Gravity Log graph Kepler 3

A bit small for the Sun - I missed a zero off should be 10 to the power of 30

Expression for g

y12 homework on Elastic Energy

Gravity questions

 (a)     arrow towards centre of planet                                                                                   1

(b)     (i)      g = GM/R2                                                                                               1

(ii)     gS/gO = R2/25R2; gS = 40/25 (= 1.6 N kg–1)                                             2

(iii)    gC/gO = R2/16R2 giving gS = 40/16 (= 2.5 N kg–1)                                    1

(iv)    average g = (2.5 + 1.6)/2 = 2.(05) (1)
Δp.e. (= mgavR) = 3.0 × 103 × 2.05 × 2.0 × 107; = 1.2 × 1011 (J) (2)       3

  (c)   g = v2/r; = 4π2(5R)/T2 (2)
1.6 = 4 × 9.87 × 1.0 × 108/T2 giving T2 = 24.7 × 108 and T = 5.0 × 104 (s) (2) 4

[12]

Springs Questions

Q1

(a)     One reading from the graph e.g. 1.0 N causes 7 mm                                               C1

Hence 5.0 (N) causes 35 ± 0.5 (mm)                                                             A1

(allow one mark for 35 ± 1 (mm)

(b)     (i)      Force on each spring is 2.5 (N)                                                             C1

extension = 17.5 (mm) allow 18 (mm) or reading from graph              A1

[allow ecf from (a)]

(ii)     strain energy = area under graph / ½ F ´ e                                            C1

                    = 2 ´ 0.5 ´ 2.5 ´ 17.5 ´ 10–3

                    = 0.044 (J)                                                                        A1

[allow ecf from (b)(i)]

Q2

(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

          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

Q3

   (a)     The extension of a spring is directly proportional to the applied force          M1
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

(ii)     extension, x =  × 80 (= 133.33) (mm)                                              C1
(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)
                                                                                      M1
m and k are constant, therefore x µ v.                                                  M1

Stretching