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IB Physics – Theme A: Space, Time and Motion

Kinematics, forces and momentum, work, energy and power, rigid body mechanics (HL).

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12 questions  ·  54 marks  ·  answers included  ·  IB Physics SL & HL (2016+ / 2023 syllabus)
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This worksheet covers:

  • Questions on kinematics
  • Forces and momentum
  • Work/energy/power
  • Rigid body mechanics (HL only, marked)

Use $g=9.81\,\text{m s}^{-2}$ and a calculator throughout; give final answers to 3 significant figures unless stated otherwise.

Questions

Section A – Kinematics

SL and HL
1

A ball is thrown vertically upwards at $15\,\text{m s}^{-1}$ from a height of $1.2\,\text{m}$ above the ground.(a) Calculate the maximum height reached above the ground.(b) Calculate the total time taken for the ball to hit the ground.

Show answer
(a) Time to top: $t=\dfrac{u}{g}=\dfrac{15}{9.81}=1.53\,\text{s}$. Maximum height $=1.2+\dfrac{15^2}{2(9.81)}=12.7\,\text{m}$.(b) Using $1.2+15t-4.905t^2=0$ gives $t=3.14\,\text{s}$.
[5]
2

A projectile is launched from level ground at $20\,\text{m s}^{-1}$ at $30\degree$ above the horizontal. Air resistance is negligible.(a) Calculate the time of flight.(b) Calculate the horizontal range.(c) Calculate the maximum height reached.

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$v_x=20\cos30\degree=17.3\,\text{m s}^{-1}$, $v_y=20\sin30\degree=10.0\,\text{m s}^{-1}$.(a) $T=\dfrac{2v_y}{g}=2.04\,\text{s}$.(b) $R=v_xT=35.3\,\text{m}$.(c) $H=\dfrac{v_y^2}{2g}=5.10\,\text{m}$.
[6]
3

Explain, in terms of forces, why a skydiver falling through air reaches a terminal velocity.

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As speed increases, air resistance (drag) increases; when drag becomes equal in magnitude to the weight, the resultant force on the skydiver is zero, so the acceleration is zero and the skydiver falls at a constant (terminal) velocity.
[3]
4

Sketch, by description, the shape of a displacement-time graph and the corresponding velocity-time graph for an object released from rest and falling freely under gravity (no air resistance) for 3 seconds.

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Displacement-time graph: a curve starting at the origin with increasing (steepening) gradient – a parabola, since $s\propto t^2$. Velocity-time graph: a straight line through the origin with constant positive gradient equal to $g$, since $v=gt$.
[3]

Section B – Forces, Momentum, Work and Energy

SL and HL
5

An Atwood machine consists of a light, inextensible string over a frictionless pulley connecting a mass $m_1=5.0\,\text{kg}$ to a mass $m_2=3.0\,\text{kg}$, both hanging freely.(a) Calculate the magnitude of the acceleration of the system.(b) Calculate the tension in the string.

Show answer
(a) $a=\dfrac{(m_1-m_2)g}{m_1+m_2}=\dfrac{2.0(9.81)}{8.0}=2.45\,\text{m s}^{-2}$.(b) $T=m_1(g-a)=5.0(9.81-2.45)=36.8\,\text{N}$ (also $T=m_2(g+a)=36.8\,\text{N}$, confirming consistency).
[5]
6

A trolley of mass $2.0\,\text{kg}$ travelling at $6.0\,\text{m s}^{-1}$ collides with, and sticks to, a stationary trolley of mass $4.0\,\text{kg}$.(a) Calculate the common velocity after the collision.(b) Calculate the kinetic energy lost in the collision.(c) State the name given to a collision in which kinetic energy is not conserved.

Show answer
(a) By conservation of momentum, $v=\dfrac{(2.0)(6.0)}{6.0}=2.0\,\text{m s}^{-1}$.(b) $E_{K,i}=\frac{1}{2}(2.0)(6.0)^2=36\,\text{J}$; $E_{K,f}=\frac{1}{2}(6.0)(2.0)^2=12\,\text{J}$; loss $=24\,\text{J}$.(c) Inelastic collision.
[6]
7

A car of mass $1200\,\text{kg}$ accelerates from rest to $25\,\text{m s}^{-1}$ in $8.0\,\text{s}$ on a level road. Friction and air resistance are negligible.(a) Calculate the work done on the car.(b) Calculate the average power output of the engine.

Show answer
(a) $W=\Delta E_K=\frac{1}{2}(1200)(25)^2=3.75\times10^5\,\text{J}$.(b) $P=\dfrac{W}{t}=\dfrac{3.75\times10^5}{8.0}=4.69\times10^4\,\text{W}$.
[4]
8

State the principle of conservation of momentum, and explain why it applies to a system of two colliding objects even when kinetic energy is not conserved.

Show answer
The total momentum of an isolated system (no external resultant force) is constant. Momentum conservation follows from Newton’s third law (equal and opposite internal forces act for equal times during the collision) and holds regardless of whether the collision is elastic or inelastic, since kinetic energy loss (e.g. to heat or sound) does not affect the vector sum of momentum.
[2]

Section C – Rigid Body Mechanics (HL)

HL only
9

A uniform rod of mass $4.0\,\text{kg}$ and length $2.0\,\text{m}$ is pivoted at one end and held horizontally, then released from rest. The moment of inertia of a uniform rod about an end is $I=\frac{1}{3}ML^2$.(a) Calculate the moment of inertia of the rod about the pivot.(b) Calculate the torque due to gravity about the pivot at the instant of release.(c) Calculate the initial angular acceleration of the rod.

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(a) $I=\frac{1}{3}(4.0)(2.0)^2=5.33\,\text{kg m}^2$.(b) The weight acts at the centre of mass, $1.0\,\text{m}$ from the pivot: $\tau=Mg\left(\frac{L}{2}\right)=4.0(9.81)(1.0)=39.2\,\text{N m}$.(c) $\alpha=\dfrac{\tau}{I}=\dfrac{39.2}{5.33}=7.36\,\text{rad s}^{-2}$.
[6]
10

A uniform disc of mass $0.50\,\text{kg}$ and radius $0.20\,\text{m}$ spins about its central axis at an angular velocity of $20\,\text{rad s}^{-1}$. The moment of inertia of a disc about its central axis is $I=\frac{1}{2}MR^2$.(a) Calculate the moment of inertia of the disc.(b) Calculate its angular momentum.(c) Calculate its rotational kinetic energy.

Show answer
(a) $I=\frac{1}{2}(0.50)(0.20)^2=0.0100\,\text{kg m}^2$.(b) $L=I\omega=0.0100(20)=0.200\,\text{kg m}^2\text{s}^{-1}$.(c) $E_K=\frac{1}{2}I\omega^2=\frac{1}{2}(0.0100)(20)^2=2.00\,\text{J}$.
[5]
11

A solid sphere (moment of inertia $I=\frac{2}{5}MR^2$) is released from rest and rolls without slipping down a slope, descending a vertical height of $0.80\,\text{m}$. Using conservation of energy, calculate the speed of the sphere’s centre of mass at the bottom of the slope.

Show answer
$Mgh=\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$ with $\omega=\dfrac{v}{R}$ gives $Mgh=\frac{7}{10}Mv^2$, so $v=\sqrt{\dfrac{10gh}{7}}=\sqrt{\dfrac{10(9.81)(0.80)}{7}}=3.35\,\text{m s}^{-1}$.
[4]
12

An ice skater spinning with moment of inertia $4.0\,\text{kg m}^2$ at an angular velocity of $2.0\,\text{rad s}^{-1}$ pulls her arms in, reducing her moment of inertia to $1.6\,\text{kg m}^2$.(a) Calculate her new angular velocity.(b) Calculate the kinetic energy before and after, and explain the source of any increase.

Show answer
(a) By conservation of angular momentum, $\omega_2=\dfrac{I_1\omega_1}{I_2}=\dfrac{4.0(2.0)}{1.6}=5.0\,\text{rad s}^{-1}$.(b) $E_{K,1}=\frac{1}{2}(4.0)(2.0)^2=8.0\,\text{J}$; $E_{K,2}=\frac{1}{2}(1.6)(5.0)^2=20.0\,\text{J}$. The kinetic energy increases because the skater does positive work pulling her arms inward against the centripetal force, converting internal (muscular/chemical) energy into rotational kinetic energy.
[5]

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More worksheets in IB Physics (SL & HL):

IB Physics – Theme B: The Particulate Nature of MatterIB Physics – Theme C: Wave BehaviourIB Physics – Theme D: FieldsIB Physics – Theme E: Nuclear and Quantum PhysicsIB Physics – Data-Based Questions (Paper 1B)

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