Question 2 - A-Level Maths

OCR M3 2007 June — Question 2 7 marks

Exam Board	OCR
Module	M3 (Mechanics 3)
Year	2007
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Impulse and momentum (advanced)
Type	Angle change from impulse
Difficulty	Standard +0.3 This is a straightforward application of the impulse-momentum theorem using vector triangles. Students need to find initial momentum (0.57 N s), final momentum (0.399 N s), apply the cosine rule to the triangle formed by initial momentum, final momentum, and impulse vectors, then solve for the angle. It's a standard M3 question requiring one main technique with clear numerical values provided, making it slightly easier than average.
Spec	6.03f Impulse-momentum: relation

2 A tennis ball of mass 0.057 kg has speed \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The ball receives an impulse of magnitude 0.6 N s which reduces the speed of the ball to \(7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Using an impulse-momentum triangle, or otherwise, find the angle the impulse makes with the original direction of motion of the ball.

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Question 2:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
[Magnitudes 0.6, 0.057 × 7, 0.057 × 10]	M1	For triangle with magnitudes shown
For magnitudes of 2 sides correctly marked	A1
For magnitudes of all 3 sides correctly marked	A1
M1	For attempting to find angle (\(\alpha\)) opposite to the side of magnitude \(0.057 \times 7\)
M1	For correct use of the cosine rule or equivalent
\(0.399^2 = 0.57^2 + 0.6^2 - 2 \times 0.57 \times 0.6\cos\alpha\)	A1ft
Angle is \(140°\)	A1	[7 marks total] \((180 - 39.8)°\)

Alternative Method

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
M1	For using \(I = \Delta mv\) parallel to the initial direction of motion or parallel to the impulse
\(-0.6\cos\alpha = 0.057 \times 7\cos\beta - 0.057 \times 10\) or \(0.6 = 0.057 \times 10\cos\alpha + 0.057 \times 7\cos\gamma\)	A1
M1	For using \(I = \Delta mv\) perpendicular to the initial direction of motion or perpendicular to the impulse
\(0.6\sin\alpha = 0.057 \times 7\sin\beta\) or \(0.057 \times 10\sin\alpha = 0.057 \times 7\sin\gamma\)	A1
M1	For eliminating \(\beta\) *or \(\gamma\)
\(0.399^2 = (0.57 - 0.6\cos\alpha)^2 + (0.6\sin\alpha)^2\) or \(0.399^2 = (0.6 - 0.57\cos\alpha)^2 + (0.057\sin\alpha)^2\)	A1ft
Angle is \(140°\)	A1	[7 marks total] \((180 - 39.8)°\)

# Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| [Magnitudes 0.6, 0.057 × 7, 0.057 × 10] | M1 | For triangle with magnitudes shown |
| For magnitudes of 2 sides correctly marked | A1 | |
| For magnitudes of all 3 sides correctly marked | A1 | |
| | M1 | For attempting to find angle ($\alpha$) opposite to the side of magnitude $0.057 \times 7$ |
| | M1 | For correct use of the cosine rule or equivalent |
| $0.399^2 = 0.57^2 + 0.6^2 - 2 \times 0.57 \times 0.6\cos\alpha$ | A1ft | |
| Angle is $140°$ | A1 | [7 marks total] $(180 - 39.8)°$ |

## Alternative Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $I = \Delta mv$ parallel to the initial direction of motion or parallel to the impulse |
| $-0.6\cos\alpha = 0.057 \times 7\cos\beta - 0.057 \times 10$ or $0.6 = 0.057 \times 10\cos\alpha + 0.057 \times 7\cos\gamma$ | A1 | |
| | M1 | For using $I = \Delta mv$ perpendicular to the initial direction of motion or perpendicular to the impulse |
| $0.6\sin\alpha = 0.057 \times 7\sin\beta$ or $0.057 \times 10\sin\alpha = 0.057 \times 7\sin\gamma$ | A1 | |
| | M1 | For eliminating $\beta$ *or $\gamma$ |
| $0.399^2 = (0.57 - 0.6\cos\alpha)^2 + (0.6\sin\alpha)^2$ or $0.399^2 = (0.6 - 0.57\cos\alpha)^2 + (0.057\sin\alpha)^2$ | A1ft | |
| Angle is $140°$ | A1 | [7 marks total] $(180 - 39.8)°$ |

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Show LaTeX source

2 A tennis ball of mass 0.057 kg has speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The ball receives an impulse of magnitude 0.6 N s which reduces the speed of the ball to $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Using an impulse-momentum triangle, or otherwise, find the angle the impulse makes with the original direction of motion of the ball.

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

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