Download Document

F4 Physics Quiz 2 Suggested Answers (10-11)
1. (a) Area under graph = (0.03 + 0.05) x 50 / 2 = 2 Ns
(1M+1A)
Area under graph represents impulse or change in momentum.
(1A)
(b) m ( v – u) = change in momentum
(1M)
0.05 ( v - - 20 ) = 2
v = 20 m s-1
(1A)
(c) Average force = change in momentum / time
Average force = 2 / 0.05 = 40 N
(1M+1A)
2. (a) By the conservation of momentum,
0 = 0.5 x 0.5 + 0.5 v
v = 0.5 m s-1
(b) (i) Total KE =
1A
1A
1
(0.5)0.5 2  2 = 0.125 J
2
1A
(ii) The energy comes from the PE originally stored in the spring.
(c) (i) 0.5 x 0.5 = (0.5+0.5) v
where v is the common speed
-1
v = 0.25 m s
(ii) Force on B = (mv – mu ) / t =0.5 (0.5 – 0.25) / 0.3 = 0.417 N
Force on A = force on B = 0.417 N
MC
1-5 A C A C B
-----------------------------------MC explanations
6-10 B B B D A
1A
1M
1A
11-14 D D B B
1. No work is done because the force applied is perpendicular to the displacement.
2. Horizontally, vx t = 1,
Vertically,
0.5 = (1/2) (10) t2 => t = 0.316 s
So
vx = 3.162 m s-1
vy2 = 2(10)(0.5) => vy2 = 10
v  v x  v y  4.47 m s-1
2
2
3. Work done = Fs = 2 x 103 + 4 x 103 x 30 = 140 kJ
4. ut = x
vY = gx/u
v  u 2  vy  u 2 
2
g 2 x2
u2
5. See answers in the exercise.
6. Initial downward velocity = 10 sin 30o = 5 m /s
vY2 = 52 + 2(10)(0.3)
=> vY = 5.57 m/s
Since the collisions are perfectly elastic, there is no energy loss. So the velocity after collision does not
change.
From Y to Z, time to reach max height t is given by
0 = 5.57 + (-10)t
=> t = 0.557 s
Time of flight = 2 x 0.557 = 1.114 s
Required distance YZ = 10 cos 30o x 1.114 = 9.64 m
8. Power = Fv = 1000 x 10 x 0.1 = 1000 W
9. Let m and M represent mass of smaller and larger fragments respectively. By conservation of momentum,
0 = mv – MV
=> mv = MV
(1) Both fragments have the same magnitude of momentum (but in opposite directions).
(2) Larger fragment has smaller speed.
(3) KE of larger fragment =
fragment. So (3) is correct.
1
1
1
1
1
MV 2  MV V  mv V  mv  v  mv 2  KE of smaller
2
2
2
2
2
10. KE + mgh = mgH = constant
KE = constant - mgh
So KE against h is a straight line with negative slope
11. By conservation of momentum, and by symmetry, the white ball and the black ball must have the same
speed so that their vertical components v sin 30o would cancel.
3 m = mv cos 30o x 2
v = 1.73 m/s
12. Since 10 % of energy is lost after each bounce, so 90 % of the energy remains after each bounce.
Thus required total energy = mgh (0.9)4 = 0.656 mgh
13. Loss in PE by Y = gain in PE of X + gain in KE by X plus Y
Gain in KE of the system = 5 (10) (2) – 4(10)(2 sin 30o) = 60 J
14. From A to B, friction acts down the incline.
Work done by friction = Fs = 2.6 x AB cos 180o = -2.6 x 10/sin 30o = - 52 J
From B to C, friction acts up the incline .
Work done by friction = 2.6 x BC cos 180o = -2.6 x 10 / sin 30o = - 52 J
Total work done by friction = -52 + -52 = -104 J
For gravity, from A to B, gravity does negative work.
Work done by gravity = 1 x 10 x sin 30o x 20 x cos 180o = -100 J
From B to C, gravity does positive work because mg sin 30o and displacement are in the same
direction.
Work done by gravity = 1 x 10 sin 30o x 20 = + 100 J
Thus total work done by gravity = -100 + 100 = 0 J