm{N}$ acts opposite to the motion. The sled’s initial speed is $1.0\ \mathrm{m\,s^{-1}}$ Ignore air resistance.

(a)

Calculate the work done on the sled by the tension in the rope.

[2]

(b)

Calculate the work done on the sled by friction.

[2]

(c)

Determine the work done by the resultant force on the sled over the displacement.

[2]

(d)

Determine the sled’s speed after it has moved through $6.0\ \mathrm{m}$

[3]

(e)

Account for the difference between the work done by the rope and the increase in the sled’s kinetic energy.

[2]

(f)

In a repeat run, the rope is at $50^\circ$ above the horizontal. The tension, displacement, initial speed and friction force are unchanged. Calculate the work done by the rope and deduce whether the sled’s final speed is greater or smaller than in the first run.

[2]

Question 7

Multiple choiceHard1 mark

An electric winch raises a 40 kg load 6.0 m up a ramp at constant speed. The rope exerts a constant 250 N force parallel to the load’s 12 m displacement. The winch receives 4.0 kJ of electrical energy. Take $g=10\ \mathrm{N\,kg^{-1}}$ and assume all energy not retained as gravitational potential energy is transferred to thermal stores.

I. The work done by the rope on the load is 3.0 kJ.
II. A Sankey diagram for the winch and load together shows 2.4 kJ of useful gravitational potential energy output and 1.6 kJ of thermal output.
III. The work done by the rope on the load equals the load’s increase in gravitational potential energy.

Which statements are correct?

Show answer and explanation

Answer: C. The rope’s work is $W=Fs=(250)(12)=3.0\times10^3\ \mathrm{J}$ so I is true. The increase in gravitational potential energy is $mg\Delta h=(40)(10)(6.0)=2.4\times10^3\ \mathrm{J}$ For the winch and load together, the remaining $4.0-2.4=1.6\ \mathrm{kJ}$ is transferred to thermal stores, so II is true; not all the rope’s work is retained as gravitational potential energy, making III false.

A: Uses the vertical rise rather than the displacement parallel to the rope force to evaluate the work in I.
B: Treats all $3.0\ \mathrm{kJ}$ of rope work as useful gravitational potential energy, leaving only $1.0\ \mathrm{kJ}$ as thermal output.
D: Equates the rope’s energy transfer with the energy retained as gravitational potential energy, overlooking the thermal transfer.

Remember: work is energy transferred by a force; it need not all be retained in one energy store.

Question 8

Multiple choiceMedium1 mark

A wheel rotates about a fixed axis. The graph shows its angular velocity; anticlockwise is defined as the positive direction. Which pair describes the wheel’s motion at $t=3.0\ \text{s}$

Show answer and explanation

Answer: C. At $t=3.0\ \text{s}$ the graph gives $\omega=+2\ \text{rad s}^{-1}$ so the wheel rotates anticlockwise. The segment has a negative slope, so $\alpha<0$ because $\omega$ and $\alpha$ have opposite signs, the angular speed is decreasing.

A: identifies the rotation direction correctly but assumes negative angular acceleration means increasing speed.
B: reverses the stated sign convention: positive angular velocity is anticlockwise.
D: reads the angular velocity as negative; at this time it is positive, while the graph’s slope is negative.
Remember: angular speed decreases when angular velocity and angular acceleration have opposite signs.

Question 9

Paper 2Medium4 marks

A horizontal turntable rotates about a fixed vertical axis. Its moment of inertia, excluding two identical small masses, is $0.40\ \mathrm{kg\,m^2}$ The masses are moved inward by radial forces through the axis. The bearing exerts negligible torque about the vertical axis. Treat the masses as point particles and the arms as having negligible mass.

Figure 1 shows the initial and final positions of the masses.

(a)

Determine the angular speed of the turntable after the masses have moved inward.

[3]

(b)

Explain why the angular momentum of the turntable–masses system about the vertical axis remains constant while the masses move.

[1]

Question 10

Multiple choiceEasy1 mark

A train moves at constant velocity to the right relative to a platform. Light flashes are emitted from its front and rear, and a passenger at the train’s midpoint receives them simultaneously. Which row gives the emission times of the flashes in the train and platform frames?

Show answer and explanation

Answer: A. The passenger is equally distant from both ends in the train frame, so simultaneous arrival means the emissions were simultaneous in that frame. Relativity of simultaneity means they are not simultaneous in the platform frame; for a train moving right, the front emission is later.

B: reverses the event order in the platform frame; the front emission is later, not the rear one.
C: treats simultaneity as unchanged between frames and assigns the front event an earlier time in the train frame.
D: also assigns simultaneity to the platform frame, and reverses the front–rear order in the train frame.
Remember: events simultaneous in one inertial frame need not be simultaneous in another.

Question 11

Paper 2Medium4 marks

In inertial frame $S$ a particle moves along the $x$ -axis as shown in Figure 1. Inertial frame $S'$ moves in the $+x$ -direction at speed $v=0.50c$ relative to $S$ The origins of the frames coincide at $t=t'=0$ Take $c=3.00\times10^8\ \mathrm{m\,s^{-1}}$

The Lorentz transformations are

x'=\gamma(x-vt),\qquad t'=\gamma\left(t-\frac{vx}{c^2}\right), \qquad \gamma=\frac{1}{\sqrt{1-v^2/c^2}}.

(a)

Determine $u/c$ for the particle in frame $S$

[1]

(b)

For two events on the particle’s motion, use the Lorentz transformations to show that

u'=\frac{u-v}{1-uv/c^2}.

[2]

(c)

Use your answers to (a) and (b) to determine $u'/c$ to 2 significant figures.

[1]

Question 12

Paper 2Medium4 marks

A trolley moves at constant velocity in the $+x$ direction relative to the ground. Let $S$ be the ground frame and $S'$ the frame of the trolley. The Galilean transformations are $x'=x-vt$ and $t'=t$ where $v$ is the velocity of $S'$ relative to $S$

(a)

Using the transformations for the positions of an object at two different times, show that its velocity $u'$ measured in $S'$ is related to its velocity $u$ measured in $S$ by $u'=u-v$

[2]

(b)

A puck moves at $+2.5\ \mathrm{m\,s^{-1}}$ relative to the ground while the trolley moves at $+4.0\ \mathrm{m\,s^{-1}}$ relative to the ground. Determine the puck’s velocity relative to the trolley, including its direction.

[2]

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

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