Question 1 - A-Level Maths

OCR MEI M1 — Question 1 18 marks

Exam Board	OCR MEI
Module	M1 (Mechanics 1)
Marks	18
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Projectiles
Type	Maximum range or optimal angle
Difficulty	Moderate -0.3 This is a standard M1 projectile motion question requiring routine application of SUVAT equations and projectile formulas. Part (i) is a straightforward derivation using basic kinematics, parts (ii-iii) involve simple substitution and comparison, and parts (iv-v) extend to constant horizontal deceleration but still use standard techniques. The multi-part structure and 14 total marks indicate moderate length, but no step requires novel insight—all are textbook applications of mechanics formulas.
Spec	1.05g Exact trigonometric values: for standard angles 1.05o Trigonometric equations: solve in given intervals 3.02i Projectile motion: constant acceleration model

Fig. 7 shows the trajectory of an object which is projected from a point O on horizontal ground. Its initial velocity is $40\text{ms}^{-1}$ at an angle of $\alpha$ to the horizontal. \includegraphics{figure_1}

Show that, according to the standard projectile model in which air resistance is neglected, the flight time, $T$ s, and the range, $R$ m, are given by $$T = \frac{80\sin\alpha}{g} \text{ and } R = \frac{3200\sin\alpha\cos\alpha}{g}.$$ [6] A company is designing a new type of ball and wants to model its flight.
Initially the company uses the standard projectile model. Use this model to show that when $\alpha = 30°$ and the initial speed is $40\text{ms}^{-1}$, $T$ is approximately $4.08$ and $R$ is approximately $141.4$. Find the values of $T$ and $R$ when $\alpha = 45°$. [3] The company tests the ball using a machine that projects it from ground level across horizontal ground. The speed of projection is set at $40\text{ms}^{-1}$. When the angle of projection is set at $30°$, the range is found to be $125$ m.
Comment briefly on the accuracy of the standard projectile model in this situation. [1] The company refines the model by assuming that the ball has a constant deceleration of $2\text{ms}^{-2}$ in the horizontal direction. In this new model, the resistance to the vertical motion is still neglected and so the flight time is still $4.08$ s when the angle of projection is $30°$.
Using the new model, with $\alpha = 30°$, show that the horizontal displacement from the point of projection, $x$ m at time $t$ s, is given by $$x = 40t\cos 30° - t^2.$$ Find the range and hence show that this new model is reasonably accurate in this case. [4] The company then sets the angle of projection to $45°$ while retaining a projection speed of $40\text{ms}^{-1}$. With this setting the range of the ball is found to be $135$ m.
Investigate whether the new model is also accurate for this angle of projection. [3]
Make one suggestion as to how the model could be further refined. [1]

Show mark scheme Show mark scheme source

Question 1:

Answer	Marks	Guidance
1	(i)	Vertical motion: initial speed 40sin

h(40sin)t1gt2

2 240sin

h0t 0 or

g

80sin

T 

Answer	Marks
g	B1

M1

Answer	Marks
E1	Correct expression for h must be seen. Condone omission of the case

t0

Perfect argument (but still condone omission of t0)

Alternative

Answer	Marks
Vertical motion: initial speed 40sin	(B1)

v40singt

T

When v0, t 

Answer	Marks	Guidance
2	(M1)	Correct expression for v must be seen

80sin

T 

Answer	Marks	Guidance
g	(E1)	Perfect argument

Horizontal motion: initial speed 40cos

R40cosT

3200sincos

R

Answer	Marks
g	B1

M1

E1

Answer	Marks
[6]	There must be evidence of intention to use T

Perfect argument

Answer	Marks
(iii)	The standard model is not accurate; 125 is
much less than 141.4	B1
[1]	The comment must be based on the figures given in the question
(iv)	Horizontal motion: sut 1at2

2 x40cos30t12t2

2 x40tcos30t2

Flight time = 4.08 s

R40cos304.08124.082

2 R124.7This is close to the experimental

Answer	Marks
result of 125 m	M1

A1

M1

E1

Answer	Marks
[4]	Use of correct formula

A comparison with 125 m is required

Answer	Marks
(v)	When 45, T 5.77

R40cos455.77125.772

2 R129.9

129.9 m is not very close to 135 m so the

Answer	Marks
model is not very accurate for this angle.	M1

A1

B1

Answer	Marks
[3]	Use of correct formula, with substitution for  and T. FT their T from

(iiii))but not 4

SC1 for substituting for T but using 30o for 

Comparison of their 129.9 with 135

If 4.08 used for T and answer 98.8 obtained for R allow FT for this

mark

Allow argument that to get to 135m takes 6.07 s which is greater than

5.77 s

Answer	Marks
(vi)	Allow for resistance in the vertical
direction as well	B1
[1]	Any sensible comment, but do not award a mark for “Allow for air

resistance” without mention of the vertical direction.

Question 1:
1 | (i) | Vertical motion: initial speed 40sin
h(40sin)t1gt2
2
240sin
h0t 0 or
g
80sin
T 
g | B1
M1
E1 | Correct expression for h must be seen. Condone omission of the case
t0
Perfect argument (but still condone omission of t0)
Alternative
Vertical motion: initial speed 40sin | (B1)
v40singt
T
When v0, t 
2 | (M1) | Correct expression for v must be seen
80sin
T 
g | (E1) | Perfect argument
Horizontal motion: initial speed 40cos
R40cosT
3200sincos
R
g | B1
M1
E1
[6] | There must be evidence of intention to use T
Perfect argument
(iii) | The standard model is not accurate; 125 is
much less than 141.4 | B1
[1] | The comment must be based on the figures given in the question
(iv) | Horizontal motion: sut 1at2
2
x40cos30t12t2
2
x40tcos30t2
Flight time = 4.08 s
R40cos304.08124.082
2
R124.7This is close to the experimental
result of 125 m | M1
A1
M1
E1
[4] | Use of correct formula
A comparison with 125 m is required
(v) | When 45, T 5.77
R40cos455.77125.772
2
R129.9
129.9 m is not very close to 135 m so the
model is not very accurate for this angle. | M1
A1
B1
[3] | Use of correct formula, with substitution for  and T. FT their T from
(iiii))but not 4
SC1 for substituting for T but using 30o for 
Comparison of their 129.9 with 135
If 4.08 used for T and answer 98.8 obtained for R allow FT for this
mark
Allow argument that to get to 135m takes 6.07 s which is greater than
5.77 s
(vi) | Allow for resistance in the vertical
direction as well | B1
[1] | Any sensible comment, but do not award a mark for “Allow for air
resistance” without mention of the vertical direction.

Show LaTeX source

Fig. 7 shows the trajectory of an object which is projected from a point O on horizontal ground. Its initial velocity is $40\text{ms}^{-1}$ at an angle of $\alpha$ to the horizontal.

\includegraphics{figure_1}

\begin{enumerate}[label=(\roman*)]
\item Show that, according to the standard projectile model in which air resistance is neglected, the flight time, $T$ s, and the range, $R$ m, are given by
$$T = \frac{80\sin\alpha}{g} \text{ and } R = \frac{3200\sin\alpha\cos\alpha}{g}.$$ [6]

A company is designing a new type of ball and wants to model its flight.

\item Initially the company uses the standard projectile model.

Use this model to show that when $\alpha = 30°$ and the initial speed is $40\text{ms}^{-1}$, $T$ is approximately $4.08$ and $R$ is approximately $141.4$.

Find the values of $T$ and $R$ when $\alpha = 45°$. [3]

The company tests the ball using a machine that projects it from ground level across horizontal ground. The speed of projection is set at $40\text{ms}^{-1}$.

When the angle of projection is set at $30°$, the range is found to be $125$ m.

\item Comment briefly on the accuracy of the standard projectile model in this situation. [1]

The company refines the model by assuming that the ball has a constant deceleration of $2\text{ms}^{-2}$ in the horizontal direction.

In this new model, the resistance to the vertical motion is still neglected and so the flight time is still $4.08$ s when the angle of projection is $30°$.

\item Using the new model, with $\alpha = 30°$, show that the horizontal displacement from the point of projection, $x$ m at time $t$ s, is given by
$$x = 40t\cos 30° - t^2.$$

Find the range and hence show that this new model is reasonably accurate in this case. [4]

The company then sets the angle of projection to $45°$ while retaining a projection speed of $40\text{ms}^{-1}$. With this setting the range of the ball is found to be $135$ m.

\item Investigate whether the new model is also accurate for this angle of projection. [3]

\item Make one suggestion as to how the model could be further refined. [1]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI M1  Q1 [18]}}

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