IB Physics HL A: Space, Time and Motion

Practise IB Physics HL mechanics through shared-core and HL work on kinematics, forces, energy, momentum, rotation and relativistic motion with data.

Covered syllabus points

A.1 Kinematics A.2 Forces and momentum A.3 Work, energy and power A.4 Rigid body mechanics A.5 Galilean and special relativity

[Maximum number: 7]

A student throws a ball towards a wall. The ball is released from a point 1.8 m above the ground and 8.0 m from the wall. The initial velocity of the ball makes an angle of $48^{\circ}$ with the horizontal. Air resistance is negligible.

The diagram shows the initial path of the ball. P is a point on the path.

Figure for Question A. Space, time and motion question 1 — IB Physics HL

(a)

Draw, on the diagram, an arrow to show

[ 2 ]

(i)

the velocity of the ball at P . Label this arrow v.

[ 1 ]

(ii)

the acceleration of the ball at P. Label this arrow a.

The ball takes 1.3 s to reach the wall.

[ 1 ]

(b)

Show that the initial speed of the ball is about $9 \mathrm{~ms}^{-1}$ .

[ 2 ]

horizontal velocity $v_{x}=\frac{8.0}{1.3} \|=6.2 \mathrm{~m} \mathrm{~s}^{-1} »$ $\text { " } \frac{v_{x}}{v}=\cos 48^{\circ} \Rightarrow \text { " } \| V=\frac{8.0}{1.3 \cos 48^{\circ}}=9.2 \text { " } \mathrm{m} \mathrm{~s}^{-1} \text { " }$

(c)

Determine the height above the ground at which the ball hits the wall.

[ 3 ]

initial vertical velocity $v_{y}=9.2 \sin 48^{\circ}$ « $=6.8 \mathrm{~m} \mathrm{~s}^{-1}$ »
$h=1.8+\left(9.2 \sin 48^{\circ} \times 1.3\right)-\frac{1}{2} \times 9.8 \times 1.3^{2}$

2.4 «m»

Marking guidance:

Award [2 max] for h=0.6 m - candidates have not taken the initial height of 1.8m into account.

Award [2 max] for h=19 m

Award [3] for BCA

[Maximum number: 7]

This question is about the forces on a skier.

A skier is pulled up a hill by a rope at a steady velocity. The hill makes an angle of $12^{\circ}$ with the horizontal. The mass of the skier and skis is 73 kg . The diagram below shows three of the forces acting on the skier.

Figure for Question A. Space, time and motion question 2 — IB Physics HL

(a)

On the diagram, draw and label one other force acting on the skier.

[ 1 ]

arrow vertically downwards labelled weight/ W / mg / gravitational
force $/ \mathrm{F}_{\mathrm{g}} / \mathrm{F}_{\text {gravitational }} /$ force of gravity; (judge by eye)

Do not allow "gravity".

(b)

Calculate the magnitude of the normal reaction acting on the skier.

[ 2 ]

$\quad(N=) m g \cos \theta /$ correct substitution;
$\left(=73 \times 9.81 \times \cos 12^{\circ}=\right) 700 \mathrm{~N}$ ;

(c)

The total frictional force acting is 65 N . Determine the tension in the rope.

[ 2 ]

tension = frictional force + component of weight parallel to slope /
tension $=65+m g \sin \theta$ ;
$214 / 210 \mathrm{~N}$ ;

(d)

Explain, using Newton's first law of motion, why the resultant force on the skier must be zero.

[ 2 ]

[Maximum number: 1]

Two sources of light produce an interference pattern on a screen.

Light of wavelength 720 nm is incident on two narrow slits that are separated by 0.12 mm . An interference pattern is observed on a screen. $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ are the points of destructive interference closest to the central maximum M .

Figure for Question A. Space, time and motion question 3 — IB Physics HL

Suggest how energy conservation is consistent with the fact that the energy at $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ is zero.

A beam of light containing all wavelengths in the range $[550 \mathrm{~nm}, 650 \mathrm{~nm}]$ is incident normally on a diffraction grating. The grating has 580 lines per mm . A diffraction pattern is observed on a screen as shown.

The energy missing at $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ is found at the maxima

[Maximum number: 9]

A student models a rotating dancer using a system that consists of a vertical cylinder, a horizontal rod and two spheres.

The cylinder rotates from rest about the central vertical axis. A rod passes through the cylinder with a sphere on each side of the cylinder. Each sphere can move along the rod. Initially the spheres are close to the cylinder.

Figure for Question A. Space, time and motion question 4 — IB Physics HL

A horizontal force of 50 N is applied perpendicular to the rod at a distance of 0.50 m from the central axis. Another horizontal force of 40 N is applied in the opposite direction at a distance of 0.20 m from the central axis. Air resistance is negligible.

(a)

Show that the net torque on the system about the central axis is approximately 30 Nm .

[ 1 ]

$\Sigma \Gamma=50 \times 0.5+40 \times 0.2$

OR

33 «Nm»

Marking guidance:

Accept opposite rotational sign convention

(b)

The system rotates from rest and reaches a maximum angular speed of $20 \mathrm{rads}^{-1}$ in a time of 5.0 s . Calculate the angular acceleration of the system.

[ 1 ]

« $\mathrm{a}=\frac{20}{5}=$ » 4 «rad s $\mathrm{s}^{-2}$ »

(c)

Determine the moment of inertia of the system about the central axis.

[ 2 ]

$\begin{aligned} I=\frac{\Gamma}{\alpha} \text { OR } 33=I \times 4 I=8.25<\mathrm{kg} \mathrm{~m}^{2}> \end{aligned}$

Allow ECF from (a) and (b)

Award [2] for a BCA

(d)

When the system has reached its maximum angular speed, the two forces are removed. The spheres now move outward, away from the central axis.

[ 5 ]

(i)

Outline why the angular speed $\omega$ decreases when the spheres move outward.

[ 2 ]

moment of inertia increases

Angular momentum is conserved

Marking guidance:

Allow algebraic expressions e.g. $\omega=\frac{L}{I}$ so $\omega$ decreases for MP2

(ii)

Show that the rotational kinetic energy is $\frac{1}{2} L \omega$ where L is the angular momentum
of the system.

[ 1 ]

$E_{\kappa} \ll=\frac{1}{2} I \omega^{2}=>\frac{1}{2}(I \omega) \omega=\frac{1}{2} L \omega \vee$

Marking guidance:

Accept equivalent methods

(iii)

When the spheres move outward, the angular speed decreases from $20 \mathrm{rad} \mathrm{s}^{-1}$ to $12 \mathrm{rads}^{-1}$ . Calculate the percentage change in rotational kinetic energy that occurs when the spheres move outward.

[ 2 ]

$\begin{aligned} & \boldsymbol{«} \mathrm{E}_{\mathrm{k}}=» \frac{1}{2} L \omega_{1}=1 / 2 L \omega_{2} & \boldsymbol{O} \boldsymbol{R} & \frac{E_{k 1}}{E_{k 2}}=\frac{\omega_{1}}{\omega_{2}} \end{aligned}$

OR

« L is constant so» $E_{\mathrm{k}}$ is proportional to $\omega$

40 \% «energy loss»

MP1 is for understanding that angular momentum is constant so change in rotational kinetic energy is proportional to change in angular velocity

Award [0] if $E=0.5 \mathrm{I} \omega^{2}$ is used with the same I value for both values of E

Award [2] for BCA

All question bank results loaded

Covered syllabus points

A. Space, time and motion question 1

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

A. Space, time and motion question 2

Question (a)

Question (b)

Question (c)

Question (d)

A. Space, time and motion question 3

A. Space, time and motion question 4

Question (a)

Question (b)

Question (c)

Question (d)

Question (i)

Question (ii)

Question (iii)