OCR MEI M1 — Question 2 19 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Marks19
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeProjectile clearing obstacle
DifficultyModerate -0.3 This is a standard M1 projectile motion question requiring routine application of SUVAT equations and trajectory formulas. While it has multiple parts and requires algebraic manipulation, all techniques are textbook-standard with no novel problem-solving insight needed. The question guides students through each step methodically, making it slightly easier than average for A-level mechanics.
Spec1.05g Exact trigonometric values: for standard angles1.05o Trigonometric equations: solve in given intervals3.02i Projectile motion: constant acceleration model
\includegraphics{figure_2} Fig. 7 shows a platform \(10\) m long and \(2\) m high standing on horizontal ground. A small ball projected from the surface of the platform at one end, O, just misses the other end, P. The ball is projected at \(68.5°\) to the horizontal with a speed of \(U\text{ms}^{-1}\). Air resistance may be neglected. At time \(t\) seconds after projection, the horizontal and vertical displacements of the ball from O are \(x\) m and \(y\) m.
  1. Obtain expressions, in terms of \(U\) and \(t\), for
    1. \(x\),
    2. \(y\). [3]
  2. The ball takes \(T\) s to travel from O to P. Show that \(T = \frac{U\sin 68.5°}{4.9}\) and write down a second equation connecting \(U\) and \(T\). [4]
  3. Hence show that \(U = 12.0\) (correct to three significant figures). [3]
  4. Calculate the horizontal distance of the ball from the platform when the ball lands on the ground. [5]
  5. Use the expressions you found in part (i) to show that the cartesian equation of the trajectory of the ball in terms of \(U\) is $$y = x\tan 68.5° - \frac{4.9x^2}{U^2(\cos 68.5°)^2}.$$ Use this equation to show again that \(U = 12.0\) (correct to three significant figures). [4]
Question 2:
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2mark notes
(i)
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(A)xUtcos68.5 B1
1
(i)
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(B)yUtsin68.54.9t2 M1
A1Allow ‘u’ = U. Allow sc. Allow g as g, ±9.8, ±9.81, ±10. Allow +2.
Accept not ‘shown’. Do not allow +2. Allow e.g + 0.5 × ( - 9.8) × t² instead of – 4.9t². Accept g not evaluated
2
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(ii)either
At D, y = 0
so Usin68.5T4.9T2 0
TUsin68.54.9T0
Usin68.5
so T = 0 (at C) or T  (at D)
4.9
or
Use (i)(A) and put x = 10 with t = T
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to get UTcos68.510M1
M1
E1
M1
M1
E1
B1
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4Equating correct y to 0 or their y to correct value.
Attempting to factorise (or solve). Allow ÷ T without comment.
Properly shown. Accept no ref to T = 0. Accept T = 0 given as well without comment.
Find time to top
Double time to the top
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(iii)Eliminating T from the results in (ii) gives
Usin68.5
Ucos68.5 10
4.9
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so U = 11.98729… so 12.0 (3 s. f.)M1
M1
E1
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3Substituting, using correct expressions or their expressions from (ii).
Attempt to solve for U ² or U.
Some evidence seen. e.g. 142.8025..U2 145.2025... with clear statement, or 11.9… seen with clear
statement or 11.98… seen. Accept 11.98… seen for full marks.
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(v)Eliminate t from (i) (B)
x
using t  from (i)(A)
Ucos68.5
4.9x2
so yxtan68.5
U2(cos68.5)2
We require y = 0 when x = 10
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so U = 11.98729… so 12.0 (3 s. f.)M1
E1
M1
E1
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4May be implied. FT their (i).
Clearly shown.
Must see attempt to solve. Or use x = 10.73… when y = – 2.
Must see evidence of fresh calculation or statement that they have now got the same expression for evaluation.
19
3 (i) y(0) = 1 B1
1
Either
(ii) 1 ( 20+5 )−5=7.5 M1 Use of symmetry e.g. use of1 ( 20+5 )
2 2
A1 12.5 o.e. seen
A1 7.5cao
or MM11 Att pt at y’ and to solve y’ = 0
A1 k(15 – 2x) where k =1 or 1
100
A1 7.5 cao, seen as final answer
y ( 7.5 )= 1 ( 100+15×7.5−7.52 ) M1 FT their 7.5
100
= 25 (1.5625) so 1.5625 m EE11 A
16
[SC2 only showing 1.5625 leads to x = 7.5]
5
PhysicsAndMathsTutor.com
AnswerMarks Guidance
hysicsAndMathsTutor.com 18 
Question 2:
2 | mark | notes
(i)
(A) | xUtcos68.5 | B1
1
(i)
(B) | yUtsin68.54.9t2 | M1
A1 | Allow ‘u’ = U. Allow sc. Allow g as g, ±9.8, ±9.81, ±10. Allow +2.
Accept not ‘shown’. Do not allow +2. Allow e.g + 0.5 × ( - 9.8) × t² instead of – 4.9t². Accept g not evaluated
2
(ii) | either
At D, y = 0
so Usin68.5T4.9T2 0
TUsin68.54.9T0
Usin68.5
so T = 0 (at C) or T  (at D)
4.9
or
Use (i)(A) and put x = 10 with t = T
to get UTcos68.510 | M1
M1
E1
M1
M1
E1
B1
4 | Equating correct y to 0 or their y to correct value.
Attempting to factorise (or solve). Allow ÷ T without comment.
Properly shown. Accept no ref to T = 0. Accept T = 0 given as well without comment.
Find time to top
Double time to the top
(iii) | Eliminating T from the results in (ii) gives
Usin68.5
Ucos68.5 10
4.9
so U = 11.98729… so 12.0 (3 s. f.) | M1
M1
E1
3 | Substituting, using correct expressions or their expressions from (ii).
Attempt to solve for U ² or U.
Some evidence seen. e.g. 142.8025..U2 145.2025... with clear statement, or 11.9… seen with clear
statement or 11.98… seen. Accept 11.98… seen for full marks.
(v) | Eliminate t from (i) (B)
x
using t  from (i)(A)
Ucos68.5
4.9x2
so yxtan68.5
U2(cos68.5)2
We require y = 0 when x = 10
so U = 11.98729… so 12.0 (3 s. f.) | M1
E1
M1
E1
4 | May be implied. FT their (i).
Clearly shown.
Must see attempt to solve. Or use x = 10.73… when y = – 2.
Must see evidence of fresh calculation or statement that they have now got the same expression for evaluation.
19
3 (i) y(0) = 1 B1
1
Either
(ii) 1 ( 20+5 )−5=7.5 M1 Use of symmetry e.g. use of1 ( 20+5 )
2 2
A1 12.5 o.e. seen
A1 7.5cao
or MM11 Att pt at y’ and to solve y’ = 0
A1 k(15 – 2x) where k =1 or 1
100
A1 7.5 cao, seen as final answer
y ( 7.5 )= 1 ( 100+15×7.5−7.52 ) M1 FT their 7.5
100
= 25 (1.5625) so 1.5625 m EE11 A
16
[SC2 only showing 1.5625 leads to x = 7.5]
5
PhysicsAndMathsTutor.com
hysic | sAndMathsTutor.com | 18
\includegraphics{figure_2}

Fig. 7 shows a platform $10$ m long and $2$ m high standing on horizontal ground. A small ball projected from the surface of the platform at one end, O, just misses the other end, P. The ball is projected at $68.5°$ to the horizontal with a speed of $U\text{ms}^{-1}$. Air resistance may be neglected.

At time $t$ seconds after projection, the horizontal and vertical displacements of the ball from O are $x$ m and $y$ m.

\begin{enumerate}[label=(\roman*)]
\item Obtain expressions, in terms of $U$ and $t$, for
\begin{enumerate}[label=(\Alph*)]
\item $x$,
\item $y$. [3]
\end{enumerate}

\item The ball takes $T$ s to travel from O to P.

Show that $T = \frac{U\sin 68.5°}{4.9}$ and write down a second equation connecting $U$ and $T$. [4]

\item Hence show that $U = 12.0$ (correct to three significant figures). [3]

\item Calculate the horizontal distance of the ball from the platform when the ball lands on the ground. [5]

\item Use the expressions you found in part (i) to show that the cartesian equation of the trajectory of the ball in terms of $U$ is
$$y = x\tan 68.5° - \frac{4.9x^2}{U^2(\cos 68.5°)^2}.$$

Use this equation to show again that $U = 12.0$ (correct to three significant figures). [4]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI M1  Q2 [19]}}
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