Here are some statistics on the second exam:
176 people in both sections took it, and grades ranged from 16 to the maximum possible
grade of 107. The average score was 61.8, and the number of people falling in different
intervals is shown below.
>90.0
16
80-90
27
70-80
26
60-70
23
50-60
27
40-50
27
30-40
18
0-30
12
The answers to the exam are given below:
Problem 1. A dart of mass m=0.05 kg moving at 10 m/s strikes a stationary block of mass M=0.45 kg
that rests on a horizontal frictionless surface.
(a) 10 pts Find the speed of the block plus dart after impact, assuming that the dart was initially
traveling horizontally just before sticking to the block
m1v1  (m1  m2 )v' Thus, (0.05)(10)  (0.05  0.45)v' and v' 1.0m / s
(b) 10 pts What average force did the dart exert on the block in (a) if the impact lasted 0.001 seconds?
F
p mv  mv0 0  (0.45)(1.0)


 450 N
t
t
0.001
(c) 5 pts Is the kinetic energy of dart plus block after impact greater than, less than, or equal to their
kinetic energy before impact? (Just state which of the 3 choices is correct – no work needed.)
KE after collision is less than before, since this is a completely inelastic collision
(d) 10 pts What would be the answer to part (a) if the dart were traveling at 10 m/s at an angle of 37
degrees below the horizontal as shown, and the block had an initial speed of +1.0 m/s initially?
Conservation of momentum, x-component equation:
m1v1 cos  m2 v2  (m1  m2 )v' Thus, (0.05)(10) cos 37  (0.45)(1.0)  (0.05  0.45)v'
and hence: v' 1.7m / s
Don’t do anything with the y-component equation, since block moves along table after impact
_____1. The maximum speed of a car around a level curve without skidding is 30 km/h. What is the maximum
speed around a curve of twice the radius, assuming the same frictional force?
(a) 42.4 km/h
(b) 45.0 km/h
(c) 60.0 km/h
(d) 120 km/h
Since f is unchanged, so is a = v^2/r, therefore if r doubles, so does v^2, and hence v itself goes up by a
factor of the square root of 2
_____2. A car makes a banked turn on a frictionless curve. The centripetal force equals the _______
component of the normal force
(a) horizontal (b) vertical
(c) net
(d) tangential
That is the component that points towards the center of the circle
_____3. A stone is attached to a string and whirled in a vertical circle of radius r. At the exact top of the circle
the tension in the string is three times the stone’s weight. The stone’s speed at this point is:
(a) 2 gr
(b)
2 gr
(c)
gr
(d) 2 gr
(e) none of them
By Newton’s 2nd Law: T + mg = ma = mv^2/r. Substituting T = 3mg yields 4mg = mv^2/r, and hence
V= 2 gr
_____4. The gravitational attractive force between two masses is F. If the masses center-to-center separation is
reduced in half, what is the gravitational force now? (a) F/4
(b) F/2
(c) 2F
(d) 4F
Since F is inversely proportional to r^2, if r is replaced by r/2, F quadruples.
_____5. The gravitational acceleration of gravity on the moon is one sixth what it is on Earth. The radius of the
moon is one fourth that of the Earth. The Earth’s mass divided by the moon’s mass is therefore
(a) 6
(b) 16
(c) 24
(d) 96
(e) none of them
g = GM/R^2, hence M = gR^2/G. On the moon g  g/6 and R R/4, hence M  (1/6)(1/4)^2 M = M/96
_____6. The planet Jupiter is 5.2 times further from the sun than the Earth is. How long does it take Jupiter to
complete one orbit around the sun? (a) 1 yr
(b) 3 yr
(c) 6 yr
(d) 12 yr
Use T^2 = KR^3. For Jupiter Rj = 5.2 Re, hence (Tj)^2 = K(5.2Re)^3 = K(5.2)^3 Re^3 = 141KRe^3
Hence Tj^2 = 141 Te^2, and so Tj = 12 Te
_____7. You lift a physics book weighing 10 N by a distance 1.0 m vertically at a constant speed. How much
work does gravity do?
(a) +10 J
(b) +98 J
(c) -10 J
(d) -98 J (e) none of them
W = Fd cos 180 = (10)(1.0)(-1) = - 10 J
_____8. How much work is required to compress a spring of force constant 20 N/m by 15 cm?
(a) 0.15 J
(b) 0.30 J
(c) 0.23 J
(d) 0.45 J
(e) none of them
W = 0.5 kx^2 = (0.5)(20)(0.15^2) = 0.23 J
_____9. A stone is dropped from rest. Momentum is conserved when the stone hits the ground only if the
(a) collision is elastic
(b) collision is inelastic
(c) Earth is considered part of the system
(d) Earth is not considered part of the system
Then the force of the Earth on the stone becomes an internal force & no change in momentum (of stone
plus Earth)
_____10. A planet orbits the sun in an elongated elliptical orbit. The kinetic energy of the planet increases
during the portion of its orbit during which it (a) approaches the sun
(b) recedes from the sun
(c) is near perihelion
(d) is near aphelion
(e) none of them
Only when planet approaches sun is the force of sun’s attraction causing planet to speed up
_____11. A ball falls from the top of a tall building. Taking air resistance into account, how does the potential
energy of the ball at the top of the building PE compare with the kinetic energy KE just before the ball hits the
ground?
(a) PE = KE
(b) PE > KE
(c) PE < KE
(d) PE = - KE
If no air resistance PE = KE, but since some air resistance is present, some of the original PE is converted
to heat by friction (with the air), and so the KE is less than the PE.
_____12. Compared to yesterday you did 3 times the work today in a third the time. In this case, your power
output today must have been ____ times your power output yesterday
(a) 1.0
(b) 1/3
(c) 3
(d) 9
(e) none of them
p = W/t. If W is tripled, and t is 1/3 as much, p goes up ninefold.
_____13. Consider two point masses 1 kg and 2 kg separated by 3 meters. How far from the 1 kg mass will
their center of mass will be located?
(a) 1.0 m
(b) 1.5 m
(c) 2.0 m
(d) 2.5 m
(e) none of them
The center of mass is the weighted average of the 2 masses. It is therefore twice as close to the 2 kg mass
as the 1 kg mass. (You could also figure this one out using the equation for x_cm)
See other side!