MARK SCHEME – AS FURTHER MATHEMATICS – 7366/2M – JUNE 2022
6
Q Marking instructions AO Marks Typical solution
1 Circles correct answer. 1.1b B1 50 J
Question total 1
Q Marking instructions AO Marks Typical solution
2 Circles correct answer. 1.1b B1 �
4
1
� ms–1
Question total 1
Q Marking instructions AO Marks Typical solution
3(a) Recalls the formula for kinetic
energy and calculates the initial
kinetic energy.
Condone missing units.
1.1b B1 KE = 1
2 ????????????????2 = 1
2 (0.75)(12)2 = 54 J
Subtotal 1
Q Marking instructions AO Marks Typical solution
3(b) Uses conservation of energy to
form an equation with PE and
their KE from part (a).
3.3 M1 mgh = 54
h =
54
(0.75)(9.8)
= 7.34..
Jeff has assumed no air resistance
to obtain h = 7.3
Gurjas includes air resistance and
so knows the ball will not reach 7. 3
metres
Solves the equation to obtain
h = 7.3
AWRT 7.3
Must have clearly rearranged
the equation to find h or obtains
correct value (7.3469..) to at
least 3 sf.
AG
1.1b A1
Makes an inference about one
or more assumptions for both
Jeff and Gurjas.
For example:
• Jeff has assumed no air
resistance, Gurjas has
taken this into account
• Jeff assumes that all
energy is conserved,
Gurjas does not.
2.2b E1
Subtotal 3
Question total 4
Version | LATEST 2022 |
Category | AQA Questions and Marking Scheme |
Release date | 2022-09-14 |
Included files | |
Authors | Qwivy.com |
Pages | 12 |
Language | English |
Tags | AQA AS FURTHER MATHEMATICS 7366/2M Paper 2 Mechanics Mark scheme June 2022 |
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High resolution | Yes |
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