Download Newton to Einstein Exercise 2 – Kinetics

PHY4116 From Newton to Einstein
Exercise Sheet 3 Answers: Dynamics
Marks out of 20
_____________________________________________________________________
These answers contain about the right amount of explanation that you should always
include in coursework and exam answers, plus examples of the kind of checking you
should routinely do. There should be diagrams too, but these are very tedious to
produce in Word, so I have omitted them.
1).
A picture frame hung against a wall is suspended from a hook by two wires
attached to its upper corners. If the two wires make the same angle with the
vertical, what must this angle be if the tension in each wire is equal to 0.75 of
the weight of the frame? (Ignore any friction between the wall and the frame.)
Answer: Using F = ma, a = 0 so horizontal components of forces sum to zero as do
vertical components. Resolve the tension of a wire into vertical and horizontal
components, Tcos and Tsin respectively. Two wires share the load, so each one
supports ½mg of vertical force. The tension T is to be 0.75mg, and Tcos and needs to
be ½mg. So,
0.75mg cos   ½ mg
0.5 2

[1]
0.75 3
2
  cos 1  48.2 
3
Check: The vertical components are 0.75 mg cos = 0.75 mg cos48.2 = 0.75 mg 
0.6665.
Similarly, the horizontal components are 0.75 mg sin = 0.75 mg sin48.2 = 0.75 mg
 0.7455.
Do these add up (by Pythagoras? (0.66652 + 0.74552) = 0.999993 ~ 1 so YES.
OR, We expect, for  = 0, the tension to be ½mg, and for  = 90, infinity. Clearly
going as tan . By inspection, for  = 45 horizontal and vertical components are
equal, so that the tension is Sqrt[2]  ½mg. Here, we get an angle just over 45, and a
tension 1½  ½mg. Since 1½ is just over Sqrt[2], this hangs together. OK.
[1]
cos  
2).
Along a certain section of road a car (mass 1400kg) will coast in neutral at a
constant speed of 72 km/h if there is no wind. Examination of a topological map
shows that for this section of road the elevation decreases by 200m for each
6000m of road. What is the total resistive force (friction plus air resistance) that
acts on the car when it is travelling at 72 km/h?
This is a block on a ramp problem. F = ma, a = 0, so vertical and horizontal force
components (or parallel and normal components) each sum to zero The normal force
and the frictional force have a resultant vertical force mg.
[1]
200

 0.0333rads  1.91 .
[1].
Road is at angle   arctan
6000
Resolving vertical force mg along road, component is mg sin  = 0.0333 mg Since
acceleration is zero, this equals resistive force, which is therefore 0.0333  10  1400
= 466 N. [Since the angle is small, you could use sin  = tan  =  radians.] [1]
Check: The work done is Fs and this should be equal to mgh. Fs is 466 N  6000 m
= 2.80 MJ; while mgh is 2.80 MJ. Good agreement, so OK.
3).
[1]
A person of weight 550N stands on a bathroom scale in an elevator. As the
elevator starts moving, the scale reads 450N. a) Find the acceleration of the
elevator (magnitude and direction). b) What is the acceleration if the scale reads
670N? c) If the scale reads zero, should the person worry? Explain.
Answer: Acceleration is due to the net force on the person. Before the elevator
moves, mg downwards equals force F exerted upwards by scales, both 550 N. As the
elevator and the person accelerate downwards, mg remains constant and F changes. [1]
The person’s acceleration downwards is thus given by (mg – F)/m.
So,
550  450
 1.82ms  2 downwards
(a) a 
[1]
550 / 9.81
550  670
(b) a 
 2.18ms  2 (i.e. upwards)
550 / 9.81
550  0
(c ) a 
 10ms  2 They should worry. The lift is in free-fall. The rope [1]
550 / 9.81
has probably broken.
Check: In the frame of reference of the person, he has got lighter (a) and heavier (b),
both by about a fifth. So he may assert that gravity has decreased (a), and
[1]
increased (b) by about a fifth, or about 2ms–2. When gravity appears to
decrease, we are accelerating downwards, as in the limiting case of free-fall, (c).
So our answers are correct.
4).
You are to lower a safe of mass 260kg at constant speed down skids 20.0m
long, from a truck 2.0m high.
a) If the coefficient of friction between the safe and the skids is 0.25, do you
need to pull the safe down or hold it back?
b) How great a force parallel to the skids is needed?
Answer: Nearly the same problem as Q.2, BUT 0.1 rad is not as small an angle so
shouldn’t use sinx = tanx = x and cosx = 1 OR 0.1 rad small so can use sinx = tanx =
x and cosx = 1 OR not clear if 20 m is horizontal distance or distance along skids, so
can use sinx = tanx = x and cosx = 1.
We need to calculate in the same way components of mg along Fpar and normal Fperp
to the ramp. Then we need the frictional force which is Fperp. Comparing this with
Fpar will tell if we need to pull or push, and by how much. [1]
2
= 5.74º.Resolving force mg along skids, component
20
is Fa = mg sin  = 0.1 mg. Resolving force normal to skids, component is Fn = mg cos
 = 0.995 mg. Frictional force is therefore Ff = Fn = 0.249 mg
(a) This is greater than Fa, so you need to pull the safe downwards. [1]
(b) You need to pull with force Fp such that Fp + Fa = Ff , since acceleration is zero,
so Fp = 0.249 mg – 0.1 mg = 0.149 mg = 387 N. [1]
Skids are at angle   arcsin
Check: Work done is approximately  mg s = 0.25  2600  20 = 13 kJ. Work done
by gravity is mgh = 5.2 kJ. So you do need to pull to supply the extra energy, threefifths of total energy. Gravity is providing about 0.1 mg force or 260 N so you need to
provide about 1½ times that, or 390N, so our answer is good. . [1]
5).
A proton with a mass 1.67 × 10-27 kg is propelled at an initial speed of 3.00 ×
105 m/s directly towards a uranium nucleus 5.00 m away. The proton is repelled by
the uranium nucleus with a force of magnitude F   / x 2 , where x is the separation
between the two objects and  = 2.12 × 10-26 Nm2. Assume that the uranium nucleus
remains at rest.
a) What is the speed of the proton when it is 8.0 ×10-10 m from the uranium
nucleus?
b) As the proton approaches the uranium nucleus, the repulsive force slows it
down until it comes momentarily to rest, after which the proton moves away
from the nucleus. How close does the proton get to the uranium nucleus?
c) What is the speed of the proton when it is again 5.00 m from the nucleus?
Answer: Force varies with distance, so acceleration varies, so very messy. Rather use
energy. Conservation of energy says that PE + KE = TE, so if we get the PE as a
function of distance, the KE is TE – PE and that gives the speed by v2 = 2 KE / m.
5m is essentially infinity, (F ~ 0) so we can put the zero of PE here.

because F = – dU/dx. (Note, opposite sign to
x
gravitational potential energy, because the force is repulsive). At start, total energy is
kinetic plus potential:

E  ½ mv 2   0.75  1016 J  4.24  10 27 J [1]
x
We may neglect the second term.
Potential energy is U 
(a) At 0.8nm from nucleus,
E  0.75  10 16 J  ½ mv 2 

 ½ mv 2  2.65  10 17 J
x
½ mv 2  4.85  10 17 J
v
2  4.85  10 17
 2.41  10 5 ms  2
 27
1.67  10
[1]
(b) At rest, all the energy is electrostatic potential energy:

 7.5  10 17 J , so
x
2.12  10 26
[1]
 2.8  10 10 m
16
1.41  10
(c) At 5 metres again, the PE and KE are each same as at the start, so v is same
x
[1]
Check: (a) 0.8 nm is an atomic-scale separation, so we should not be surprised that
some velocity has been lost – i.e. our answer is in the right ballpark, 0 < v < initial
velocity.
(b) 2.8  10–10m is 0.28 nm which is >0 and < the distance of part (a), so again in the
right ballpark.
(c) This is a symmetry argument – no energy has been dissipated, so by time-reversal
symmetry the outgoing kinetic energy and hence speed at 5m must be the same as the
incoming.
[Note – it would be very tedious to check by doing a force – acceleration calculation.
Can you think of any other ways to check your results?]
[1]
[1]