Download Second Midterm Exam Solutions

PHYS 2013, General Physics I, Spring 2015
Name:
ID:
Second Exam
Problem Max. Points
1
5
2
5
3
5
4
5
5
10
6
10
7
20
8
20
9
20
10 (bonus)
20
Score
Total:
Comments:
* You must clearly show your work in order to receive full credit on any
problem.
* You are allowed to use a graphing calculator on this exam. You may NOT
use a smartphone or any other device with internet capabilities.
* Turn in your equation sheet with your exam.
* If you need more space for calculations, a blank page is included at the
end of the exam.
* You will lose points for numerical answers with missing or incorrect units.
* You will lose points for inappropriate use of sig-figs.
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Problem 1 (5 points): A block sits at rest on a ramp (“Ramp 1”) that is inclined X
degrees above the horizontal. The same block is then placed on a second ramp (“Ramp 2”)
that makes the same angle with the horizontal, but which has a larger coefficient of static
friction than Ramp 1. If the block is released from rest on Ramp 2, which of the following
statements is true?
Circle ONE of the following options.
A. The friction force between the block and Ramp 2 is greater than that of Ramp 1.
B. The friction force between the block and Ramp 2 is less than that of Ramp 1.
C. The friction force between the block and Ramp 2 is equal to that of Ramp
1.
D. It is impossible to tell without more information.
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Problem 2 (5 points): The graph below shows the net force as a function of position,
acting on a car that moves in a straight line.
(a) If the car moves from point A to point B, how does its speed at those two points
compare?
• vA > vB
• vA = vB
• vA < vB
Integral of the function from point A to point B (and thus work) is positive and therefore
the kinetic energy (and thus velocity) at point B must be larger,
vA < vB .
(b) Would your answer to part (a) have been different if the car were moving from point B
to point A instead?
• Yes
• No
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In this case the integral of the function from point A to point B (and thus work) is negative
and therefore the kinetic energy (and thus velocity) at point A must be smaller, but we still
have
vA < vB .
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Problem 3 (5 points): A superball and putty-ball strike a wall with the same velocity.
The superball bounces off the wall. The putty-ball sticks to the wall. Which ball exerts a
greater impulse on the wall?
• Superball
• Putty-ball
• Neither (the impulse is the same in both cases)
Since the momentum of superball changes by a larger amount it must exert a larger impulse
on the wall.
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Problem 4 (5 points): A block is suspended from the ceiling by a non-linear spring.
Below is the potential energy diagram for the block+spring+earth system when the y-axis
is directed upwards. Indicate the portions of the graph where the system would experience
upwards acceleration (ay > 0).
Force and also acceleration would point upward (in positive y direction) when the derivative of potential is negative, i.e. for
y < y0
or
y1 < y < y 2 .
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Problem 5 (10 points): Consider the modified Atwood machine shown below. Blocks
A and B are connected to each other by an ideal (massless, flexible, inextendible) string that
passes over an ideal (massless, frictionless) pulley. There is no friction between block A and
the table that it slides along. Air resistance is also negligible.
For each system listed in the top row of the table below, indicate whether the quantities
to the left are positive (“ + ”), negative (“ - ”), or zero (“ 0 ”) after the system is released
from rest.
system=block A (alone) system = blocks B (alone)
(a) Change in kinetic energy of the system
+
+
(b) Change in potential energy of the system
0
-
(e) Change in total mechanical energy
+
-
(d) Work done on the system by internal forces
0
-
(c) Work done on the system by external forces
+
+
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Problem 6 (10 points): Block A slides along a frictionless horizontal surface and
collides with block B. Block B is held fixed by a nail. The velocities of block A before and
after the collision are shown.
For each system listed in the top row of the table below, draw a free-body diagram of the
system during the collision, indicate whether the change in momentum is positive (“+”),
negative (“-”), or zero (“0”), and briefly explain why momentum is, or is not, conserved
during the collision by making explicit reference to the free-body diagram for that system.
System = Block A (alone) System = Block B (alone)
(a) Draw the system’s free-body
diagram during collisions:
(b) What is the change in
momentum of the system?
(positive, negative, or zero)
Negative
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Zero
Problem 7 (20 points): Bob is riding on a sled down a snowy hill that is inclined
at an angle of 30◦ above the horizontal, as shown below. The coefficient of kinetic friction
between the sled and the hill is 0.75. The combined mass of Bob and the sled is 50 kg.
(a) Calculate the total work done on Bob and the sled after descending a vertical distance
of 3.0 m.
(b) If the sled starts with speed 5.0 m/s, what is its speed after descending a vertical
distance of 3.0 m?
(c) Assuming the hill continues at the same angle and with the same coefficient of friction,
how far down the hill will Bob and the sled have traveled when they finally come to rest?
a) Let’s choose x-axis to point along sled’s motion and y-axis in the direction of normal
force, then The second law for the sled says
mg sin(30◦ ) − f = ma
−mg cos(30◦ ) + N = 0
and from definition of frictional force
f = µk N.
Thus we have three equations with three unknowns f , N and a which can be solved to get
a = g(sin 30◦ − µk cos 30◦ ) = 9.8 m/s2 (sin 30◦ − 0.75 cos 30◦ ) = −1.5 m/s2
and thus work done by all forces is
W = F s = (50 kg) −1.5 m/s
10
2
3.00 m sin 30◦
= −440 J
b) If the initial speed is vi = 5.0 m/s then the final speed vf is given by
1 2 1 2
mv = mvi + W
2 f
2
and thus
s
vf =
2W
=
vi2 +
m
s
5.02 −
2 · 440
= 2.7 m/s.
50
c) If the sled comes to rest then its final energy is zero and thus from the work energy
theorem
1
(50 kg) (5.0 m/s)2 + (50 kg) (−1.5 m/s) X = 0
2
or
X = 8.3 m.
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Problem 8 (20 points): A cart of mass 1.0 kg and block of mass 5.0 kg collide with
each other on a frictionless horizontal surface. The velocities before and after the collision
are shown in the figure below (shown as magnitudes and directions).
(a) Is the collision inelastic or elastic? Justify your answer with a calculation.
The total energy before and after collisions is not the same
1
1
1
1
(1.0 kg) (5.0 m/s)2 + (5.0 kg) (2.0 m/s)2 6= (1.0 kg) (5.0 m/s)2 + (5.0 kg) (0.0 m/s)2
2
2
2
2
and thus the collision is inelastic.
(b) In the graph below, sketch the force that each object exerts on the other during the
collision, as a function of time. (Your sketch only needs to be qualitatively correct, not
quantitatively accurate). Clearly indicate (label!) which curve belongs to which force.
Due to Newton’s third law the forces that the cart and block exert on each other must be
equal in magnitude and opposite in direction at each moment of time.
(c) If the collision lasts for 3.0 ms, what is the magnitude of the average force that the
block exerts on the cart?
Since the change in momentum of the cart equals to impulse
J = ∆p = pf − pi = (1.0 kg) (−5.0 m/s) − (1.0 kg) (−5.0 m/s) = 10.0 kg·m/s
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the average force is
F =
10.0 kg·m/s
∆p
=
= 3300 N.
∆t
3.0 × 10−3 s
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Problem 9 (20 points): Two curling stones collide on a frictionless horizontal surface.
Stone B has mass of 10 kg and is initially at rest. Stone A has mass of 20 kg and is initially
moving east toward stone B at a speed of 2.0 m/s. After the collision, stone A moves with
speed 1.0 m/s at an angle of 30◦ south of east.
Find the magnitude and direction of the final velocity of stone B.
Conservation of momenta along x and y axis implies
(20 kg) (2.0 m/s) + 0 = (20 kg) (1.0 m/s) cos 30◦ + (10 kg) vx
0 + 0 = − (20 kg) (1.0 m/s) sin 30◦ + (10 kg) vy
and thus
vx = 2.3 m/s
vy = 1.0 m/s.
Therefore the magnitude is
v=
√
2.32 + 12 = 2.5 m/s
and angle is
θ = arctan
north of east.
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1.0
= 23◦
2.3
(Bonus) Problem 10 (20 points): A cart of mass m is pushed against an ideal spring
(obeying Hooke’s law) of spring constant k. The cart is then released and free to glide along
the frictionless track shown. The spring is compressed by a distance d just before the cart
is released. When the cart reaches point D along the loop-the-loop of radius R, what is the
magnitude of the normal force that the track exerts on the cart? Your answer may contain
m, k, d, R and any pure numbers or physical constants.
The initial (elastic) potential energy of spring is
1 2
kd
2
which is converted into (gravitational) potential energy
mg(2R)
and kinetic energy
1 2
mv .
2
From conservation of energy
1 2
1
kd = 2mgR + mv 2
2
2
and thus
s
v=
kd2
− 4gR.
m
Since the cart is in circular motion the centripetal acceleration must be
v2
a=
=
R
kd2
m
− 4gR
kd2
=
− 4g.
R
mR
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and from Newton’s second law along y axis (with positive direction pointing downwards)
N + mg = ma
or
kd2
kd2
N = m(a − g) = m
− 4g − g =
− 5mg.
mR
R
!
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