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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 cube of mass \( m \) is accelerated upwards along a smooth vertical surface by a force \( F \) applied at an angle \( \theta \) to the vertical. What is the magnitude of the cube’s acceleration?
 
 
 
 
 
 
 
 
 
 
 
 
(A) \(\frac{F \cos \theta – mg}{m}\)
(B) \(\frac{F \sin \theta – mg}{m}\)
(C) \(\frac{F \cos \theta – g}{m}\)
(D) \(\frac{F \sin \theta – g}{m}\)
▶️ Answer/Explanation
Detailed solution

Resolving the force \( F \) into components:
– Vertical component: \( F \cos \theta \) (upwards)
– Horizontal component: \( F \sin \theta \) (perpendicular to the surface)

Since the surface is frictionless and vertical, only the vertical forces affect the upward acceleration.
Using Newton’s second law in the vertical direction:
\( F \cos \theta – mg = ma \)
Rearranging for acceleration \( a \):
\( a = \frac{F \cos \theta – mg}{m} \)
✅ Answer: (A)

Question

A ball attached to a string is made to rotate with constant speed along a horizontal circle. The string is attached to the ceiling and makes an angle \( \theta \) with the vertical. The tension in the string is \(T\).
A.2 Forces and momentum HL Paper 1
What is correct about the horizontal component and vertical component of the net force acting on the ball?
A.2 Forces and momentum HL Paper 1
▶️ Answer/Explanation
Detailed solution

The ball moves in a horizontal circle with constant speed.

Therefore, the vertical component of the tension balances the weight of the ball, so the net vertical force is zero.

The horizontal component of the tension provides the centripetal force required for the circular motion.

✅ Answer: (D)

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)

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