PHYS 211 – MT3 Fall 2012 Sample 3 Solutions

PHYS 211 – MT3
Fall 2012
Sample 3 Solutions
1. The following true-false questions are all based on the following set-up and each is
worth 2 points (so these 5 questions together are worth 2 of the multiple choice
questions, 10 points)
A particle moving in one dimension is subject to the potential energy curve shown in the
plot.
A. T or F: At point xa, the force on the particle is in the negative x direction.
Mathematically: The force is the negative slope (slope is negative so force is to the
right). Conceptually: Objects like to roll down hill (to lower potential energy)
B. T or F: If the particle has total energy of 3 Joules at point xc and its velocity is in the
positive x direction, it will eventually reach point xa.
It won’t be able to make it over the hump between xb and xc
C. T or F: If the particle has kinetic energy of 3 Joules at point xc it can reach point xb.
The total energy, 4 J, is enough to make it over the hump
D. T or F: If the total energy of the particle at point xb is 3 Joules it can reach point xc.
For the same reason as case B
E. T or F: If the particle is released from rest at xc, it will travel in the positive x
direction.
The force will be to the left (downhill)
Page 1 of 10
PHYS 211 – MT3
Fall 2012
Sample 3 Solutions
2. A boy walks on level ground at a constant speed of 2.0 m/s while holding a 40 N box
horizontally at arm's length. His arm is 1.5m above the ground. What is the work done
by the force of the boy on the box while he walks with it?
A.
B.
C.
D.
E.
0J
27 J
40 J
60 J
90 J
No change in energy, no work done (F perpendicular to distance)
3. You and your friend are both standing on a horizontal frictionless surface. To get your
friend's attention, you throw a ball due west at your friend. The ball bounces elastically
off your friend's back. Which one of the following statements is true about this situation?
A. Your friend will remain stationary because the ball bounces elastically and does
not impart its momentum to her.
B. Your friend will remain stationary because the kinetic energy of the ball is
conserved in an elastic collision.
C. Your friend will remain stationary due to conservation of both linear momentum
and kinetic energy.
D. You will remain stationary after you throw the ball due to conservation of both
linear momentum and kinetic energy.
E. You will move east after you throw the ball due to conservation of linear
momentum. Ball goes W, you go E. Your friend will go W (in all conditions)
F. More than one of the above
4. A large car and a hard-headed bird collide, with the bird bouncing off the windshield.
Which of the following statements is true about the vector impulse on the bird ( J bird ) and
the vector impulse on the car ( J car ).
A. J bird  J car
B. J bird  J car
C. J bird  J car
Equal and opposite forces, same time
D. J bird  J car
E. None of the above
Page 2 of 10
PHYS 211 – MT3
Fall 2012
Sample 3 Solutions
5. Two bodies, A and B, have equal kinetic energies. The mass of A is nine times that of
B. The ratio of the momentum of A to that of B is:
A.
B.
C.
D.
E.
1:3
1:1
1:9
9:1
3:1
1
K A p A2 2mA p A2 mB p A2 mB
p2
p
 2
 2
 2
 A2  9  A  3
K B pB 2mB pB mA pB 9mB
pB
pB
6. Each of the figures shown below indicates the direction and magnitude of a force F
acting on a system which is moving with a velocity v.
Which figure represents the most negative (or least positive if none are negative) power
being done by the force F at the instant pictured ?
Need F & v to point in
opposite directions to be
negative
7. A student standing on a third floor balcony throws a water balloon onto the ground
below. The work done on the balloon by gravity (during the fall) and the ground (as the
balloon smashes into it), respectively, are:
A.
B.
C.
D.
E.
Negative, Negative
Positive, Negative
Negative, Positive
Positive, Positive
None of the above
The ground doesn’t move so it does no work.
Gravity does positive work (force and motion
both down)
Page 3 of 10
PHYS 211 – MT3
Fall 2012
Sample 3 Solutions
8. A 2500 kg car traveling down I-80 at 100 km/hr smacks into a 0.001 kg fly traveling in
the opposite direction at 1 km/hr. Which of the following statements is true?
A.
B.
C.
D.
The car exerts a much bigger force on the fly than does the fly on the car.
The fly exerts no force on the car whatsoever.
If the fly sticks to the car, it exerts a greater force on it than if it bounces off it.
The magnitude of the change in momentum of the fly is the same as the
magnitude of the change in momentum of the car, and points in the same
direction.
E. The magnitude of the change in momentum of the fly is the same as the
magnitude of the change in momentum of the car, and points in the opposite
direction. Equal & opposite force, impulse and hence change in momentum




9. A particle of mass 5 kg is acted on by two forces, F1  3ˆi  4ˆj N and F2  2ˆi  ˆj N
 


and is moved from ri  ˆi  ˆj m to rf  5ˆi  4ˆj m . What is the net work done on the
particle?
A.
B.
C.
D.
E.
-21 J
-11 J
-9 J
-1 J
7J



  
r  r  r   5ˆi  4ˆj m   ˆi  ˆj m   4ˆi  5ˆj m
W =F  r   ˆi  5ˆj N   4ˆi  5ˆj m  4  25J  21J
Fnet  F1  F2  3ˆi  4ˆj N  2ˆi  ˆj N= ˆi  5ˆj N
f
net
i
net
10. A paratrooper whose chute fails to open land in snow; he is hurt slightly. Had he
landed on bare ground, the stopping time would have been 10 times shorter. Does the
presence of the snow increase, decrease, or leave unchanged the values of the
impulse stopping the paratrooper and the force stopping the paratrooper?
A.
B.
C.
D.
E.
impulse unchanged, force unchanged
impulse unchanged, increased force
impulse unchanged, decreased force
decreased impulse, decreased force
decreased impulse, force unchanged
Same momentum to
stop, so same impulse.
But longer time means
smaller force.
Page 4 of 10
PHYS 211 – MT3
Fall 2012
Sample 3 Solutions
11. A firecracker at rest on the ground explodes into just two pieces (A and B):
Piece A: mass M
Piece B: mass 3M
What can we say about the magnitude of momentum (p) and the kinetic energy (K) of
these two pieces?
A. Both pieces have the same magnitude of momentum, but A has 3 times the
kinetic energy of B. p conserved so same, small m  big K= p 2 2m
B. Both pieces have the same magnitude of momentum, but B has 3 times the
kinetic energy of A.
C. A has three times the magnitude of momentum and three times the kinetic
energy of B.
D. B has three times the magnitude of momentum and three times the kinetic
energy of A.
E. Both pieces have the same kinetic energy, but B has three times the magnitude
of momentum of A.
F. None of the above
12. 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)
A stock person at the local grocery store has a job consisting of the following five
segments. During each segment determine whether she is doing:
A. Positive work
B. Negative work
C. No work
D. Either positive or negative work (not enough info to determine)
1.
2.
3.
4.
5.
picking up boxes of tomatoes from the stockroom floor (A: F, s up)
accelerating to a comfortable speed (A: F, s forward)
carrying the boxes to the tomato display at constant speed (C: const. v  no F)
decelerating to a stop (B: F backward, s forward)
lowering the boxes slowly to the floor. (B: F up, s down)
Page 5 of 10
PHYS 211 – MT3
Fall 2012
Sample 3 Solutions
13. A mass M moving in the x-direction with a velocity of 10 m/s ˆi collides with a mass
m (with half the mass of M) that has a velocity -5 m/s ˆi . Immediately afterwards, M is
observed to have a velocity of 0 m/s ˆi and m is observed to have a velocity of 15 m/s ˆi .
What can we say about this collision?
A. It is an elastic collision.
B. It is an inelastic (but not completely inelastic) collision.
C. It is a completely inelastic collision.
D. It is possible only if some other energy was released in the collision.
E. It is not possible.
1
2
?
M 10 ms   12  12 M   5 ms   12 M  0 ms   12  12 M  15 ms 
2
2
2
2
Multiply by 4/M and the units, and we get
?
?
2 10    5  15  225  2  100  25 YES!
2
2
2
14. A single force, shown below on the graph, acts on a 2 kg object. If the object's
speed at x = 0 m is 2 m/s, what is its speed at 8 m?
A. 2 m/s


W   Fdx 
1
2
 4N  4m    4N  2m   12  4N  2m   20J
B. 2 1  5 m/s
Start with Kinetic energy, add this, get new kinetic energy:
C. 2 6 m/s
KE f  12 mv 2f  KE0  W
D. 20 m/s
E. 24 m/s
F. None of the above
 v f  v02  m2 W 
 2 ms 
2
 2 2kg 20J  24
m
s
 2 6 ms
Page 6 of 10
PHYS 211 – MT3
Fall 2012
Sample 3 Solutions
15. A large care package, attached securely to a bungee cord, is released from rest
from a stationary helicopter. The helicopter is 30 meters above the ground. The bungee
cord has an unstretched length of 20 meters and a spring constant of 100 Newtons per
meter. The mass of the package is 50 kg. Unfortunately the helicopter is too low and the
package hits the ground. With approximately what velocity does it hit the ground?
As always, use g = 10 m/s2 and ignore air resistance
A. 0 m/s
B. 2 10 m/s
C. 20 m/s
D. 20 2 m/s
E. 20 6 m/s
F. None of the above
Use conservation of energy, with the initial energy at the helicopter and the final at
the ground. We’ll call the ground zero height for gravitational potential. The bungee
cord only pulls, so it has no energy in the helicopter (as opposed to a compression
spring, which would)
E0  U 0  mgy0  E f  U g , f  U B , f  K f  0  12 k  s   12 mv 2
2
v  2 gy0 


k
100 N
2
2
 s   2 10 sm2  30 m   m  30m - 20m   600  200
m
50 kg
m
s
 20 ms
Start with Kinetic energy, add this, get new kinetic energy:
KE f  12 mv 2f  KE0  W
 v f  v02  m2 W 
 2 ms 
2
 2 2kg 20J  24
m
s
 2 6 ms
Page 7 of 10
PHYS 211 – MT3
Fall 2012
Sample 3 Solutions
16. A massless spring connects a ball of mass M to a vertical post on a horizontal
frictionless table. The spring has a spring constant k and an unstretched length of Lo. If
the post is rotated at a constant rate, the ball travels at constant speed in a horizontal
circle on the table (so gravity plays no part in this problem). The spring stretches when
the ball is moving in such a circle. At what speed, v, should the ball be moving so that
the spring is stretched to twice its unstretched length (i.e. so that it has a length of 2Lo)?
FBD
Acc
aC
A. L0 4k m
This problem is a dynamics problem. Draw the FBD & acc diagrams
(above). Then solve:
B. L0 2k m
C. L0 k m
D. L0 k 2m
Fspring  k s  k  2 L0  L0   mac  m
v  2 L0
v2
2 L0
k
k
L0  L0 2
m
m
E. L0 k 4m
Page 8 of 10
PHYS 211 – MT3
Fall 2012
Sample 3 Solutions
17. Two rods are connected to make an "L" shape. Each rod is uniform, and they have
the same length l, but one is twice as massive as the other. If the rods are placed as
shown below, with the more massive rod on the y-axis, what is the coordinate of the
center of mass of the "L"?
To find the center of mass of the system, we can work from the CM
of the two rods. We have 2m @ (0,l/2) and m @ (l/2,0)
1 
1 l
rCM 
2m 0, l  m l ,0  
,l  l , l
2
2
2
6 3


m  2m
3

A.
B.
C.
D.
E.
 


 

(l/2, l/2)
(l/6, l/3)
(l/6, l/2)
(l/2, l/3)
(l, l)
Page 9 of 10
PHYS 211 – MT3
Fall 2012
Sample 3 Solutions
18. A 1 kg box is given a quick push (from rest) up a ramp angled at 30° to the
horizontal, using a spring launcher of spring constant k = 4000 N/m, initially compressed
by 0.1 m. The box slides up the ramp with a coefficient of kinetic friction of 1
3 until it
comes to a halt. What total distance D along the slide does the box travel before it
stops?
FBD
Acc
f
A. 2 m
B. 4
We use conservation of energy, remembering that the work
done will add (negatively) to the initial energy. We start with
spring potential energy (we’ll define the height to be zero at the
spring) and end with gravitational potential energy (since it
comes to rest).
We have to draw the FBD and acc diagrams to get the friction
force right: N  mg cos  0; f  k N  k mg cos
3m
C. 4 m
D. 4 3 m
E. 4 3
F.
4
k  1 3
a
5
m
1  3 
G. 4 3

H. 4 3

m
Now we do energy/work:
E0  W  12 k  s   fD  E f  mgy f  mgD sin 
2

3 1

3 1
m
m
1
2
k  s   k mg cos  D  mgD sin 
2
Dmg  k cos   sin    k  s 
1
2
2
k  s 
D
mg  k cos   sin  
1
2
2
Now we just plug in numbers:
D
1
2
 4000 mN  101 m 
2
1kg  10 sm    13  23    12 
 2m
2
Page 10 of 10