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 = Pt = 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 1025.0 102 = 60N;
5
(ii)v2 = 2as = 2 5.0 102 6.4 103;
to give v = 25ms1;2
(e)5.6 102 N;1