Download 2009 JC1 H2 Physics

2009 JC1 H2 Physics
Summative Assessment 1: Forces and Dynamics
Instructions:
 Answer all the questions using a dark blue or black pen.
 All workings must be shown clearly.
 Marks will be penalised for inappropriate no. of s.f., wrong or no units or for using g = 10 m s-2.
1. A block of mass 6.0 kg, is resting on a compressed spring along a smooth slope as
shown in Fig. 1. The spring constant for the spring is 400 N m-1, and the angle of
elevation of the slope is 40˚ above the horizontal.
40˚
Fig. 1
(a) Calculate the magnitude of the elastic force exerted on the spring due to the block.[2]
When the mass is resting on the spring, it is in equilibrium, i.e. Fn = 0
N
F
x'
W
Along the slope: ∑Fx’ = 0
F + ( - W sin 40°) = 0
F = 6.0 × 9.81 sin 40°
= 37.8 N
[1]
[1]
(b) Hence, calculate the compression of the spring.
F=kx
37.83 = 400 x
x = 0.0946 m
2
[2]
[1]
[1]
(a) State the two conditions for the equilibrium of a body which is acted upon by a
number of forces.
[2]
1. There is no resultant force in any direction.
2. There is no resultant torque at any point.
[1]
[1]
1
(b) Figure 2 shows a crane consisting of a uniform arm, with a length of 4.0 m and a
mass of 2.0 x 10³ kg. The arm is held stationary at an angle of 50 to the vertical, by
a horizontal cable attached to the arm at 1.0 m from the top.
The weight of the arm is W, the tension in the horizontal cable is T and the force
exerted on the arm at the hinge is R.
(i) Calculate the tension T in the horizontal cable.
The arm is held stationary, i.e. it is in equilibrium.
There is no resultant force and no resultant torque acting on the arm.
Since there are two unknown forces, T and R, taking pivot at the hinge can help
eliminate the turning effect of R.
Take pivot at the hinge,
 = 0
total clockwise moment = total anticlockwise moment
T (3.0 cos 50) = W (2.0 sin 50)
T (3.0 cos 50) = (2.0  103 ) (9.81) (2.0 sin 50)
T = 1.56  104 N (to 2 s.f)
[1]
[1]
(ii) Draw a vector triangle to represent the relationship of the forces W, T and R.
Label the forces clearly. (Scale drawing is not necessary.)
[1]
Since the arm is in equilibrium, the resultant force is zero, i.e. the vector triangle
of W, T and R will form a closed loop.
(Note: the angle 50° is the angle between the arm and the vertical, not the angle
of R with respect with the vertical (look at the diagram carefully)
T
R
W
or
R
W
[1]
T
2
(iii) Hence, or otherwise, determine the magnitude and the direction of the force R. [3]
Since the vector triangle is a right-angle triangle, trigonometric function can be
used. (In general, use sine and cosine rule)
Using Pythagoras theorem,
R2 = T2 + W2
R2 = 155882 + [(2.0  103 ) (9.81)]2
[1]
4
R = 2.51  10 N
[1]
W
T
(2.0  103 )(9.81)
tan 
15588
  51.5
tan 
[1]
above the negative x-direction
(iv) If the angle between the arm and the vertical is increased to 70, deduce if
tension T would increase, decrease or remain unchanged.
[1]
Assuming that the direction of T remains horizontal,
T
W
T  W tan 
tan  
When  increases, T also increases. [1]
3
A car is hanging on a helicopter by two cables as shown in Fig. 3. The mass of the car is
950 kg.
Boundary cut by
the cable
Boundary cut by
the cable
(a) Draw a free-body diagram of the car. Label all the forces clearly (i.e. write out the
names of the forces explicitly).
[1]
tension
tension
weight
3
(b) Neglecting air resistance, determine the tension in each of the two cables connected
to the car.
T cos 30°
T cos 30°
Tsin30°
The same FBD
can be used for
(i), (ii) and (iii)
Tsin30°
W
(i) when the helicopter is stationary.
[2]
When the helicopter is stationary, it is in equilibrium, i.e. Fnet = 0
Consider the vertical direction,
T cos 30 + T cos 30 + ( - W) = 0
2 T cos 30 + (950 × 9.81) = 0
T = 5380 N
[1]
[1]
(ii) when the helicopter is ascending vertically with a constant speed.
When the helicopter is ascending vertically with a constant speed, it is moving
with constant velocity. Thus, even though it is moving, it is also in equilibrium, i.e.
Fnet = 0.
Consider the vertical direction,
Fnet = 0
T = 5380 N
[1]
(iii) when the helicopter is ascending vertically with a constant acceleration of
1.4 m s-2.
Since the helicopter is ascending vertically with a constant acceleration, there
must be a resultant force acting on the helicopter.
Consider the vertical direction,
Fnet = ma
T cos 30 + T cos 30 + (- W) = m a
2 T cos 30 - (950 × 9.81)= (950 × 1.4)
T = 6150 N
4
[1]
[1]
(a) (i) Define linear momentum.
[1]
Linear momentum is the product of its mass and its linear velocity. [1]
(ii) State whether linear momentum is a vector or a scalar quantity.
[1]
Vector [1]
(b) State the principle of conservation of momentum.
[1]
It states that the total momentum of a system of interacting bodies remains constant
provided no net external force act on the system. [1]
4
(c) The principle in (b) can be applied in different types of interaction described below.
(i) Elastic collision: a neutron of mass m travelling with a velocity 6.5  105 m s-1
collides head-on with a stationary carbon atom of mass 12 m, as shown in
Fig. 4.1. The carbon atom moves off in the same direction with velocity
1.0  105 m s-1.
mass
m
mass
12 m
Fig. 4.1
1. Calculate the velocity of the neutron after the collision.
By conservation of momentum,
Taking rightwards as positive,
m (+6.5  105) + 12 m (0) = m (v1) + 12 m (+1.0  105)
v1 = – 5.5  105 m s-1
v1 = 5.5  105 m s-1 (to the left)
[2]
[1]
[1]
2. State what happens to the total kinetic energy as a result of this collision.
[1]
The total kinetic energy remains the same (It is conserved). [1]
(ii) Inelastic collision (perfectly inelastic collision): a fast moving neutron of mass m
collides head-on with a stationary atom of nitrogen of mass 14 m as shown in
Fig. 4.2. The neutron is captured by the atom to form a “heavy” isotope of
nitrogen. The speed of the neutron before capture is 6.0 × 107 m s-1.
mass
m
mass
14 m
Fig. 4.2
1. Calculate the speed of the “heavy” nitrogen atom.
By conservation of momentum,
Taking rightwards as positive,
m (+6.0  107) + 14 m (0) = (m + 14 m) (v)
v = + 4.0 106 m s-1
[2]
[1]
[1]
2. Calculate the change in the total kinetic energy (if any) as a result of this
collision. The value of m is 1.66  10 - 27 kg.
 ke = ke final - ke initial
 ke = ½×15×1.66  10 – 27×(4.0 106)2 – [½×1.6610 – 27×(6.0  107)2 + 0] [1]
 ke = - 2.79  10-12 J
[1]
5
Bonus Question
Two boxes of fruits of mass 10 kg and 20kg are connected by a light string on a frictionless
horizontal surface as shown in the figure. A force of 50 N is applied to the 20 kg box.
20 kg
10 kg
50 N
Determine the acceleration of each box and the tension in the string.
[2]
Having two boxes moving along a horizontal surface, the direction of their acceleration
should be along the horizontal plane. Thus the resultant force should also be acting in the
horizontal direction.
N2
Consider the two boxes as a system,
N1
Fnet = m a
50 = (10 + 20) a
a = 1.67 m s-2
[1]
10 kg
W1
Consider the 10 kg box,
Fnet = m a
T = 10 (1.667)
T = 16.7 N
[1]
(Check your answer by looking at the FBD of 20kg box)
20 kg
50 N
W2
N1
10 kg
T
W1
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