OCR Further Mechanics Specimen — Question 3 5 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
SessionSpecimen
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImpulse and momentum (advanced)
TypeImpulse from variable force (then find velocity)
DifficultyStandard +0.3 This is a straightforward application of impulse-momentum theorem with integration. Students must integrate a polynomial-exponential function (standard A-level technique) and apply I = mv - mu. The integration is routine for Further Maths students, and the two-part structure guides them through the solution methodically.
Spec3.02a Kinematics language: position, displacement, velocity, acceleration6.03f Impulse-momentum: relation

3 A body, \(Q\), of mass 2 kg moves in a straight line under the action of a single force which acts in the direction of motion of \(Q\). Initially the speed of \(Q\) is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). At time \(t \mathrm {~s}\), the magnitude \(F \mathrm {~N}\) of the force is given by $$F = t ^ { 2 } + 3 \mathrm { e } ^ { t } , \quad 0 \leq t \leq 4$$
  1. Calculate the impulse of the force over the time interval.
  2. Hence find the speed of \(Q\) when \(t = 4\).

Question 3:
AnswerMarks Guidance
3(i) p
S
I (cid:32)(cid:179)t2(cid:14)3etdt(cid:32)1t3(cid:14)3et
3
4
I (cid:32)(cid:179) Fdt(cid:32)55(cid:14)3e4 (cid:32)182, so impulse is 182Ns
AnswerMarks
0 3M1
A1
A1
AnswerMarks
[3]3.3
1.1
AnswerMarks Guidance
1.1Integrating one term of F correctly
BC182.1277…
3(ii) 2(v(cid:16)5)(cid:32)55(cid:14)3e4
3
AnswerMarks
Speed is 96.1ms-1M1
A1
AnswerMarks Guidance
[2]3.4
1.1Use of I (cid:32)mv(cid:16)mu 96.0638…
Question 3:
3 | (i) | p
S
I (cid:32)(cid:179)t2(cid:14)3etdt(cid:32)1t3(cid:14)3et
3
4
I (cid:32)(cid:179) Fdt(cid:32)55(cid:14)3e4 (cid:32)182, so impulse is 182Ns
0 3 | M1
A1
A1
[3] | 3.3
1.1
1.1 | Integrating one term of F correctly
BC | 182.1277…
3 | (ii) | 2(v(cid:16)5)(cid:32)55(cid:14)3e4
3
Speed is 96.1ms-1 | M1
A1
[2] | 3.4
1.1 | Use of I (cid:32)mv(cid:16)mu | 96.0638…
3 A body, $Q$, of mass 2 kg moves in a straight line under the action of a single force which acts in the direction of motion of $Q$. Initially the speed of $Q$ is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At time $t \mathrm {~s}$, the magnitude $F \mathrm {~N}$ of the force is given by

$$F = t ^ { 2 } + 3 \mathrm { e } ^ { t } , \quad 0 \leq t \leq 4$$

(i) Calculate the impulse of the force over the time interval.\\
(ii) Hence find the speed of $Q$ when $t = 4$.

\hfill \mbox{\textit{OCR Further Mechanics  Q3 [5]}}