Name: ________________________________ (
)
CG: ___________
Physics JC1 2009
Assignment 4: Dynamics
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1. A model rocket of initial mass 1.3 kg is fired vertically into the air. Its mass decreases at a
constant rate of 0.23 kg s-1 as the fuel burns. The final mass of the rocket is 0.38 kg. The
rocket rises to a height such that, during the flight, the gravitational field strength of the
Earth may be considered to have the constant value of 9.8 m s-2.
(a) The variation with time t of the upward force on the rocket during the first 3 seconds
after firing is shown in Fig. 1.1.
12.7
3.7
Fig. 1.1
(i) On Fig. 1.1, use the same scales to draw a line to represent the variation with
time t of the total weight of the rocket during the first 5 seconds after firing.
[2]
Winitial = mg = 1.3 (9.8) = 12.7 N
Wfinal = mg = 0.38 (9.8) = 3.7 N
Amount of time that fuel lasted = 1.3  0.38  4.0 s
0.23
The weight of the rocket decreases at a constant rate (indicated by a downwardsloping straight line from 12.7 N to 3.7 N in a time period of 4.0 s) [1]. After which,
the weight of the rocket remains constant at 3.7 N (a horizontal straight line) [1].
(ii) Hence, read off from Fig 1.1 the time delay between firing the rocket and lift-off. [1]
At the point of lift off, there must be a resultant force acting on the rocket. Thus
we are interested at the moment just before the lift-off where Fn is still equal to 0.
i.e. upward force = weight
intersection point of graphs is at t = 0.50 s [1]
(b) (i) Write down an equation to represent the relation between the resultant force F on
a body, the time t for which the force acts and the change in momentum p of the
body.
[1]
p = F t
[1]
1
(ii) On Fig 1.1, shade the area of the graph which represents the change in
momentum of the rocket during the first 3 seconds after the rocket is fired.
[1]
Since p = F t, the change in momentum must be equal to the area under a F-t
graph. But the graph shown in Fig 1.1 is the upward force and the weight acting
on the rocket, not the resultant force. Hence to find the change in momentum of
the rocket due to the effects of upward force and weight, we should be looking at
the region bounded by the upward force and the weight after lift-off. [1]
Note: before lift-off, the rocket is not moving yet, thus there is no change in
momentum, even though there is a difference in the upward force and weight.
Can you think of the force that could exist then?
2. (a) Momentum is conserved when two objects collide or interact. State the condition
under which momentum is conserved.
[1]
No net external force acts on the system.
[1]
(b) Fig. 2.1 shows momentum against time graphs for two colliding trucks A and B.
Fig. 2.1
The masses of trucks A and B are 2000 kg and 4000 kg respectively.
(i) Explain why the gradients of the graphs during the collision have opposite sign. [1]
The gradient of the graph (which is equivalent to rate of change of momentum)
corresponds to the force acting on that object.
The gradients have opposite sign because during the collision, the force on A by B is
in opposite direction to the force on B by A. (Newton’s 3rd Law)
[1]
(ii) Show that momentum is conserved when the two trucks collide.
[1]
Total momentum before collision = (20 + 16) × 103 = 3.6 × 104 kg m s-1
Total momentum after collision = (24 + 12) × 103 = 36 × 104 kg m s-1
[1]
Since total momentum before collision is equal to the total momentum after collision,
momentum is conserved.
2
(iii) Calculate the force acting on truck B during the collision.
From the graph, truck B experiences a change in momentum during 1.5 s to 3.0 s.
This indicate that the collision occurs between this time frame. Hence the force can
be computed using values from the graph.
3
dp  24  20   10
[1]
F

dt
3.0  1.5
F = 2.7 × 103 N
[1]
(iv) Calculate the change in the kinetic energy of the trucks and state the type of
collision.
1 m2v 2 1  mv 
p2
Ek  ½mv 


2 m
2 m
2m
2
2
pA 2
pB 2

Initial Ek of the trucks 
2mA 2mB
16  10    20  10 

3
2
3
2  2000
2
2  4000
= 1.14  10 J
5
Final Ek of the trucks 
[1]
pA '2 pB '2

2mA 2mB
12  10    24  10 

3
2
3
2  2000
2
2  4000
= 1.08  10 J
5
[1]
Change in Ek = final Ek – initial Ek
= (1.08 – 1.14) × 105
= - 6000 J
[1]
Since the total KE is not conserved, the collision is inelastic. [1]
3. (a) A sphere of mass m travelling in a straight line with speed u collides head-on with a
sphere that has a speed u2. The speed u of the incoming sphere is greater than u2
(see Fig. 3.1). The collision is perfectly elastic.
Fig. 3.1
The incoming sphere of kinetic energy E may lose an amount of kinetic energy W.
3
Fig. 3.2 shows how W/E, the fractional energy lost by the incoming sphere, depends
on the ratio u2 /u.
0.85
Fig. 3.2
(i) What happens to the kinetic energy lost by the incoming sphere?
[2]
As the collision is elastic, the total kinetic energy of the spheres must be conserved.
[1]. Hence, the loss of kinetic energy of the incoming sphere goes to the gain of
kinetic energy of the bombarded sphere. [1]
(ii) Given that the initial energy of the incoming sphere is 1.6 × 10 -13 J, calculate the
energy lost by this sphere in the perfectly elastic collision when u2 /u = 0.40.
[2]
From graph, for
u2
 0.40,
u
W
 0.85
E
[1]
Since E  1.6  10 -13 J
W
 0.85
1.6  10 13
energy loss, W  1.4  10 13 J
[1]
(iii) Use the graph of Fig. 3.2 to suggest why paraffin wax, which has a high number
density of protons, is a good absorber of high speed neutrons.
[2]
The graph can be used on the ground that the proton and neutron have the about the
same mass. The protons in the paraffin wax are virtually not moving within the nucleus
hence they have very low initial velocities compared to the fast-moving neutrons,
hence the ratio
u2
is small. [1]
u
W
ratio will be high, meaning the neutrons would lose most or all
E
of its kinetic energy during collision [1]. The incoming neutrons will be absorbed or
stopped at its tracks by the paraffin wax.
From the graph, the
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