International Physics Olympiad

Conducted By: Organized annually by the host country
For: Pre-university students
Team Size: Up to five students per country
Papers: One theoretical and one experimental exam
Duration: Two 5-hour exams on separate days
Question Format: Long-form problems requiring detailed solutions and analysis
Frequency: Annually
Awards: Gold, Silver, and Bronze medals; Honourable Mentions

The International Physics Olympiad (IPhO) is the world's premier physics competition for secondary school students. Each year, participating countries send national teams of up to five students, who are selected through a rigorous process of national-level olympiads. The event is designed to challenge and celebrate the most talented young physicists, encouraging an interest in the subject and fostering international collaboration among future scientists.

The competition consists of two main parts held on separate days: a theoretical exam and an experimental exam. Both are five hours long. The problems are exceptionally difficult, requiring a deep conceptual understanding and mathematical fluency that goes far beyond a standard high school curriculum. Questions often ask students to model complex physical situations from first principles, combining ideas from different areas of physics.

The practical difficulty of the IPhO lies in its demand for both creativity and stamina. The theoretical problems require novel approaches, not just the application of known formulas. The experimental exam tests a student's ability to handle apparatus with precision, manage time effectively, and perform careful data analysis to derive a result. The sheer length and intensity of the exams make it a significant test of concentration and mental endurance.

How a question gets worked here

From a physics paper rather than International Physics Olympiad, worked the way every question on this site is. International Physics Olympiad questions are being added.

JEE Advanced 2009

A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg. Taking $g = 10 m/s^{2},$ find the work done (in joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest.

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Step 1: Identify the given parameters.

The masses of the two blocks and the acceleration due to gravity are given.

m_1 = 0.36\ \text{kg}

m_2 = 0.72\ \text{kg}

g = 10\ \text{m/s}^2

The system is released from rest, so the initial velocity is zero.

u = 0

Step 2: Write the equations of motion for the blocks.

Since $m_2 > m_1$ , block $m_2$ accelerates downwards and block $m_1$ accelerates upwards with the same acceleration $a$ . Let $T$ be the tension in the string.

For the block of mass $m_1$ :

T - m_1 g = m_1 a

For the block of mass $m_2$ :

m_2 g - T = m_2 a

Step 3: Calculate the acceleration of the system.

Add the two equations of motion to eliminate $T$ .

(m_2 - m_1)g = (m_1 + m_2)a

a = \frac{(m_2 - m_1)g}{m_1 + m_2}

Substitute the given values into the expression.

a = \frac{(0.72 - 0.36) \times 10}{0.36 + 0.72}

a = \frac{10}{3}\ \text{m/s}^2

Step 4: Calculate the tension in the string.

Express tension $T$ using the equation of motion for $m_1$ .

T = m_1(g + a)

Substitute the values of $m_1$ , $g$ , and $a$ .

T = 0.36 \left(10 + \frac{10}{3}\right)

T = 4.8\ \text{N}

Step 5: Calculate the displacement of the $0.36\ \text{kg}$ block.

Use the kinematic equation for displacement during the first second ( $t = 1\ \text{s}$ ).

s = \frac{1}{2}at^2

s = \frac{1}{2} \times \frac{10}{3} \times (1)^2

s = \frac{5}{3}\ \text{m}

Step 6: Calculate the work done by the tension on the $0.36\ \text{kg}$ block.

Since the tension force and the displacement are both in the upward direction, the work done is positive.

W = T \cdot s

W = 4.8 \times \frac{5}{3}

W = 8\ \text{J}

Answer: The work done by the string on the block of mass $0.36\,\text{kg}$ during the first second is $8\ \text{J}$ .

Get a question like this solved →