Download Exercise 14 Wave Motion

Exercise 3 (Forces and motion) Suggested Solutions
1. (a) (i) v2 = u2 + 2as
152 = 0 + 2 a (75)
a = 1.5 m s-2
(ii) There are two forces acting on the helicopter. Refer to the figure. F is
the uplifting force and mg is the weight of the helicopter.
F – mg = ma
F = m(a + g) = 1500 (1.5 + 10) = 17250 N
(b) (i) At time t = 0, v = -45 m/s
(ii) The area represents the displacement of the object in its upward
motion.
(iii) Height = area of graph below time axis from t = 3s to 6s.
= (15+45) x 3 / 2 = 90 m
(iv) Statement 1 is incorrect.
The acceleration of the object is always
10 m s-2 downward for the whole journey. At t = 1.5s,
the object reaches the highest point, only its velocity
is zero but the acceleration is still 10 m s-2 downward.
Statement 2 is correct. The acceleration of the
heavier object will still be 10 m s-2.
2. (a) (i) (1) Speed at A = length of card / time =
0.03 / 0.05 = 0.6 m s-1.
(2) Speed at B = 0.03 / 0.025 = 1.2 m s-1.
(ii) (1) By v2 = u2 + 2as
1.22 - 0.62 = 2a (0.4)=> a = 1.35 m s-2.
(2) Tension in string T = ma = 1.5 (1.35)
= 2.025 N
(b) A motion sensor is facing the back of the
trolley. The trolley is given a push and runs down the
runway. The velocity-time graph obtained is examined. If the runway is friction-compensated, the v-t graph
should be a horizontal straight line.
(c) The trolley will travel along the runway with a uniform speed.
3. (a) (i) F = ma = 60 x (40/4) = 60 x 10 = 600 N
(ii) F = ma = 60 x 0 = 0 N
(b) The parachutist opens the parachute at t = 4s because after opening the parachute, the air resistance is
greatly increased and greater than his weight, as a result, the net force is upwards, and therefore he starts to
decelerate from t = 4s.
(c) Distance traveled = area under graph = 4 x 40 /2 = 80 m
(d) Distance traveled in t = 10s to t = 15s
= 5 x 5 = 25 m
Distance required = 205 - 25 - 80 = 100 m
(e) (i) F = ma = 60 x (5-0)/0.5 = 600 N
(ii) Bending the knee will increase the contact time with the ground. The decrease in velocity on landing
is unchanged, independent of whether the knee is bent or not, so increasing the deceleration time would reduce
the force acting on the parachutist, causing less damage.
4
(a)
normal reaction
friction
weight
(b) (i)
(ii) Acceleration of block
=
=
slope of the straight line / 2
4/2 = 2 m s-2
(iii) mg sin θ – f = ma
0.4×10 × sin 30o – f = 0.4 × 2 => f = 1.2 N
so frictional force required is 1.2 N
5. (a) a  6  0
20
= 3 m s2 (downwards)
(b)
A: 380 N B: 510 N C: 614 N
In stage B, balance reading = weight,
mg = m × 9.81 m s2 = 510 N
m = 52.0 kg
(c) (i) For stage C, by Newton’s 2nd law,
F = ma
(510  614) N = (52.0 kg) a
a = 2 m s2
Thus, a  0  5  2 m s  2
T  12
T = 12.5 s
(ii) Height  displacement of lift = area under graph 
(10  2)  12.5
 5  51.3 m
2
weight of container = mg = 22000 × 9.81 = 2.16 × 105 N
Tension (= ¼ mg) = 5.40 × 104 N
(b) moment(=force×distance)=22000g × 32 = 6.91 × 106 N m
(c) The counterweight provides a (sufficiently large) anticlockwise moment (about Q) or moment in
opposite direction (to that of the container to prevent the crane toppling clockwise)
6. (a)
OR
Left hand pillar pulls (down) and provides anticlockwise moment
OR
The centre of mass of the crane (frame and the counterweight) is between the two pillars which prevents the
crane toppling clockwise/to right
7. (a) From t = 0.5 to 1.5s, the block decelerates uniformly.
From t = 1.5 to 3.5s, it then moves with uniform acceleration down the plane.
(b) (i) a = - (2-0) / (1.5-0.5) = -2 m s-2.
(ii)
(c)
mg sin + f = ma
(1)(9.81)sin + f = (1)(2) ----(1)
Moving downward: mg sin - f = ma
(1)(9.81)sin - f = (1)(0.5) ----(2)
On solving, we have
2f = 1.5
f = 0.75 N
(d) Moving upward :
8.
(a) (i) From t = 0 to 2 s, it moves in uniform speed
From t 2.0 to 2.5 s, it decelerates uniformly to zero.
(ii) Air cushion cannot support the weight and the puck touches the ground. Friction causes the
deceleration.
(iii) Distance = area under the graph
= (2+2.5) x 1.5 / 2
= 3.38 m
(b)
20
= 0.4 m s–2
24  12  14
(b) For the 12 kg block,
9. (a)
a=
20 – T1 = 12  0.4
 T1 = 15.2 N
(c) For the 24 kg block,
T2 = 24  0.4 = 9.6 N
The tension in string 1 is larger than that in string 2.
(d) Train compartments are similar to this system.
MC
1-5 D D B C C
21-25 B B C C C
6-10 B D B A E
26-27 A D C
11-15 B B A A D
16-20 D D C D E
1. The weight does not vary even the object is inside an accelerating lift. Only the apparent weight varies.
2. For 2 kg mass, 6 – T = 2a
For 3 kg mass, T = 3a
Solving the above equations, a = 1.2 m s-2, T = 3(1.2) = 3.6 N
3. By inertia, the velocity of the stone at the moment of release is 10 m s-1 upwards.
4. If the apparent weight is reduced, R < mg, so the net force is downward, hence the acceleration must be
downward. Both (2) and (3) have acceleration directed downward.
5. When the force is 10 N, the block travels with constant velocity, this implies that the frictional force must
be also 10 N. When the external force is 14 N,
F = ma
14 – 10 = 2 a
=> a = 2 m s-2.
6. When the car is moving down the inclined plane with constant velocity, friction acts upward (the plane
forms a friction-compensated runway). The net force is zero. So
friction f = mg sin 
When the car is moving up the inclined plane with constant velocity, friction acts downward. The net force
is again zero. So
external force required = mg sin  + friction = 2 mg sin  = 2 (0.5)(10) sin 20o = 3.4 N
7. Moving downwards with deceleration means that acceleration is upward. Hence, normal reaction > weight.
8. (1) is wrong. The friction is mg sin , not zero.
(2) is wrong. When the trolley runs upward, the friction acts up the incline and is impossible to balance the
friction by mg sin .
10. When the block is moving upwards,
net force = friction + mg sin  = ma ---(1)
When the block is moving downwards,
net force = mg sin  - friction = ma’ ---(2)
From (1) and (2), it can be concluded that
a’ < a
So upward deceleration (velocity positive) < downward acceleration (velocity negative). The answer is E.
11. The ball will not press on the wall, otherwise, if the surface of the ground is smooth, the ball will move to
LHS by reaction. So the only forces acting on the ball are weight and normal reaction.
12. Acceleration a = F / m = 2 / 2 = 1 m s-2. After 2 seconds, the force is removed and by Newton’s first law,
the object will travel with constant velocity v given by
v = u + at = 2 + 1 (2) = 4 m s-1.
13. Treat the F direction as x-axis. Then the tension is vertical direction.
Consider equilibrium in x-direction,
F = 20 sin 25o = 8.45 N
15. Refer to Newton’s first law.
16. Mass is a measure of inertia.
17. After cancellation, there are only 3 net forces, each of 3 N, one is in OD direction, one is in OE direction,
and one is in OF direction.
Resultant force between the force in OE and OF = 2 x 3 cos 30o = 3 3
Thus resultant force =
 
32  3 3
2
 6N
18. For friction-compensated runway,
mg sin  = friction
Hence, net force = F + mg sin  - friction = ma => F = ma => a 
1
F
m
A plot of a against F is a straight with slope = 1/m
If the runway is inclined at a greater angle ’, then mg sin ’ > friction
net force = F + mg sin ’ – friction = F + F = ma where F = mg sin ’ – friction = positive value
a
1
F
F
m
m
A plot of a against F will be a straight line with slope = (1/m) and positive intercept F / m
19. Resolving components in horizontal and vertical direction.
In horizontal direction,
horizontal component of F1 + horizontal component of F2 = 2 + 2 = 4 N
In vertical direction,
vertical component of F1 + vertical component of F2 = 3 + (-1) = 2 N
Resultant =
20.
4 2  2 2  4.47 N
2  3  4  0 => 3  4  2
21. Taking moments about A,
moment produced by weight of plank + moment produced by boy’s weight = moment produced by RB
100 x 0.6 + 400 x 0.9 = RB x 1.2
RB = 350 N
RA + RB = 100 + 400
=> RA = 150 N
22. Taking moments about Y, at the point of tilting, RX = 0
clockwise moment = anticlockwise moment
450 x 2.5 = (675 + 6n) x 1.5 where n is the no. of weights in the basket
n = 12.5
Thus the maximum number of weights that can carry without tilting is 12.
23. Taking moments about P,
F x 40 = 200 x 5
F = 25 N
24. F1 – T = ma
T – F2 = 2ma
On solving, T =
2F1  F2
3
26. Friction must be 12-2 = 10 N
At the instant the 12 N force is removed, the net force acting on the block is 2 + 10 = 12 N
27. Taking moments about P,
T sin 45o (1.5) = Mg (1)
T = 0.943 Mg
After the gangplank is raised, its center of gravity is closer to P, hence the moment arm is shorter (i.e., less
than 1 m, thus T is decreased