arthham

Exam

IB Physics HL

Conducting Body
International Baccalaureate (IB)
Who it is for
Final-year IB Diploma students
Papers
Three written papers
Question Types
Multiple-choice, structured, and data-based questions
Total Duration
4.5 hours
Syllabus Focus
Core, Additional Higher Level (AHL), and one Option topic
Frequency
Twice a year (May and November)
Assessment
External exams and an internal assessment (IA)

The IB Physics Higher Level (HL) exam is the final assessment for students completing the two-year IB Diploma Programme course in Physics HL. It is designed for students with a strong interest in the physical sciences, often those planning to pursue university degrees in fields like engineering, physics, or medicine. The exam evaluates a student's mastery of a broad and advanced physics curriculum, and the resulting grade is a significant component of their final IB Diploma score, which is used for university applications globally.

The examination is split into three papers. Paper 1 consists of multiple-choice questions covering the entire syllabus, while Paper 2 contains extended-response questions that require detailed, structured answers. Paper 3 assesses data analysis skills and knowledge of a specialised option topic chosen by the school. The primary difficulty lies in the vast scope of the syllabus and the conceptual depth required. Questions demand more than just formula recall; they test the ability to apply principles to new and complex scenarios, often combining ideas from different topics. Managing time effectively across the papers, particularly in the long-answer sections, is a common challenge for students.

How a question gets worked here

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

JEE Advanced 2019

A ball is thrown from ground at angle θ with horizontal and with an initial speed u0.u_{0}. For the resulting projectile motion, the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is V1.V_{1}. After hitting the ground, the ball rebounds at the same angle θ but with a reduced speed of u0/α.u_{0}/α. Its motion continues for a long time as shown in figure. If the magnitude of average velocity of the ball for entire duration of motion is 0.8V1,0.8 V_{1}, the value of α is ____
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: Analyze the first projectile motion.

The horizontal range R1R_1 and the time of flight T1T_1 for the first projectile motion are given by the standard formulas.

R1=u02sin⁡2θgR_1 = \frac{u_0^2 \sin 2\theta}{g} T1=2u0sin⁡θgT_1 = \frac{2u_0 \sin\theta}{g}

Step 2: Calculate the average velocity for the first flight.

The magnitude of the average velocity V1V_1 is the ratio of the horizontal displacement to the time of flight.

V1=R1T1V_1 = \frac{R_1}{T_1} V1=u02sin⁡2θg2u0sin⁡θgV_1 = \frac{\frac{u_0^2 \sin 2\theta}{g}}{\frac{2u_0 \sin\theta}{g}} V1=u0cos⁡θV_1 = u_0 \cos\theta

Step 3: Calculate the total horizontal displacement.

Since the speed is reduced by a factor of α\alpha after each bounce, the range of successive bounces decreases by a factor of α2\alpha^2.

X=∑n=1∞Rn=R1+R1α2+R1α4+…X = \sum_{n=1}^{\infty} R_n = R_1 + \frac{R_1}{\alpha^2} + \frac{R_1}{\alpha^4} + \dots X=R11−1α2X = \frac{R_1}{1 - \frac{1}{\alpha^2}} X=u02sin⁡2θg⋅α2α2−1X = \frac{u_0^2 \sin 2\theta}{g} \cdot \frac{\alpha^2}{\alpha^2 - 1}

Step 4: Calculate the total time of flight.

The time of flight for successive bounces decreases by a factor of α\alpha.

T=∑n=1∞Tn=T1+T1α+T1α2+…T = \sum_{n=1}^{\infty} T_n = T_1 + \frac{T_1}{\alpha} + \frac{T_1}{\alpha^2} + \dots T=T11−1αT = \frac{T_1}{1 - \frac{1}{\alpha}} T=2u0sin⁡θg⋅αα−1T = \frac{2u_0 \sin\theta}{g} \cdot \frac{\alpha}{\alpha - 1}

Step 5: Determine the average velocity for the entire motion.

The average velocity for the entire duration is the ratio of the total displacement to the total time.

Vavg=XTV_{\text{avg}} = \frac{X}{T} Vavg=u02sin⁡2θg⋅α2α2−12u0sin⁡θg⋅αα−1V_{\text{avg}} = \frac{\frac{u_0^2 \sin 2\theta}{g} \cdot \frac{\alpha^2}{\alpha^2 - 1}}{\frac{2u_0 \sin\theta}{g} \cdot \frac{\alpha}{\alpha - 1}} Vavg=u0cos⁡θ⋅αα+1V_{\text{avg}} = u_0 \cos\theta \cdot \frac{\alpha}{\alpha + 1}

Step 6: Solve for α\alpha.

Equate the total average velocity to 0.8 V10.8\,V_1.

Vavg=0.8 V1V_{\text{avg}} = 0.8\,V_1 u0cos⁡θ⋅αα+1=0.8 u0cos⁡θu_0 \cos\theta \cdot \frac{\alpha}{\alpha + 1} = 0.8\,u_0 \cos\theta αα+1=0.8\frac{\alpha}{\alpha + 1} = 0.8 α=4\alpha = 4

Answer: 4

Get a question like this solved →