Edexcel M2 2016 June — Question 1 8 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2016
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImpulse and momentum (advanced)
TypeEnergy change from impulse
DifficultyModerate -0.3 This is a straightforward two-part mechanics question requiring standard application of impulse-momentum theorem and kinetic energy formula. Part (a) involves vector addition and magnitude calculation, while part (b) requires computing KE before and after—both routine M2 procedures with no conceptual challenges or novel problem-solving required.
Spec6.03e Impulse: by a force6.03f Impulse-momentum: relation

  1. A particle of mass 3 kg is moving with velocity \(( 3 \mathbf { i } + 5 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\) when it receives an impulse \(( - 4 \mathbf { i } + 3 \mathbf { j } ) \mathrm { N }\) s.
Find
  1. the speed of the particle immediately after receiving the impulse,
  2. the kinetic energy gained by the particle as a result of the impulse.

1a
AnswerMarks
Impulse-momentum equation: \((-4i + 3j) = 3(v - (3i + 5j))\)M1A1
\(v = \frac{5}{3}i + 6j\)A1
Find the magnitude: speed \(= \sqrt{(\frac{5}{3})^2 + 6^2} = 6.23\) (m s\(^{-1}\)) (6.2 or better)M1A1 (5)
Notes: First M1 for \(+(-4i + 3j) = 3(v - (3i + 5j))\) (M0 if 3 omitted or wrong mass used or term omitted); First A1 for a correct equation; Second A1 for a correct \(v\); Second M1 for finding the magnitude of their \(v\); Third A1 for \(\frac{\sqrt{349}}{3}\), 6.2 or better.
1b
AnswerMarks Guidance
Gain in KE \(= \frac{m}{2}(v ^2 -
\(= 7.17\) (J) (7.2 or better) (must be +ve)A1 (3) [8]
Notes: M1 for \(\pm\frac{3}{2}((their 6.23)^2 - (3^2 + 5^2))\) (M0 if 3 omitted or wrong mass used or term omitted); Also M0 for \(\pm\frac{3}{2}[(\frac{5}{3} + 6j)^2 - (3i + 5j)^2]\) unless it becomes \(\pm\frac{3}{2}[(\frac{5}{3})^2 + 6^2) - (3^2 + 5^2)]\); First A1ft on their \(v\) for a correct expression; Second A1 for 43/6 oe, 7.2 or better.
## 1a
| Impulse-momentum equation: $(-4i + 3j) = 3(v - (3i + 5j))$ | M1A1 |
| $v = \frac{5}{3}i + 6j$ | A1 |
| Find the magnitude: speed $= \sqrt{(\frac{5}{3})^2 + 6^2} = 6.23$ (m s$^{-1}$) (6.2 or better) | M1A1 (5) |
| **Notes:** First M1 for $+(-4i + 3j) = 3(v - (3i + 5j))$ (M0 if 3 omitted or wrong mass used or term omitted); First A1 for a correct equation; Second A1 for a correct $v$; Second M1 for finding the magnitude of their $v$; Third A1 for $\frac{\sqrt{349}}{3}$, 6.2 or better. |

## 1b
| Gain in KE $= \frac{m}{2}(|v|^2 - |u|^2) = \frac{3}{2}((their 6.23)^2 - (3^2 + 5^2))$ | M1A1 ft |
| $= 7.17$ (J) (7.2 or better) (must be +ve) | A1 (3) [8] |
| **Notes:** M1 for $\pm\frac{3}{2}((their 6.23)^2 - (3^2 + 5^2))$ (M0 if 3 omitted or wrong mass used or term omitted); Also M0 for $\pm\frac{3}{2}[(\frac{5}{3} + 6j)^2 - (3i + 5j)^2]$ unless it becomes $\pm\frac{3}{2}[(\frac{5}{3})^2 + 6^2) - (3^2 + 5^2)]$; First A1ft on their $v$ for a correct expression; Second A1 for 43/6 oe, 7.2 or better. |

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\begin{enumerate}
  \item A particle of mass 3 kg is moving with velocity $( 3 \mathbf { i } + 5 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ when it receives an impulse $( - 4 \mathbf { i } + 3 \mathbf { j } ) \mathrm { N }$ s.
\end{enumerate}

Find\\
(a) the speed of the particle immediately after receiving the impulse,\\
(b) the kinetic energy gained by the particle as a result of the impulse.\\

\hfill \mbox{\textit{Edexcel M2 2016 Q1 [8]}}