Home / IBDP Physics- A.2 Forces and momentum- IB Style Questions For HL Paper 1A

IBDP Physics- A.2 Forces and momentum- IB Style Questions For HL Paper 1A -FA 2025

IBDP Physics HL Paper 1 -FA 2025- All Chapters

Question 

A force is applied to a mass of \(3\,\mathrm{kg}\). The graph shows the variation with time \(t\) of the acceleration \(a\) of the mass.

What is the average force acting on the mass?

(A) \(5\,\mathrm{N}\)
(B) \(6\,\mathrm{N}\)
(C) \(12\,\mathrm{N}\)
(D) \(24\,\mathrm{N}\)
▶️ Answer/Explanation

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

The average acceleration is found from the area under the acceleration-time graph:

\(\mathrm{average\ acceleration}=\dfrac{\text{area under graph}}{\text{total time}}\)

From \(t=0\) to \(t=2\,\mathrm{s}\), the graph is a rectangle:

\(A_1=2\times1=2\,\mathrm{m\,s^{-1}}\)

From \(t=2\,\mathrm{s}\) to \(t=6\,\mathrm{s}\), the graph is a trapezium with parallel sides \(1\,\mathrm{m\,s^{-2}}\) and \(4\,\mathrm{m\,s^{-2}}\):

\(A_2=\dfrac{1+4}{2}\times4=10\,\mathrm{m\,s^{-1}}\)

Therefore, the total area is

\(A=2+10=12\,\mathrm{m\,s^{-1}}\)

The average acceleration is

\(a_{\mathrm{avg}}=\dfrac{12}{6}=2\,\mathrm{m\,s^{-2}}\)

Using Newton’s second law,

\(F=ma\)

\(F_{\mathrm{avg}}=3\times2\)

\(F_{\mathrm{avg}}=6\,\mathrm{N}\)

Thus, the average force acting on the mass is

\( \boxed{6\,\mathrm{N}} \)

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

Question 

An object of mass \(m\), moving with a speed \(v\), collides with a stationary object of mass \(m\). The objects stick together.

What is the change in kinetic energy in the collision?

(A) Zero
(B) \(\dfrac{mv^2}{8}\)
(C) \(\dfrac{mv^2}{4}\)
(D) \(\dfrac{3mv^2}{8}\)
▶️ Answer/Explanation

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

Since the two objects stick together, the collision is perfectly inelastic. Momentum is conserved:

\(mv=(m+m)v_{\mathrm{f}}\)

\(mv=2mv_{\mathrm{f}}\)

\(v_{\mathrm{f}}=\dfrac{v}{2}\)

The initial kinetic energy is

\(K_{\mathrm{i}}=\dfrac{1}{2}mv^2\)

The final kinetic energy of the combined mass is

\(K_{\mathrm{f}}=\dfrac{1}{2}(2m)\left(\dfrac{v}{2}\right)^2\)

\(K_{\mathrm{f}}=\dfrac{mv^2}{4}\)

Therefore, the kinetic energy lost in the collision is

\(\Delta K_{\mathrm{lost}}=K_{\mathrm{i}}-K_{\mathrm{f}}\)

\(\Delta K_{\mathrm{lost}}=\dfrac{1}{2}mv^2-\dfrac{1}{4}mv^2=\dfrac{1}{4}mv^2\)

Thus, the change in kinetic energy in magnitude is

\( \boxed{\dfrac{mv^2}{4}} \)

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

Question

A block of mass \(2.0\,\text{kg}\) is placed on a trolley of mass \(5.0\,\text{kg}\) moving horizontally. A force of \(8.0\,\text{N}\) is applied to the block, which slides on the surface of the trolley. The frictional force between the trolley and the ground is zero.

The trolley accelerates at a rate of \(1.0\,\text{m s}^{-2}\).

What is the coefficient of dynamic friction between the block and the trolley?

(A) \(0.05\)
(B) \(0.15\)
(C) \(0.25\)
(D) \(0.35\)
▶️ Answer/Explanation
Detailed solution

Since the trolley accelerates at \(1.0\,\text{m s}^{-2}\) and the friction with the ground is zero, the only horizontal force acting on the trolley is the friction force due to the block.
Hence, the friction force is \( F = ma = 5.0 \times 1.0 = 5.0\,\text{N} \).

The normal reaction between the block and the trolley is \( N = mg = 2.0 \times 10 = 20\,\text{N} \).

Using \( F = \mu_k N \), \( \mu_k \times 20 = 5.0 \).
Therefore, \( \mu_k = 0.25 \).

✅ Answer: (C)

More A.2 Forces and momentum HL Paper 1 Questions..
Scroll to Top