# PHYS 211 – MT3 Fall 2012 Sample 2

PHYS 211 – MT3
Fall 2012
Sample 2
1. Danielle and Susan stand on a frictionless cart on one side of the room while Austin
stands on a second frictionless cart on the opposite side of the room. Both carts are
initially at rest, and a long rope is held between them. The girls pull as hard as they can
on the rope, while Austin merely holds on to his end. A short while later the carts collide.
Which of the following must be true?
A. Austin's cart would have moved a greater distance if he'd been the one pulling on
the rope.
B. Danielle & Susan's cart would have moved a greater distance if Austin had been
the one pulling on the rope.
C. The point of collision will be approximately halfway between the initial separation.
D. The point of collision depends upon who pulls the rope.
E. The point of collision will be approximately the location of center of mass of the
system.
Treating the two as a system, their center of mass can’t change as there is no external
force.
2. A woman lifts a barbell 2.0 m in 5.0 s. If she lifts it the same distance in 10 s, the work
done by her is:
A.
B.
C.
D.
E.
one-fourth as great
one-half as great
the same
two times as great
four times as great
Power changes with time, not work
3. A skydiver of mass 80 kg falls through a distance of 1000 m at a constant terminal
velocity of 100 m/s. What is the net work done on the skydiver? (Use g = 10 m/s2)
A.
B.
C.
D.
E.
F.
G.
-8 MJ
-800 kJ
-80 kJ
80 kJ
800 kJ
8 MJ
None of the above
No change in velocity  none in kinetic energy 
no net work
Page 1 of 8
PHYS 211 – MT3
Fall 2012
Sample 2
4.
The graph above shows the force on an arrow due to a bow string as a function of
distance as the arrow is released.
What is the kinetic energy of the arrow just after it has left the bow? You may ignore
friction and assume that the bow is fired horizontally.
A.
B.
C.
D.
E.
60 J
80 J
100 J
120 J
160 J
W   Fdx  400N  40cm 
1m
 160J
100cm
5. Two boxes, A and B, slide along a frictionless floor with equal magnitude momenta
(pA = pB). If the mass of A one fourth the mass of B (mA = ¼mB), how do their kinetic
energies compare?
A.
B.
C.
D.
E.
KA = KB
KA = 2 KB
KA = 4 KB
KA = 8 K B
KA = 16 KB
K A p A2 2mA p A2 mB
m
 2
 2
 1 B 4
K B pB 2mB pB mA 4 mB
Page 2 of 8
PHYS 211 – MT3
Fall 2012
Sample 2
6. Suppose rain falls straight down into an open bucket sliding along a frictionless,
horizontal surface. The rain slowly fills up the moving bucket, and after it is full, the
bucket is still moving.
The empty bucket has an initial momentum po and an initial kinetic energy Ko. The full
bucket (including the water it contains) has a final momentum pf and a final kinetic
energy Kf. Which of the following is true?
A.
B.
C.
D.
E.
F.
po = pf and Ko < Kf
po < pf and Ko = Kf
po < pf and Ko < Kf
po = pf and Ko = Kf
po = pf and Ko > Kf
None of the above
No external force  no change in momentum
(it will slow down as it has to accelerate the rain)
KE 
p2
will shrink, as m increases but p doesn't change
2m
7. The following 5 “matching” questions are all based on the following set-up and are
each worth two points (so these 5 questions together are worth 2 of the other
multiple choice questions, 10 points)
Below are 5 types of problems that you have been asked to solve. Note that the
problems aren’t well enough defined to actually get a solution. However, you should be
able to determine which of the below approaches you would use if you were given the
necessary information. For each problem pick one:
A.
B.
C.
D.
E.
Only conservation of momentum
Only conservation of mechanical energy
Work-Energy Theorem
Conservation of both momentum and energy
None of the above can be used
a) Playing billiards, a cue ball hits another ball off center (elastically), and they both
go off at given angles. How fast were they moving? D (elastic)
b) A bullet is fired at a wood block, hanging from a string. It embeds in the block. To
what angle does the block swing upward. D (momentum 1st, then energy)
c) A bullet is fired at a wood block, hanging from a string. It passes through the
block, and the block swings upward a known amount. How fast does the bullet
exit the block? D (momentum 1st, then energy)
d) You push a block with a given constant force across a rough table. How fast is it
going when it gets to the other side of the table? C (work is change in KE)
e) A ball slides along horizontal surface and collides with a second identical ball.
Given the final speed of the first ball, what is the final speed of the second ball?
A (have to use only momentum , aren’t told it is elastic)
Page 3 of 8
PHYS 211 – MT3
Fall 2012
Sample 2
8. A potential energy curve is shown below.
If a 4 kg object has a speed of 2 m/s at x = 12 m, what is the largest x value it can
reach?
A.
B.
C.
D.
E.
F.
8m
38 m
40 m
42 m
44 m
46 m
At x = 12 m it has a potential energy of 28 J. It has a kinetic energy
K  12 mv 2 
1
2
 4kg   2 ms 
2
 8J  E mech  36J
This corresponds to the red horizontal line on the plot, which intersects
the potential energy curve at x = 44 m
Page 4 of 8
PHYS 211 – MT3
Fall 2012
Sample 2
9. The figure below shows the force exerted on a particle as a function of time. What is
the impulse?
A.
B.
C.
D.
E.
20 Ns
100 Ns
200 Ns
300 Ns
400 Ns
J   F dt 
1
2
 20N  35s-5s   300Ns
10. About how long will it take a 2000 W motor to lift a 300 kg piano from the ground to
a window 20 m above the ground?
A.
B.
C.
D.
E.
10 s
m
W
W mg y  300kg  10 s2   20m 
P
 t 


 30s
30 s
t
P
P
2000W 

100 s
133 s
None of the above
Page 5 of 8
PHYS 211 – MT3
Fall 2012
Sample 2
11. Each of the following true-false questions is worth 2 points (so these 5 questions
together are worth 2 of the multiple choice questions, 10 points)
A. T or F: If the dot product of two nonzero vectors is zero, the vectors must be
perpendicular to each other.
B. T or F: If two nonzero vectors point in the same direction, their dot product must
be zero. It will be their length squared
C. T or F: The value of the dot product of two vectors depends on the particular
coordinate system being used. ABcos no matter the coordinates
D. T or F: A  B  B  A Dot products are based on multiplication, so commutative
E. T or F: If a force always acts perpendicular to an object's direction of motion,
that force cannot change the object's kinetic energy.
The work is zero so it can’t change the kinetic energy
12. A 2,000. kg car sits on frictionless, horizontal road surface when it is struck from
behind by a 4,000. kg truck. If the two vehicles' bumpers lock so they stick together and
travel at 2.0 m/s immediately after the collision, how fast was the truck traveling when it
hit the car?
Conservation of momentum in this inelastic collision:
A. 2.0 m/s
B.
6 m/s
p0  mt v0  p f   mc  mt  v f
 v0 
C. 3.0 m/s
D. 6.0 m/s
E. None of the above
 mc  mt  v
mt
f

 6000kg  2 m  3 m
s
 4000kg  s
13. A ball of mass m is thrown with speed vo
at a block of mass M, which sits on a
frictionless surface. In which situation is the
average force on the block from the ball the
greatest?
A. The ball sticks to the block, the collision lasts a time 4T.
B. The ball sticks to the block, the collision lasts a time T.
C. The ball bounces back off the block with about the same speed it originally had,
the collision lasts a time 4T.
D. The ball bounces back off the block with about the same speed it originally had,
the collision lasts a time T. Bouncing delivers more impulse (more p).
Shorter time means larger average force (J=Ft)
E. The ball goes through the block (e.g., like a bullet going through a block of
styrofoam), the collision lasts a time T.
Page 6 of 8
PHYS 211 – MT3
Fall 2012
Sample 2
14. A shell, at rest, explodes into two fragments, one fragment 25 times heavier than the
other. If any gas from the explosion has negligible mass, then
A. the momentum change of the lighter fragment is 25 times as great as the
momentum change of the heavier fragment.
B. the momentum change of the heavier fragment is 25 times as great as the
momentum change of the lighter fragment.
C. the momentum change of the lighter fragment is exactly the same as the
momentum change of the heavier fragment. (NO, p25m  pm )
D. the kinetic energy change of the heavier fragment is 25 times as great as the
kinetic energy change of the lighter fragment.
E. the kinetic energy change of the lighter fragment is 25 times as great as the
kinetic energy change of the heavier fragment. p 2 2m , same p, small m  big K
F. More than one of the above
15. A man sits in the back of a small boat in still water. He then moves forward to the
front of the boat and sits there. Afterwards the boat:
A.
B.
C.
D.
E.
is forward of its original position and moving forward.
is forward of its original position and moving backward.
is forward of its original position and not moving.
is behind its original position and moving backward.
is behind its original position and not moving. No external F, xcm stays same
16. A 1/2 kg block is sliding down a frictionless incline which makes an angle of  =
30.0° with the horizontal. It then slides on a frictionless horizontal surface until it slides
into a relaxed horizontal spring with a spring constant k = 250. N/m. The block becomes
attached to the spring and continues moving, compressing the spring 1/2 m before
momentarily stopping.
What is the speed of the block just before it hits the spring? (Use g = 10 m/s2)
Kinetic energy before hit transforms to spring potential energy after hit
A.
5 m/s
B.
10 m/s
1
2
mv 2  12 k  s   v 
2
k
250 N
s  1 m 12 m 
m
2 kg
1 m
2 s
5  100  5 5 ms
C. 5 5 m/s
D. 10 5 m/s
E.
115 m/s
Page 7 of 8
PHYS 211 – MT3
Fall 2012
Sample 2
17. Two carts on a frictionless air track, each of mass 0.5 kg, are held together at rest
with a spring compressed between them. The spring has a spring constant k = 2500
N/m and is compressed 0.1 m from its relaxed length. The carts are released
simultaneously. Find the speed of one cart with respect to the air track.
Same magnitude momentum and same mass  same velocity, same K
A. 5 m/s
2 K f  2 12 mv 2  12 k  s 
B. 5 2 m/s
C. 10 5 m/s
v
D. 10 10 m/s
E. None of the above
2
k
2500 mN
s 
0.1m  50 1s  0.1m  5 ms
2m
2  12 kg
18. Sarah Jane Smith rides a frictionless rollercoaster with an initial hill of height H (on
top of which the car is barely moving). At the bottom, where the track has a radius of
curvature R, the car has a speed V and her apparent weight is 25% heavier than her
actual weight
FBD
Acc
aC
If the hill were four times as tall (i.e., 4H), the magnitude of the normal force acting on
her at the bottom would be, in terms of the force of gravity acting on her Fg:
A.
B.
C.
D.
E.
1.5Fg
2Fg
2.5Fg
4Fg
5Fg
“Apparent Weight” = Normal Force! Draw FBD, Acc diagrams (above) to
see that N  FG  macentripetal  R1 mv 2 
2 1
R 2
mv 2  R2 K
If we start out 4 times higher, we will have 4 times the KE at the bottom
and hence require 4 times the net force. So instead of N  FG  0.25FG
we will have N  FG  FG  N  2FG
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