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Physics 8.012 Fall 2008
Exam 2
NAME : ______________ _ _ _ _ _ _ _ _ _ _ __________________^ SOLUTIONS
MIT ID number: __________________________________________________
Instructions:
- Do all FIVE (5) problems. You have 90 minutes.
- SHOW ALL WORK. Be sure to circle your final answer.
- Read the questions carefully.
- All work must be done in this booklet in workspace provided. Extra blank pages are provided at the end if needed.
- NO books, notes, calculators or computers are permitted. A sheet of useful equations is provided on the last page.
Your Scores
Problem Maximum Score Grader 1 10 2 20 3 20 4 25 5 25 Total 100
Problem 1: Quick Multiple Choice Questions [10 pts]
For each of the following questions circle the correct answer. You do not need to show any work.
(a) Which of the following is the result of an elastic
collision between the two isolated objects at right? m^ 2m
v
v
m 2m
v/3 (^) m 2m 2v/3 v
v m 2m m 2m v/
This can be verified by examining conservation of momentum and energy (since it is an elastic collision).
(b) Which experiences the largest impulse?
Fully elastic
collision of 10 kg
ball dropped
from 10 m
Fully inelastic
collision of 20 kg
ball dropped
from 10 m
Fully elastic
collision of 20 kg
ball dropped
from 5m
These all have
the same
impulse
The first two choices experience the same impulse, which equals ( 1 or 2). The 3 rd^ option experiences the largest impulse by a factor of
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Problem 2: Collision [20 pts]
m
2m
m
2m
!
v
v
v
U
Two balls, one of mass m and one of mass 2m, approach from orthogonal directions with identical speeds v and collide. After the collision, the more massive ball moves with the same speed v but downward (orthogonal to its original direction) and the less massive ball moves with speed U at an angle θ with respect to horizontal. Assume that no external forces act during the collision.
(a) [10 pts] Calculate the final speed U of the less massive ball and the angle θ.
(b) [10 pts] Determine how much kinetic energy is lost or gained by the two balls during the collision. Is this collision elastic, inelastic or superelastic?
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[SOLUTION]
(a) Momentum in both horizontal (x) and vertical (y) directions must be conserved, hence:
so:
so
(b) The change in kinetic energy is just the change in the kinetic energy of the small mass (since the big mass has the same KE before and after the collision):
the collision is superelastic.
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[SOLUTION]
(a) Only conservative forces are present, so total mechanical energy is conserved. Hence the KE of the block comes simply from extracting gravitational potential energy:
(b) The block loses contact when the normal force acting on it disappears. Using polar coordinates and the force diagram at right, we can write down the radial equation of motion:
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Problem 4: Tabletop Rocket [25 pts]
M 0 /
vex
M 0 /
A rocket of total mass M 0 , half of which is fuel, starts at rest on a long horizontal table. The coefficient of friction between the rocket and table surfaces is μ. At time t = 0, the rocket is ignited, ejecting fuel out at a constant rate γ = |dM/dt| with velocity vex relative to the rocket. Constant gravitational acceleration g acts downward.
(a) [10 pts] What condition must be met for the rocket to start moving at t = 0?
(b) [10 pts] Assuming that the rocket satisfies this requirement, what is the maximum speed VMAX achieved by the rocket?
(c) [5 pts] How far does the rocket go after it runs out of fuel? You can express your answer in terms of VMAX.
(d) [BONUS 5 pts] How far does the rocket travel in total? For this you will need to make use of the following integral:
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tmax is the time it takes for the fuel to run out, which is (M 0 /2)/ γ. Hence,
(c) After the rocket runs out of fuel, the problem reduces to a simple mass accelerated by a constant friction force:
We can also solve this using energy, since the work done on the rocket to reduce its kinetic energy to 0 is simply the friction force acting over the distance the rocket travels:
(d) For those who attempted it, we simply use the expression of velocity as a function of time from our derivation of the rocket equation above, and integrate this to get the distance traveled:
where we now explicitly put in the time dependence of the rocket mass. To integrate the first term on the right, we make the substitution:
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Using our substitution for tmax for the limits of u and solving the left side and second term on the right side:
The total distance traveled is then
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[SOLUTIONS]
(a) The potential can be derived from the force law:
In this case we know that the constant C = 0 since U(∞) = 0. Hence:
Note the m is a factor in this and other expressions (mistakenly neglected in the question).
The kinetic energy is
but we can substitute this using the force law in polar coordinates:
the total mechanical energy E = K + U is then:
(b) The work required to effectively bring the planet to r = ∞ is simply that needed to climb the potential well to U = 0. In other words:
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So the kinetic energy must double, which means the speed must increase by a factor of.
(c) The loss of mechanical energy simply comes from the work done by the frictional force over one orbit (note: this is formally true since the potential energy doesn’t change – work only acts to change kinetic energy). Hence:
where the substitution for mv^2 from above is used
(d) We can also write the change in energy as:
to compute a velocity, we note that this change in r occurs over one orbit period for which:
Thus:
Note that the radial velocity is negative, so the planet falls in to the star.
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