Edexcel M2 2018 June — Question 2 7 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2018
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImpulse and momentum (advanced)
TypeImpulse from velocity change
DifficultyStandard +0.3 This is a standard M2 impulse-momentum question requiring vector resolution in two perpendicular directions and basic trigonometry. The setup is straightforward with given values, requiring application of impulse-momentum principle and Pythagoras/trigonometry to find magnitude and direction—routine techniques for M2 students with no novel insight needed.
Spec6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form
2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{88731f1c-5177-4096-841b-cd9c3f87782b-06_314_1118_219_427} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} The points \(A , B\) and \(C\) lie on a smooth horizontal plane. A small ball of mass 0.2 kg is moving along the line \(A B\) with speed \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). When the ball is at \(B\), the ball is given an impulse. Immediately after the impulse is given, the ball moves along the line \(B C\) with speed \(7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The line \(B C\) makes an angle of \(35 ^ { \circ }\) with the line \(A B\), as shown in Figure 1.
  1. Find the magnitude of the impulse given to the ball.
  2. Find the size of the angle between the direction of the impulse and the original direction of motion of the ball.
Question 2:
Part 2a:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\begin{pmatrix} I\cos\theta \\ I\sin\theta \end{pmatrix} = 0.2\begin{pmatrix} 7\cos 35 \\ 7\sin 35 \end{pmatrix} - 0.2\begin{pmatrix} 4 \\ 0 \end{pmatrix}\)M1 Must be subtracting. Need both components. Could consider components separately
\(= \begin{pmatrix} 0.347 \\ 0.803 \end{pmatrix}\)A1 Correct unsimplified equation. Accept +/-
\(I = \sqrt{0.347^2 + 0.803^2}\)
\(I = 0.875\)
(4 marks total)
Alternative using vector triangle:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Vector triangle with sides 1.4, 0.8 and angle \(35°\)M1 Allow with velocities rather than impulse/momentum
Cosine rule: \(I ^2 = 1.4^2 + 0.8^2 - 2\times1.4\times0.8\times\cos 35°\)
\(I = 0.875\) (N s) (0.87 or better)
(4 marks total)
Part 2b:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\tan\theta = \frac{0.803}{0.347}\) \(\left(\text{or } \cos\theta = \frac{0.347}{0.875}\right)\)M1 Trig ratio of a relevant angle (using velocities or impulse/momentum)
A1ftCorrect expression for correct \(\theta\). Ft on values from (a). Do not ISW
\(\theta = 66.6°\ (67)\)A1 Or better from correct work
(3 marks total)
Alternative using sine rule:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{\sin\theta}{1.4} = \frac{\sin 35}{I }\)
\(\theta = 66.6°\ (67)\)A1 Or better
(3 marks total) 
# Question 2:

## Part 2a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix} I\cos\theta \\ I\sin\theta \end{pmatrix} = 0.2\begin{pmatrix} 7\cos 35 \\ 7\sin 35 \end{pmatrix} - 0.2\begin{pmatrix} 4 \\ 0 \end{pmatrix}$ | M1 | Must be subtracting. Need both components. Could consider components separately |
| $= \begin{pmatrix} 0.347 \\ 0.803 \end{pmatrix}$ | A1 | Correct unsimplified equation. Accept +/- |
| $|I| = \sqrt{0.347^2 + 0.803^2}$ | DM1 | Use Pythagoras to find magnitude. Dependent on previous M1 |
| $|I| = 0.875$ | A1 | 0.87 or better |
| **(4 marks total)** | | |

**Alternative using vector triangle:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Vector triangle with sides 1.4, 0.8 and angle $35°$ | M1 | Allow with velocities rather than impulse/momentum |
| Cosine rule: $|I|^2 = 1.4^2 + 0.8^2 - 2\times1.4\times0.8\times\cos 35°$ | DM1 A1 | Dependent on previous M1 |
| $|I| = 0.875$ (N s) (0.87 or better) | A1 | |
| **(4 marks total)** | | |

## Part 2b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\theta = \frac{0.803}{0.347}$ $\left(\text{or } \cos\theta = \frac{0.347}{0.875}\right)$ | M1 | Trig ratio of a relevant angle (using velocities or impulse/momentum) |
| | A1ft | Correct expression for correct $\theta$. Ft on values from (a). Do not ISW |
| $\theta = 66.6°\ (67)$ | A1 | Or better from correct work |
| **(3 marks total)** | | |

**Alternative using sine rule:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\sin\theta}{1.4} = \frac{\sin 35}{|I|}$ | M1 A1ft | |
| $\theta = 66.6°\ (67)$ | A1 | Or better |
| **(3 marks total)** | | |

---
2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{88731f1c-5177-4096-841b-cd9c3f87782b-06_314_1118_219_427}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

The points $A , B$ and $C$ lie on a smooth horizontal plane. A small ball of mass 0.2 kg is moving along the line $A B$ with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. When the ball is at $B$, the ball is given an impulse. Immediately after the impulse is given, the ball moves along the line $B C$ with speed $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The line $B C$ makes an angle of $35 ^ { \circ }$ with the line $A B$, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the impulse given to the ball.
\item Find the size of the angle between the direction of the impulse and the original direction of motion of the ball.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2018 Q2 [7]}}
This paper (7 questions)
View full paper
Q1 8 Q2 7 Q3 8 Q4 10 Q5 13 Q6 14 Q7 15