Moderate -0.3 This is a straightforward mechanics problem requiring standard SUVAT equations (vΒ² = uΒ² + 2as) to find acceleration, then F = ma to find force magnitude, followed by scaling a unit vector. All steps are routine AS-level mechanics with no conceptual challenges or novel problem-solving required, making it slightly easier than average.
A particle of mass 0.1 kg is initially stationary.
A single force \(\mathbf{F}\) acts on this particle in a direction parallel to the vector \(7\mathbf{i} + 24\mathbf{j}\)
As a result, the particle accelerates in a straight line, reaching a speed of \(4\text{ m s}^{-1}\) after travelling a distance of 3.2 m
Find \(\mathbf{F}\). [5 marks]
Question 14:
14 | Uses appropriate constant
acceleration equation to find the
acceleration | 3.4 | M1 | 2 2
Using π£π£ = π’π’ +2 ππaππnd
ππ =3 .2,π’π’ = 0 π£π£ = 4
ππ = 2.5
πΉπΉ = ππππ βΉ πΉπΉ = 0.25
2 2
οΏ½7 +24 = 25
F
1
= 100(7π’π’+24π£π£)
Obtains correct value for | 1.1b | A1
ππ
Uses their value with Newtonβs
second Law to determine
magnitude ππof force | 3.4 | M1
Calculates the magnitude of the
given vector | 1.1b | B1
Deduces, using given information,
correct vector form of the force | 2.2a | A1
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
A particle of mass 0.1 kg is initially stationary.
A single force $\mathbf{F}$ acts on this particle in a direction parallel to the vector $7\mathbf{i} + 24\mathbf{j}$
As a result, the particle accelerates in a straight line, reaching a speed of $4\text{ m s}^{-1}$ after travelling a distance of 3.2 m
Find $\mathbf{F}$. [5 marks]
\hfill \mbox{\textit{AQA AS Paper 1 2020 Q14 [5]}}