OCR MEI M1 — Question 4 7 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeTwo-particle meeting or overtaking
DifficultyModerate -0.3 This is a standard M1 kinematics problem involving two objects moving towards each other. Part (i) requires straightforward application of SUVAT equations (s=ut for P, s=ut+½at² for Q). Part (ii) involves setting up an equation where distances sum to 90m, then solving a quadratic - all routine techniques for M1 with clear guidance ('show that'). Slightly easier than average due to the structured parts and standard setup.
Spec3.02d Constant acceleration: SUVAT formulae
Two cars, P and Q, are being crashed as part of a film 'stunt'. At the start
  • P is travelling directly towards Q with a speed of \(8\) ms\(^{-1}\),
  • Q is instantaneously at rest and has an acceleration of \(4\) ms\(^{-2}\) directly towards P.
P continues with the same velocity and Q continues with the same acceleration. The cars collide \(T\) seconds after the start.
  1. Find expressions in terms of \(T\) for how far each of the cars has travelled since the start. [2]
At the start, P is 90 m from Q.
  1. Show that \(T^2 + 4T - 45 = 0\) and hence find \(T\). [5]
Question 4:
AnswerMarks
4(i)For P: the distance is 8T
For Q: the distance is 14T2
AnswerMarks
2B1
B1
AnswerMarks
2Allow – ve. Allow any form.
Allow – ve. Allow any form.
AnswerMarks
(ii)Require 8T14T2 90
2
so 8T + 2T² – 90 = 0
so T² + 4T – 45 = 0
This gives
(T – 5)(T + 9) = 0
AnswerMarks
so T = 5 since T > 0M1
A1
E1
M1
A1
AnswerMarks
5For linking correct expressions or their expressions from (i) with 90. Condone sign errors and use of
displacement instead of distance. Condone ‘= 0’implied.
The expression is correct or correctly derived from their (i). Reason not required.
Must be established. Do not award if their ‘correct expression’ comes from incorrect manipulation.
Solving to find +ve root. Accept (T + 5)(T – 9).
Condone 2nd root not found/discussed but not both roots given.
7
5(i)
(ii)
AnswerMarks
(iii)a=6t−12
We need  3 (3t2−12t+14)dt
1
= t3−6t2+14t 3
 
1
either
= (27 – 54 + 42) – (1 – 6 + 14)
= 15 – 9 = 6 so 6 m
or
s=t3−6t2+14t+C
s = 0 when t = 1 gives
0 = 1 – 6 + 14 + C so C = – 9
Put t = 3 to give
s = 27 – 54 + 42 – 9 = 6 so 6 m.
v > 0 so the particle always travels in the
same (+ve) direction
As the particle never changes direction,
the final distance from the starting point
AnswerMarks
is the displacement.M1
A1
M1
A1
M1
A1
M1
AA11
E1
AnswerMarks
E1Differentiating
cao
Integrating. Neglect limits.
At least two terms correct. Neglect limits.
Dep on 1st M1. Use of limits with attempt at
subtraction seen.
cao
Dep on 1st M1. An attempt to find C using s(1) = 0
and then evaluating s(3).
ca
Only award if explicit
AnswerMarks
Complete argument2
4
2
8
Question 4:
--- 4(i) ---
4(i) | For P: the distance is 8T
For Q: the distance is 14T2
2 | B1
B1
2 | Allow – ve. Allow any form.
Allow – ve. Allow any form.
(ii) | Require 8T14T2 90
2
so 8T + 2T² – 90 = 0
so T² + 4T – 45 = 0
This gives
(T – 5)(T + 9) = 0
so T = 5 since T > 0 | M1
A1
E1
M1
A1
5 | For linking correct expressions or their expressions from (i) with 90. Condone sign errors and use of
displacement instead of distance. Condone ‘= 0’implied.
The expression is correct or correctly derived from their (i). Reason not required.
Must be established. Do not award if their ‘correct expression’ comes from incorrect manipulation.
Solving to find +ve root. Accept (T + 5)(T – 9).
Condone 2nd root not found/discussed but not both roots given.
7
5(i)
(ii)
(iii) | a=6t−12
We need  3 (3t2−12t+14)dt
1
= t3−6t2+14t 3
 
1
either
= (27 – 54 + 42) – (1 – 6 + 14)
= 15 – 9 = 6 so 6 m
or
s=t3−6t2+14t+C
s = 0 when t = 1 gives
0 = 1 – 6 + 14 + C so C = – 9
Put t = 3 to give
s = 27 – 54 + 42 – 9 = 6 so 6 m.
v > 0 so the particle always travels in the
same (+ve) direction
As the particle never changes direction,
the final distance from the starting point
is the displacement. | M1
A1
M1
A1
M1
A1
M1
AA11
E1
E1 | Differentiating
cao
Integrating. Neglect limits.
At least two terms correct. Neglect limits.
Dep on 1st M1. Use of limits with attempt at
subtraction seen.
cao
Dep on 1st M1. An attempt to find C using s(1) = 0
and then evaluating s(3).
ca
Only award if explicit
Complete argument | 2
4
2
8
Two cars, P and Q, are being crashed as part of a film 'stunt'.

At the start
\begin{itemize}
\item P is travelling directly towards Q with a speed of $8$ ms$^{-1}$,
\item Q is instantaneously at rest and has an acceleration of $4$ ms$^{-2}$ directly towards P.
\end{itemize}

P continues with the same velocity and Q continues with the same acceleration. The cars collide $T$ seconds after the start.

\begin{enumerate}[label=(\roman*)]
\item Find expressions in terms of $T$ for how far each of the cars has travelled since the start. [2]
\end{enumerate}

At the start, P is 90 m from Q.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that $T^2 + 4T - 45 = 0$ and hence find $T$. [5]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI M1  Q4 [7]}}
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