OCR Further Mechanics Specimen — Question 1 9 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
SessionSpecimen
Marks9
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Mark schemeDownload PDF ↗
TopicAdvanced work-energy problems
TypeWork done by vector force displacement
DifficultyStandard +0.8 This Further Mechanics question requires differentiation of vector functions to find force, calculation of power using F·v, and application of the work-energy theorem with integration. While the techniques are standard for Further Maths, the multi-step vector calculus and careful algebraic manipulation place it moderately above average difficulty.
Spec3.02g Two-dimensional variable acceleration6.02d Mechanical energy: KE and PE concepts6.02l Power and velocity: P = Fv
1 A body, \(P\), of mass 2 kg moves under the action of a single force \(\mathbf { F } \mathrm { N }\). At time \(t \mathrm {~s}\), the velocity of the body is \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\), where $$\mathbf { v } = \left( t ^ { 2 } - 3 \right) \mathbf { i } + \frac { 5 } { 2 t + 1 } \mathbf { j } \text { for } t \geq 2$$
  1. Obtain \(\mathbf { F }\) in terms of \(t\).
  2. Calculate the rate at which the force \(\mathbf { F }\) is working at \(t = 4\).
  3. By considering the change in kinetic energy of \(P\), calculate the work done by the force \(\mathbf { F }\) during the time interval \(2 \leq t \leq 4\).
Question 1:
AnswerMarks Guidance
1(i) dv 10
a(cid:32) (cid:32)2ti(cid:16) j
dt (2t(cid:14)1)2
20
F(cid:32)ma(cid:32)4ti(cid:16) j
AnswerMarks
(2t(cid:14)1)2M1
A1
A1FT
AnswerMarks
[3]1.1
1.1
AnswerMarks
3.3Differentiating one term of v correctly
FT their a
AnswerMarks Guidance
1(ii) F.v(cid:32)4t (cid:11) t2 (cid:16)3 (cid:12) (cid:16) 100
(cid:11)2t(cid:14)1(cid:12)3
AnswerMarks
Power is 208WM1
A1
A1
AnswerMarks
[3]3.4
1.1
AnswerMarks
1.1n
Attempt at scalar product
e
m
AnswerMarks
207.86282…OR
M1 at t = 4
A1F(cid:152)v(cid:32) (cid:11) 16i(cid:16)20j (cid:12) (cid:152) (cid:11) 13i(cid:14)5j (cid:12)
81 9
AnswerMarks Guidance
1(iii) T (cid:32) 1 mv.v(cid:32) (cid:11) t2 (cid:16)3 (cid:12)2 (cid:14) 25
2 (cid:11)2t(cid:14)1(cid:12)2
p
AnswerMarks
Work done is 167JB1
e
M1
A1
AnswerMarks
[3]i
c
3.4
1.1
AnswerMarks
1.1T(4)(cid:16)T(2)
167.308642…OR
B1 for value of T at 2 or 4
M1 T(4)(cid:16)T(2)
Question 1:
1 | (i) | dv 10
a(cid:32) (cid:32)2ti(cid:16) j
dt (2t(cid:14)1)2
20
F(cid:32)ma(cid:32)4ti(cid:16) j
(2t(cid:14)1)2 | M1
A1
A1FT
[3] | 1.1
1.1
3.3 | Differentiating one term of v correctly
FT their a
1 | (ii) | F.v(cid:32)4t (cid:11) t2 (cid:16)3 (cid:12) (cid:16) 100
(cid:11)2t(cid:14)1(cid:12)3
Power is 208W | M1
A1
A1
[3] | 3.4
1.1
1.1 | n
Attempt at scalar product
e
m
207.86282… | OR
M1 at t = 4
A1F(cid:152)v(cid:32) (cid:11) 16i(cid:16)20j (cid:12) (cid:152) (cid:11) 13i(cid:14)5j (cid:12)
81 9
1 | (iii) | T (cid:32) 1 mv.v(cid:32) (cid:11) t2 (cid:16)3 (cid:12)2 (cid:14) 25
2 (cid:11)2t(cid:14)1(cid:12)2
p
Work done is 167J | B1
e
M1
A1
[3] | i
c
3.4
1.1
1.1 | T(4)(cid:16)T(2)
167.308642… | OR
B1 for value of T at 2 or 4
M1 T(4)(cid:16)T(2)
1 A body, $P$, of mass 2 kg moves under the action of a single force $\mathbf { F } \mathrm { N }$. At time $t \mathrm {~s}$, the velocity of the body is $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$, where

$$\mathbf { v } = \left( t ^ { 2 } - 3 \right) \mathbf { i } + \frac { 5 } { 2 t + 1 } \mathbf { j } \text { for } t \geq 2$$

(i) Obtain $\mathbf { F }$ in terms of $t$.\\
(ii) Calculate the rate at which the force $\mathbf { F }$ is working at $t = 4$.\\
(iii) By considering the change in kinetic energy of $P$, calculate the work done by the force $\mathbf { F }$ during the time interval $2 \leq t \leq 4$.

\hfill \mbox{\textit{OCR Further Mechanics  Q1 [9]}}
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