OCR M3 2007 January — Question 2 7 marks

Exam BoardOCR
ModuleM3 (Mechanics 3)
Year2007
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImpulse and momentum (advanced)
TypeImpulse from velocity change
DifficultyStandard +0.3 This is a standard M3 impulse-momentum question requiring resolution of velocities in two perpendicular directions, application of impulse = change in momentum in each direction, then combining using Pythagoras and trigonometry. It's slightly above average difficulty due to the 2D nature and need to work with components, but follows a well-practiced procedure with no novel insight required.
Spec6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form
2 \includegraphics[max width=\textwidth, alt={}, center]{f334f6e4-2a60-4647-8b37-e48937c85747-2_231_971_539_587} When a tennis ball of mass 0.057 kg bounces it receives an impulse of magnitude \(I \mathrm {~N} \mathrm {~s}\) at an angle of \(\theta\) to the horizontal. Immediately before the ball bounces it has speed \(28 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a direction of \(30 ^ { \circ }\) to the horizontal. Immediately after the ball bounces it has speed \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a direction of \(30 ^ { \circ }\) to the horizontal (see diagram). Find \(I\) and \(\theta\).
Question 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(28\cos30° - 10\cos30°\) \([=\Delta v_H = (I/m)\cos\theta]\)
\(10\sin30° + 28\sin30°\) \([=\Delta v_V = (I/m)\sin\theta]\)
\([X = -I\cos\theta = -0.8885,\ Y = I\sin\theta = 1.083]\)M1 For using mv change for component or resultant
M1For using \(I^2 = X^2 + Y^2\)
\(I = 1.40\)A1
\([\tan\theta = 1.083/0.8885 \text{ or } 19/15.588]\)M1 For using \(\theta = \tan^{-1}(Y/-X)\) or \(\tan^{-1}(
\(\theta = 50.6°\)A1 Total: 7
Alternatively:
AnswerMarks Guidance
Answer/WorkingMark Guidance
M1For using cosine rule in correct triangle
\((I/m)^2 = 28^2 + 10^2 - 2\times28\times10\cos60°\) \([=604]\)A1
\([I = 0.057\sqrt{604}]\)M1 For using \(I =\) mv change
\(I = 1.40\)A1
M1For using sine rule in correct triangle
\((I/m)/\sin60° = 10/\sin(\theta-30°)\) or \(28/\sin(150°-\theta)\)A1
\(\theta = 50.6°\)A1 Total: 7 
# Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $28\cos30° - 10\cos30°$ $[= |\Delta v_H|= (I/m)\cos\theta]$ | B1 | |
| $10\sin30° + 28\sin30°$ $[= |\Delta v_V| = (I/m)\sin\theta]$ | B1 | |
| $[X = -I\cos\theta = -0.8885,\ Y = I\sin\theta = 1.083]$ | M1 | For using mv change for component or resultant |
| | M1 | For using $I^2 = X^2 + Y^2$ |
| $I = 1.40$ | A1 | |
| $[\tan\theta = 1.083/0.8885 \text{ or } 19/15.588]$ | M1 | For using $\theta = \tan^{-1}(Y/-X)$ or $\tan^{-1}(|\Delta v_V|/|\Delta v_H|)$ |
| $\theta = 50.6°$ | A1 | **Total: 7** |

**Alternatively:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using cosine rule in correct triangle |
| $(I/m)^2 = 28^2 + 10^2 - 2\times28\times10\cos60°$ $[=604]$ | A1 | |
| $[I = 0.057\sqrt{604}]$ | M1 | For using $I =$ mv change |
| $I = 1.40$ | A1 | |
| | M1 | For using sine rule in correct triangle |
| $(I/m)/\sin60° = 10/\sin(\theta-30°)$ or $28/\sin(150°-\theta)$ | A1 | |
| $\theta = 50.6°$ | A1 | **Total: 7** |

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2\\
\includegraphics[max width=\textwidth, alt={}, center]{f334f6e4-2a60-4647-8b37-e48937c85747-2_231_971_539_587}

When a tennis ball of mass 0.057 kg bounces it receives an impulse of magnitude $I \mathrm {~N} \mathrm {~s}$ at an angle of $\theta$ to the horizontal. Immediately before the ball bounces it has speed $28 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a direction of $30 ^ { \circ }$ to the horizontal. Immediately after the ball bounces it has speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a direction of $30 ^ { \circ }$ to the horizontal (see diagram). Find $I$ and $\theta$.

\hfill \mbox{\textit{OCR M3 2007 Q2 [7]}}
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