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

Question

A stone is released from rest and falls vertically. Air resistance is negligible. What is correct about the stone during each consecutive second of its motion?

(A) The change in velocity is constant.
(B) The change in displacement is constant.
(C) The change in acceleration decreases.
(D) The change in speed increases.

▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{A}} \)

The stone is in free fall, so it has a constant acceleration due to gravity:

\(a=g\)

Acceleration is the rate of change of velocity:

\(a=\frac{\Delta v}{\Delta t}\)

Since \(g\) is constant, the velocity changes by the same amount during every equal time interval.

For each consecutive \(1\,\mathrm{s}\) interval,

\(\Delta v=g(1)=g\)

Therefore, the change in velocity is constant.

The displacement during each second is not constant because the speed increases continuously.

Hence, the correct answer is \( \boxed{\mathrm{A}} \).

Question

A projectile is launched horizontally from the top of a cliff with a speed of \(10\,\mathrm{m\,s^{-1}}\). The projectile hits the ground at a distance of \(30\,\mathrm{m}\) from the base of the cliff. Air resistance is negligible.

What is the height of the cliff?

(A) \(15\,\mathrm{m}\)
(B) \(30\,\mathrm{m}\)
(C) \(45\,\mathrm{m}\)
(D) \(90\,\mathrm{m}\)

▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{C}} \)

The horizontal motion is at constant velocity because there is no horizontal acceleration.

Using

\(x=v_xt\)

we find the time of flight:

\(30=10t\)

\(t=3.0\,\mathrm{s}\)

The initial vertical velocity is zero. Therefore, the vertical displacement is

\(y=u_yt+\frac{1}{2}gt^2\)

Since \(u_y=0\),

\(y=\frac{1}{2}gt^2\)

Taking \(g=10\,\mathrm{m\,s^{-2}}\),

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

\(y=45\,\mathrm{m}\)

Thus, the height of the cliff is

\( \boxed{45\,\mathrm{m}} \)

Hence, the correct answer is \( \boxed{\mathrm{C}} \).

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?

(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)

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

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