Edexcel M3 2013 June — Question 2 7 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2013
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAdvanced work-energy problems
TypeWork done by non-constant force integration
DifficultyStandard +0.8 This M3 question requires integrating force with respect to time to find acceleration, then integrating again to find velocity, and finally using the work-energy theorem. The fractional power (t^{1/2}) adds algebraic complexity beyond standard M1/M2 questions, and the multi-step integration process with careful application of initial conditions makes this moderately challenging but still within standard M3 scope.
Spec6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.06a Variable force: dv/dt or v*dv/dx methods
2. A particle of mass 4 kg is moving along the horizontal \(x\)-axis under the action of a single force which acts in the positive \(x\)-direction. At time \(t\) seconds the force has magnitude \(\left( 1 + 3 t ^ { \frac { 1 } { 2 } } \right) \mathrm { N }\).
When \(t = 0\) the particle has speed \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the positive \(x\)-direction. Find the work done by the force in the interval \(0 \leqslant t \leqslant 4\)
Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(F = 1 + 3t^{\frac{1}{2}} = m\frac{dv}{dt} = 4\frac{dv}{dt}\)B1
\(4v = \int 1 + 3t^{\frac{1}{2}}\,dt = t + 2t^{\frac{3}{2}} (+C)\)M1A1
\(v = \frac{1}{4}(t + 2t^{1.5}) + 2\)A1
\(t = 4,\ v = \frac{1}{4}(4 + 16) + 2 = 7\ (\text{m s}^{-1})\)A1ft
Work done = gain in KE \(= \frac{1}{2}mv^2 - \frac{1}{2}mu^2\)M1 their \(v\)
\(= \frac{1}{2}\times 4\times 7^2 - \frac{1}{2}\times 4\times 2^2 = 90\ \text{(J)}\)A1 (7) 
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $F = 1 + 3t^{\frac{1}{2}} = m\frac{dv}{dt} = 4\frac{dv}{dt}$ | B1 | |
| $4v = \int 1 + 3t^{\frac{1}{2}}\,dt = t + 2t^{\frac{3}{2}} (+C)$ | M1A1 | |
| $v = \frac{1}{4}(t + 2t^{1.5}) + 2$ | A1 | |
| $t = 4,\ v = \frac{1}{4}(4 + 16) + 2 = 7\ (\text{m s}^{-1})$ | A1ft | |
| Work done = gain in KE $= \frac{1}{2}mv^2 - \frac{1}{2}mu^2$ | M1 | their $v$ |
| $= \frac{1}{2}\times 4\times 7^2 - \frac{1}{2}\times 4\times 2^2 = 90\ \text{(J)}$ | A1 | **(7)** |

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2. A particle of mass 4 kg is moving along the horizontal $x$-axis under the action of a single force which acts in the positive $x$-direction. At time $t$ seconds the force has magnitude $\left( 1 + 3 t ^ { \frac { 1 } { 2 } } \right) \mathrm { N }$.\\
When $t = 0$ the particle has speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the positive $x$-direction. Find the work done by the force in the interval $0 \leqslant t \leqslant 4$\\

\hfill \mbox{\textit{Edexcel M3 2013 Q2 [7]}}
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