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