1. (a) The
extension of a spring is directly proportional to the applied force M1
as long as the elastic limit is not exceeded) A1
as long as the elastic limit is not exceeded) A1
(b) (i) Correct pair of values read from the graph
force constant = 12/0.080 C1
force constant = 150 (N m–1) A1
force constant = 12/0.080 C1
force constant = 150 (N m–1) A1
(ii) extension, x = 20/12 × 80 (= 133.33) (mm) C1
(E = ½ Fx)
energy = ½ × 20 × 133.33 × 10–3
energy = 1.33 (J) A1
(E = ½ Fx)
energy = ½ × 20 × 133.33 × 10–3
energy = 1.33 (J) A1
(iii) The
spring has not exceeded its elastic limit B1
(iv) (elastic potential energy = kinetic energy)
1/2kx2 = 1/2mv2 M1
m and k are constant, therefore x µ v. M1
21/2kx2 = 1/2mv2 M1
m and k are constant, therefore x µ v. M1
(a) (i) F
= kx / k is the gradient of the graph C1
k = 2.0 / 250 ´ 10–3 = 8.0 A1
Correct unit for
value given in (a)(i)
i.e.
0.008 or 8 ´
10–3 requires N mm–1.
Allow
N m–1 / kg s–2 if no working in (a)(i).
Do
not allow unit mark if incorrect physics in part (a)(i) B1
(ii) W = ½
(F ´
extension) / area under the graph C1
= ½
´ 2.0 ´ 0.250
= 0.25 (J) A1
(b) (i) F = 8 ´ 0.15 = 1.2 (N) A1
(ii) Hooke’s law continues to be obeyed / graph
continues as a straight
line
/ k is constant / elastic limit has not been reached B1
(c) (i) 1. correct
time marked on the graph with a V (t = 0.75 s or 1.75 s) B1
2. tangent in the correct place for downward
velocity or implied
by values B1
by values B1
value between 0.95 to 1.1(m s–1) A1
(ii) 1. X
marked in a correct place (maximum or minimum on graph) M1
2. relates the extension / compression to F
= kx to explain why the
force is a maximum or maximum
extension gives max force or
maximum extension gives max acceleration A1
maximum extension gives max acceleration A1
3
(a) The extension of a spring is directly
proportional to the applied force M1
as long as the elastic limit is not exceeded) A1
as long as the elastic limit is not exceeded) A1
(b) (i) Correct
pair of values read from the graph
force constant = 12/0.080 C1
force constant = 150 (N m–1) A1
force constant = 12/0.080 C1
force constant = 150 (N m–1) A1
(ii) extension, x = 20/12 × 80 (= 133.33) (mm) C1
(E = ½ Fx)
energy = ½ × 2/ × 133.33 × 10–3
energy = 1.33 (J) A1
(E = ½ Fx)
energy = ½ × 2/ × 133.33 × 10–3
energy = 1.33 (J) A1
(iii) The
spring has not exceeded its elastic limit B1
(iv) (elastic potential energy = kinetic energy)
1/2kx2 = 1/2mv2 M1
m and k are constant, therefore x µ v. M1
1/2kx2 = 1/2mv2 M1
m and k are constant, therefore x µ v. M1