Home / IBDP Physics- A.1 Kinematics- IB Style Questions For HL Paper 1A

IBDP Physics- A.1 Kinematics- IB Style Questions For HL Paper 1A -FA 2025

Question

Which of the following best defines instantaneous velocity?
(A) Total displacement divided by total time taken
(B) The rate at which an object’s position changes
(C) Total distance traveled divided by total time taken
(D) The rate at which an object’s distance from a point changes
▶️ Answer/Explanation
Detailed solution

Instantaneous velocity is defined as the rate of change of position with respect to time at a specific instant.
Option (A) defines average velocity, (C) defines average speed, and (D) refers to the rate of change of distance (speed), not velocity.
Answer: (B)

Question

A car moves clockwise around a circular track of radius \(R\). What is the magnitude of the displacement from point \(X\) to point \(Y\)?
Circular track displacement
(A) \( R \dfrac{3\pi}{2} \)
(B) \( R \dfrac{\pi}{2} \)
(C) \( R\sqrt{2} \)
(D) \( R \)
▶️ Answer/Explanation
Detailed solution

Displacement is defined as the straight-line distance between the initial and final positions, independent of the path taken.

Points \(X\) and \(Y\) lie on the circumference of the circle and are separated by a quarter turn, so the displacement is the chord joining \(X\) and \(Y\).

The chord subtends an angle of \(90^\circ\) at the center of the circle. Using geometry, the length of the chord is \[ \sqrt{R^2 + R^2} = R\sqrt{2}. \]

Chord displacement

Answer: (C)

Question

A stone of mass \(m\) is projected vertically upwards with speed \(u\) from the top of a cliff. The speed of the stone just before it reaches the ground is \(v\).

What is the magnitude of the change in momentum of the stone?

Stone projected from cliff
(A) \( m\left(\dfrac{v+u}{2}\right) \)
(B) \( m\left(\dfrac{v-u}{2}\right) \)
(C) \( m(v+u) \)
(D) \( m(v-u) \)
▶️ Answer/Explanation
Detailed solution

Take the upward direction as positive.

Initial momentum of the stone is \(p_{\text{initial}} = +mu\).
Final momentum of the stone just before hitting the ground is \(p_{\text{final}} = -mv\).

The change in momentum is given by \(\Delta p = p_{\text{final}} – p_{\text{initial}}\).
Substituting values, \(\Delta p = (-mv) – (mu) = -m(v + u)\).

Therefore, the magnitude of the change in momentum is \(|\Delta p| = m(v + u)\).

Answer: (C)

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