OCR MEI Further Mechanics Major Specimen — Question 2 3 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
SessionSpecimen
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeVector impulse: find velocity or speed after impulse
DifficultyModerate -0.8 This is a straightforward impulse-momentum question requiring only vector manipulation: find the unit vector in the given direction, multiply by magnitude to get impulse vector, apply impulse = change in momentum, then solve for final velocity. It's routine application of a standard formula with no conceptual difficulty beyond basic 3D vector arithmetic.
Spec1.10d Vector operations: addition and scalar multiplication6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form
A particle of mass 5 kg is moving with velocity \(2\mathbf{i} + 5\mathbf{j}\) m s\(^{-1}\). It receives an impulse of magnitude 15 N s in the direction \(\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\). Find the velocity of the particle immediately afterwards. [3]
Question 2:
AnswerMarks
2Impulse is 5(i + 2j –2k)
5(i + 2j –2k) = 5(v – 2i – 5j)
AnswerMarks
v = 3i + 7j –2kB1
M1
A1
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[3]1.1
3.4
AnswerMarks
1.1N
E
Use
Impulse = change in momentum
AnswerMarks Guidance
22 0
3i1 0
3ii2 0
4i1 0
4ii1 0
4iii1 0
5i2 1
5ii2 0
6i3 1
6ii4 0
6iii0 1
7i1 2
7iiA0 1
1
AnswerMarks Guidance
7iiB1 1
8i1 0
12
8ii3 1
8iii3 0
9i2 1
03 6
9ii4 1
9iii0 0
01 1
10i1 0
10ii2 C
00 0
10iiiA1 0
10iiiBE
01 1
10iv1 0
10v0 0
11iP
43 0
11ii3 1
S
AnswerMarks Guidance
11iii2 1
12i3 1
12iiA2 1
12iiB1 0
12iii2 0
12iv0 3
12v0 0
Totals60 21
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18
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Question 2:
2 | Impulse is 5(i + 2j –2k)
5(i + 2j –2k) = 5(v – 2i – 5j)
v = 3i + 7j –2k | B1
M1
A1
[3] | 1.1
3.4
1.1 | N
E
Use
Impulse = change in momentum
2 | 2 | 0 | 0 | 1 | 3
3i | 1 | 0 | 0 | 1 | 2
3ii | 2 | 0 | 0 | 1 | 3
4i | 1 | 0 | 0 | 1 | 2
4ii | 1 | 0 | 0 | 1 | 2
4iii | 1 | 0 | 1 | 0 | 2
5i | 2 | 1 | 0 | 1 | 4
5ii | 2 | 0 | 0 | 2 | 4
6i | 3 | 1 | 0 | 0 | 4
6ii | 4 | 0 | 0 | 1 | 5
6iii | 0 | 1 | 0 | 0 | 1
7i | 1 | 2 | 0 | 1 | 4
7iiA | 0 | 1 | 0 | 0 | N
1
7iiB | 1 | 1 | 1 | 1 | 4
8i | 1 | 0 | 0 | E
1 | 2
8ii | 3 | 1 | 1 | 2 | 7
8iii | 3 | 0 | 3 | 1 | 7
9i | 2 | 1 | M
0 | 3 | 6
9ii | 4 | 1 | 1 | 1 | 7
9iii | 0 | 0 | I
0 | 1 | 1
10i | 1 | 0 | 0 | 3 | 4
10ii | 2 | C
0 | 0 | 0 | 2
10iiiA | 1 | 0 | 0 | 1 | 2
10iiiB | E
0 | 1 | 1 | 0 | 2
10iv | 1 | 0 | 2 | 0 | 3
10v | 0 | 0 | 0 | 1 | 1
11i | P
4 | 3 | 0 | 0 | 7
11ii | 3 | 1 | 0 | 0 | 4
S
11iii | 2 | 1 | 1 | 1 | 5
12i | 3 | 1 | 0 | 1 | 5
12iiA | 2 | 1 | 0 | 0 | 3
12iiB | 1 | 0 | 0 | 0 | 1
12iii | 2 | 0 | 0 | 0 | 2
12iv | 0 | 3 | 0 | 0 | 3
12v | 0 | 0 | 0 | 1 | 1
Totals | 60 | 21 | 11 | 28 | 120
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A particle of mass 5 kg is moving with velocity $2\mathbf{i} + 5\mathbf{j}$ m s$^{-1}$. It receives an impulse of magnitude 15 N s in the direction $\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}$. Find the velocity of the particle immediately afterwards. [3]

\hfill \mbox{\textit{OCR MEI Further Mechanics Major  Q2 [3]}}
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