OCR M3 2013 January — Question 1 5 marks

Exam BoardOCR
ModuleM3 (Mechanics 3)
Year2013
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImpulse and momentum (advanced)
TypeAngle change from impulse
DifficultyStandard +0.3 This is a standard M3 impulse-momentum problem requiring resolution in two perpendicular directions and use of given trigonometric information. While it involves multiple steps (resolving components, using Pythagoras, finding angles), the method is routine for this topic with no novel insight required. Slightly easier than average due to the given sin α value simplifying calculations.
Spec6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form

1 \includegraphics[max width=\textwidth, alt={}, center]{dfe477d4-eae6-40e1-b704-1a97485f4c7e-2_477_534_261_770} A ball of mass 0.6 kg is moving with speed \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a straight line. It is struck by an impulse \(I \mathrm { Ns }\) acting at an acute angle \(\theta\) to its direction of motion (see diagram). The impulse causes the direction of motion of the ball to change by an acute angle \(\alpha\), where \(\sin \alpha = \frac { 8 } { 17 }\). After the impulse acts the ball is moving with a speed of \(3.4 \mathrm {~ms} ^ { - 1 }\). Find \(I\) and \(\theta\).

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
Use of cosine ruleM1 Condone \(+\) for \(-\); missing 2/ missing '0.6'; angle as '\(\theta\)' for M1
\(I^2 = 2.04^2 + 0.9^2 - 2 \times 2.04 \times 0.9 \times \frac{15}{17}\)A1 Condone \(+\) for \(-\)
\(1.32\) (N)A1 CAO (1.3159)
\(46.8°\) with initial direction of ballM1, A1 Can be in terms of \(I\), \(\alpha\) and \(\theta\); (46.8476) (0.8176 rads); Accept 46.7 from using \(I = 1.32\)
OR: \(0.9 + I\cos\theta = 0.6 \times 3.4 \times \frac{15}{17}\)M1 Allow missing 0.6 and/or sign or trig error
\(I\sin\theta = 0.6 \times 3.4 \times \frac{8}{17}\); square and add to find \(I^2\); or divide to find \(\theta\)M1
\(I, \theta\)A1, A1 CAO
[5]
# Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of cosine rule | M1 | Condone $+$ for $-$; missing 2/ missing '0.6'; angle as '$\theta$' for M1 |
| $I^2 = 2.04^2 + 0.9^2 - 2 \times 2.04 \times 0.9 \times \frac{15}{17}$ | A1 | Condone $+$ for $-$ |
| $1.32$ (N) | A1 | CAO (1.3159) |
| $46.8°$ with initial direction of ball | M1, A1 | Can be in terms of $I$, $\alpha$ and $\theta$; (46.8476) (0.8176 rads); Accept 46.7 from using $I = 1.32$ |
| OR: $0.9 + I\cos\theta = 0.6 \times 3.4 \times \frac{15}{17}$ | M1 | Allow missing 0.6 and/or sign or trig error |
| $I\sin\theta = 0.6 \times 3.4 \times \frac{8}{17}$; square and add to find $I^2$; or divide to find $\theta$ | M1 | |
| $I, \theta$ | A1, A1 CAO | |
| **[5]** | | |

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\includegraphics[max width=\textwidth, alt={}, center]{dfe477d4-eae6-40e1-b704-1a97485f4c7e-2_477_534_261_770}

A ball of mass 0.6 kg is moving with speed $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a straight line. It is struck by an impulse $I \mathrm { Ns }$ acting at an acute angle $\theta$ to its direction of motion (see diagram). The impulse causes the direction of motion of the ball to change by an acute angle $\alpha$, where $\sin \alpha = \frac { 8 } { 17 }$. After the impulse acts the ball is moving with a speed of $3.4 \mathrm {~ms} ^ { - 1 }$. Find $I$ and $\theta$.

\hfill \mbox{\textit{OCR M3 2013 Q1 [5]}}