OCR MEI M1 — Question 4 6 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeProjectile passing through given point
DifficultyStandard +0.3 This is a standard projectile motion problem requiring students to find the time when the stone is at a given height (solving a quadratic), then calculate horizontal distance using constant velocity. While it involves multiple steps and careful arithmetic with the given vertical displacement formula, it follows a routine method with no conceptual surprises. The provision of the vertical displacement formula and cos α simplifies what would otherwise be standard M1 work, making it slightly easier than average.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=13.02i Projectile motion: constant acceleration model

Sandy is throwing a stone at a plum tree. The stone is thrown from a point O at a speed of \(35\text{ms}^{-1}\) at an angle of \(\alpha\) to the horizontal, where \(\cos\alpha = 0.96\). You are given that, \(t\) seconds after being thrown, the stone is \((9.8t - 4.9t^2)\) m higher than O. When descending, the stone hits a plum which is \(3.675\) m higher than O. Air resistance should be neglected. Calculate the horizontal distance of the plum from O. [6]

Question 4:
AnswerMarks Guidance
4Mark Comment
either
We need 3.675=9.8t−4.9t2
Solving 4t2−8t+3=0
gives t = 0.5 or t = 1.5
or
Time to greatest height
0=35×0.28−9.8t so t = 1
Time to drop is 0.5

total is 1.5 s

then
Horiz distance is 35×0.96t
AnswerMarks
So distance is 35×0.96×1.5=50.4 m*M1
M1*
A1
F1
M1
A1
A1
A1
B1
AnswerMarks
F1Equating given expression or their attempt at y to
±3.675. If they attempt y, allow sign errors,
g = 9.81 etc. and u = 35.
Dependent. Any method of solution of a 3 term
quadratic.
cao. Accept only the larger root given
Both roots shown and larger chosen provided
both +ve. Dependent on 1st M1.
[Award M1 M1 A1 for 1.5 seen WW]
Complete method for total time from motion in
separate parts. Allow sign errors, g = 9.81 etc.
Allow u = 35 initially only.
Time for 1st part
Time for 2nd part
cao
Use of x=ucosαt. May be implied.
AnswerMarks
FT their quoted t provided it is positive.6
6
Question 4:
4 | Mark | Comment | Sub
either
We need 3.675=9.8t−4.9t2
Solving 4t2−8t+3=0
gives t = 0.5 or t = 1.5
or
Time to greatest height
0=35×0.28−9.8t so t = 1
Time to drop is 0.5
total is 1.5 s
then
Horiz distance is 35×0.96t
So distance is 35×0.96×1.5=50.4 m | *M1
M1*
A1
F1
M1
A1
A1
A1
B1
F1 | Equating given expression or their attempt at y to
±3.675. If they attempt y, allow sign errors,
g = 9.81 etc. and u = 35.
Dependent. Any method of solution of a 3 term
quadratic.
cao. Accept only the larger root given
Both roots shown and larger chosen provided
both +ve. Dependent on 1st M1.
[Award M1 M1 A1 for 1.5 seen WW]
Complete method for total time from motion in
separate parts. Allow sign errors, g = 9.81 etc.
Allow u = 35 initially only.
Time for 1st part
Time for 2nd part
cao
Use of x=ucosαt. May be implied.
FT their quoted t provided it is positive. | 6
6
Sandy is throwing a stone at a plum tree. The stone is thrown from a point O at a speed of $35\text{ms}^{-1}$ at an angle of $\alpha$ to the horizontal, where $\cos\alpha = 0.96$. You are given that, $t$ seconds after being thrown, the stone is $(9.8t - 4.9t^2)$ m higher than O.

When descending, the stone hits a plum which is $3.675$ m higher than O. Air resistance should be neglected.

Calculate the horizontal distance of the plum from O. [6]

\hfill \mbox{\textit{OCR MEI M1  Q4 [6]}}