Energy questions: Markscheme

1.(a)

(i)acceleration;1 max

(ii)velocity;1 max

(b)(i)KE = mv2 = 0.08 × 225 = 18 J;1 max

(ii)18 J;1 max

(iii)loss in PE = mgh = 40 J;1 max

(iv)totalKE = 58 J;
v = = 27 m s–1;2 max

[7]

2.(a)mass × velocity;1

(b)(i)momentum before = 800 × 5 = 4 000 N s;
momentum after = 2 000v;
conservation of momentum gives v = 2.0 m s–1;3

(ii)KE before = 400 × 25 = 10 000 J KE after = 1 000 × 4 = 4 000 J;
loss in KE = 6 000 J;2

(c)transformed / changed into;
heat (internal energy) (and sound);2

Do not accept “deformation of trucks”.

[8]

3.(a)Note: for part (i) and (ii) the answers in brackets are those arrived at if 19.3 is
used as the value for the height.

(i)height raised = 30 sin 40 = 19 m;
gain in PE = mgh = 700 × 19 = 1.3 × 104 J (1.4 × 104 J);2

(ii)48 × 1.3 × 104 J = 6.2 × 105 J (6.7 × 105 J);1

(iii)the people stand still / don’t walk up the escalator
their average weight is 700 N / ignore any gain in KE of the people;1 max

(b)power required = = 10 kW (11kW);
Eff = , Pin = ;
Pin= 14 kW (16 kW);3

[7]

4.(a)momentum of object = 2 × 103 × 6.0;
momentum after collision = 2.4 × 103 × v;
use conservation of momentum, 2 × 103 × 6.0 = 2.4 × 103 × v;
to get v = 5.0 m s–1;4

Award [2 max] for mass after collision = 400 kg and v=30ms–1.

(b)KE of object and bar + change in PE = 1.2 × 103 × 25 + 2.4 × 103 × 10 × 0.7533;
use E = Fd, 4.8 × 104 = F × 0.75;
to give F = 64 kN;

Award [2 max] if PE missed F = 40 kN.

or
a = ;
F – mg = ma;
to give F = 64kN;3

Award [2 max] if mg missed.

[7]

5.(a)(i)F = Mg sin θ
= 960 × 9.8 × 0.26;
2.4 × 103 N2

(ii)KE = = (480 × 81) = 3.9 × 104 J;1

(b)KE = Fs;
to give F = 2.6×103 N; 2
Award [1 max] if v2= 2 asis used.

(c)recognize that KE = mass × spht × rise in temperature;
;
= 4.2 Κ;
Award full marks for bald correct answer.

no energy / heat loss to the surroundings / energy distributed evenly in
shoe and drum;4

[9]

6.Mechanical power

(a)(i)

= 2

(ii)W = mgh = 1.2  104 300 = 3.6  106 J;1

(iii)work done against friction = 4.8  103 5.0  102;

total work done = 2.4  106 + 3.6  106;

total work done = Pt = 6.0  106;

to give 4

(d)(i)sin =

weight down the plane = W sin = 1.2  104 0.047 = 5.6  102N;

net force on car F = 5.6  1025.0  102 = 60N;

5

(ii)v2 = 2as = 2  5.0  102 6.4  103;

to give v = 25ms1;2

(e)5.6  102 N;1