Moderate -0.8 This is a straightforward application of Newton's second law with friction on a horizontal surface. Students need to find the normal reaction (equals weight), calculate friction force (μR), then apply F=ma with net force D-F=ma. All steps are standard and routine with no problem-solving insight required.
14 A car has an initial velocity of \(1 \mathrm {~ms} ^ { - 1 }\)
A particle, \(Q\), moves in a straight line across a rough horizontal surface.
A horizontal driving force of magnitude \(D\) newtons acts on \(Q\)
\(Q\) moves with a constant acceleration of \(0.91 \mathrm {~ms} ^ { - 2 }\)
\(Q\) has a weight of 0.65 N
The only resistance force acting on \(Q\) is due to friction.
The coefficient of friction between \(Q\) and the surface is 0.4
Find \(D\)
Substitutes \(v=10\), \(t=3\) into \(v=kt^3-kt^2+t+1\); must include at least one more intermediate step; AG
## Question 14:
$v = \int a\, dt$ | M1 | Integrates $a$ with at least one term correct
$v = kt^3 - kt^2 + t + c$; $v=1$ when $t=0$ so $c=1$ | A1 | Fully correct expression for $v$; condone omission of constant; ACF
$v=10$, $t=3$: uses initial conditions to find constant of integration (must be done before substituting $v=10$, $t=3$) | M1 |
$10 = 27k - 9k + 3 + 1 \Rightarrow 18k = 6 \Rightarrow k = \frac{1}{3}$ | A1 | Substitutes $v=10$, $t=3$ into $v=kt^3-kt^2+t+1$; must include at least one more intermediate step; AG
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14 A car has an initial velocity of $1 \mathrm {~ms} ^ { - 1 }$
A particle, $Q$, moves in a straight line across a rough horizontal surface.\\
A horizontal driving force of magnitude $D$ newtons acts on $Q$\\
$Q$ moves with a constant acceleration of $0.91 \mathrm {~ms} ^ { - 2 }$\\
$Q$ has a weight of 0.65 N\\
The only resistance force acting on $Q$ is due to friction.\\
The coefficient of friction between $Q$ and the surface is 0.4
Find $D$
\hfill \mbox{\textit{AQA Paper 2 2023 Q14 [4]}}