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F.5S Physics Test 5 (09-10) Marking Scheme
(a)
(b)
(c)
(d)
Potential energy = mgh
= 75  10  3
= 2250 J
The diver needs 2250 J to climb to the springboard.
(i) By the law of conservation of energy,
energy stored in the springboard = PE gained by the diver
= mgh = 75  10  1
= 750 J
Elastic potential energy of 750 J is stored in the springboard.
(ii) Let v be the take-off velocity of the diver.
By the law of conservation of energy,
gain in kinetic energy loss = loss in elastic potential energy
1
 75  v 2 = 750
2
v = 4.47 m s1
The take-off velocity of the diver is 4.47 m s 1 .
Take the upward direction as positive.
For the whole journey,
1
by s = ut + at2,
2
1
3 = 4.47  t +  (10)  t2
2
t = 1.34 s or 0.447 s (rejected)
The time taken between the diver taking off and entering water is 1.34 s.
The diver can step on the springboard harder, so that more elastic
potential energy is stored in the springboard.
Then by the law of conservation of energy, the diver can jump higher
and he will have longer time to finish his motion.
1M
1A
1M
1A
1M
1A
1M
1A
1A
1A
2.
(a)
Change in PE of the cart
= mgh = 1500  10  (50  40)
= 150 000 J
OR, change in PE = final PE – initial PE = 1500 x 10 x (40-50) = -150000 J
(b) Work done against friction
= Fs = 400  200
= 80 000 J
(c) Let v be the speed of the cart at B.
By the law of conservation of energy,
PE lost + KE at A = KE at B + work done against friction
1M
1A
1M
1A
1M
1
1
150 000 + × 1500 × 0.52 =  1500  v2 + 80 000
2
2
v = 9.67 m s–1
The speed of the cart at B is 9.67 m s–1.
1A
3. (a)
(b) If ball A is earthed, it becomes neutral. Balls A and B attract and fall down. When they touch
each other, they share their charges and become like charge again. They will repel again.
4A+ 1C
MC
1-5 A A A A A
6-10 B B B B A
11-15 D A C A C
16-18 D A C D B
Explanation to mc
1. The electrons in the rod are not free, even they are free in metals, they cannot go outside.
3. The force on the negative charge is opposite to the force of the electric field and does not depend
on motion of the charge.
4. Positive charges are induced on the LHS of X while negative charges are induced on RHS of Z.
After earthing, the negative charges of Z escape to the earth. So Y and Z are neutral.
5. From highest point to A,
Loss in PE = gain in KE
mgh 
1 2
1
2
mv  mg (h  0.5h)  mv A  v A  gh
2
2
From A to B,
mg (0.5h) 
1
1
1
2
2
2
mvB  mv A  m(v B  gh)  v B  2 gh  2v A
2
2
2
So statement 1 is incorrect.
6. v2= u2 + 2as = 0 + 2 x 10 x 2.5 x 8 => v = 20 m s-1
R – mg = ma = m v / t
R – 0.05 x 10 = 0.05 x 20 / 0.01 => R = 100.5 N
7. KE loss =


1
 0.04 330 2  208 2  1310 J
2
8. Output power = mgh / t = mgv = 1000 x 10 x 0.1 = 1000 W
9. Increase of momentum from t = 5s to 7s = area of triangle = (7-5) x 10/2 = 10 Ns
Decrease of momentum from t = 7s to 9s = area of triangle = (9-7) x 10 / 2 = 10 Ns
So net increase of momentum = area of rectangle = 10 x 5 = 50 Ns
10. Statement (1). The definition of inelastic collision is that the total kinetic energy is decreased
after collision. Completely inelastic collision (not inelastic collision) is a collision such that the
colliding objects stick together and move off with a common speed.
Statement (2). The linear momentum of each body is NOT conserved. But the total momentum is
conserved.
11. 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 =
1
1
1
1
1
MV 2  MV V  mv V  mv  v  mv 2  KE of
2
2
2
2
2
smaller fragment. So (3) is correct.
12. KE + mgh = mgH = constant
KE = constant - mgh
So KE against h is a straight line with negative slope
13. By conservation of momentum,
2 x 4 = (2+2) v =>v = common speed = 2 m/s
Loss in KE = ½ (2) (4)2 = ½ (4)(2)2 = 8 J
14. Work done by friction = fs cos = friction x distance moved cos 180o = -3 x 1.5 = - 4.5 J
Note that work done against friction = f s = 3 x 1.5 = 4.5 J
15. Quickest method: s = (u + v) t /2
0.1 = (600 + 450) t /2
t = 0.190 ms
Alternatively, Work done against friction = loss in KE =


1
 0.1  600 2  450 2  7875 J
2
fs = f x 0.1 = 7875 => Friction f = 78750 N
Deceleration a = f/m = 78750 / 0.1 = 787500
Alternatively, v2 = u2 + 2as => 4502 = 6002 + 2a(0.1) => a = -787500 m s-2
v = u + at => 450 = 600 + (-787500)t
=> t = 0.190 ms
16. Statement (1) is correct because an alpha source would produce a lot of ion pairs, therefore the
ions would neutralize the charges of the spheres, as a result, the spheres become neutral shortly and
fall down due to gravity.
Statement (2) is correct because the force acting on A is of the same magnitude as the force
acting on B as they are an action and reaction pair. A raises lower, showing that it is heavier.
Statement (3) is correct as mentioned above.
17. Apparent weightless is due to the lack of reaction force or supporting force.
18. When the cord attains its maximum length, the jumper is at the lowest point, so he must be
momentarily at rest.
Statement (2) is incorrect because the net force is upward (elastic force or restoring force is
greater than the weight) so that he will move upward later.
Statement (3), PE lost = gain in KE + gain in elastic PE. At the lowest point, there is no gain in
KE, so PE lost = gain in elastic PE.
19. Total momentum is conserved in all types of collisions. KE is conserved only in elastic collision.
20. Potential energy can be negative, depending on where we choose to be the reference line.