AQA M3 2013 June — Question 1 6 marks

Exam BoardAQA
ModuleM3 (Mechanics 3)
Year2013
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeTime to reach given speed
DifficultyStandard +0.3 This is a straightforward M3 variable force question requiring application of F=ma with a time-dependent force, integration to find velocity, and solving a quadratic equation. The setup is clear, the integration is routine (polynomial), and the method is standard textbook procedure. Slightly easier than average due to the direct application of a well-practiced technique with no conceptual obstacles.
Spec1.08d Evaluate definite integrals: between limits3.03c Newton's second law: F=ma one dimension6.03e Impulse: by a force6.03f Impulse-momentum: relation

1 A stone, of mass 2 kg , is moving in a straight line on a smooth horizontal sheet of ice under the action of a single force which acts in the direction of motion. At time \(t\) seconds, the force has magnitude \(( 3 t + 1 )\) newtons, \(0 \leqslant t \leqslant 3\). When \(t = 0\), the stone has velocity \(1 \mathrm {~ms} ^ { - 1 }\).
When \(t = T\), the stone has velocity \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find the value of \(T\).
(6 marks)

Question 1:
AnswerMarks Guidance
Working/AnswerMark Guidance
Using impulse-momentum: \(\int_0^T (3t+1)\, dt = m(v-u)\)M1 Attempt to integrate force, equate to change in momentum
\(\left[\frac{3t^2}{2} + t\right]_0^T = 2(5-1)\)A1 Correct integral expression
\(\frac{3T^2}{2} + T = 8\)A1 Correct equation
\(3T^2 + 2T - 16 = 0\)M1 Form quadratic and attempt to solve
\((3T + 8)(T - 2) = 0\)A1 Correct factorisation or equivalent
\(T = 2\)A1 Accept positive root only, with \(0 \leq T \leq 3\) confirmed
# Question 1:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Using impulse-momentum: $\int_0^T (3t+1)\, dt = m(v-u)$ | M1 | Attempt to integrate force, equate to change in momentum |
| $\left[\frac{3t^2}{2} + t\right]_0^T = 2(5-1)$ | A1 | Correct integral expression |
| $\frac{3T^2}{2} + T = 8$ | A1 | Correct equation |
| $3T^2 + 2T - 16 = 0$ | M1 | Form quadratic and attempt to solve |
| $(3T + 8)(T - 2) = 0$ | A1 | Correct factorisation or equivalent |
| $T = 2$ | A1 | Accept positive root only, with $0 \leq T \leq 3$ confirmed |

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1 A stone, of mass 2 kg , is moving in a straight line on a smooth horizontal sheet of ice under the action of a single force which acts in the direction of motion. At time $t$ seconds, the force has magnitude $( 3 t + 1 )$ newtons, $0 \leqslant t \leqslant 3$.

When $t = 0$, the stone has velocity $1 \mathrm {~ms} ^ { - 1 }$.\\
When $t = T$, the stone has velocity $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

Find the value of $T$.\\
(6 marks)

\hfill \mbox{\textit{AQA M3 2013 Q1 [6]}}