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Physics 215 Fall 2008 Exam 3 Version D (751464)
Instructions
Be sure to answer every question. Follow the rules shown on the screen for filling in the Scantron form. Each problem is worth
10% of the exam. When you are finished, check with Dr. Mike or his TA to be sure you have finished the scantron correctly.
1. Momentum and Energy [714509]
Bailey D. Wonderdog is flying in a straight line at a speed of v0. When she sees trouble ahead, she increases her speed to vf
while still taveling in the same direction. What is the ratio of a change in her kinetic energy to a change in the magnitude of her
momentum?
*03*c* v2 / 2
*10*a*(vf2 - v02) / 2 v
*05*b* v / 2
*03*d*v0 / 2
*01*e*v02 / 2
Solution or Explanation
Her change in kinetic energy is
K = 1/2 m (vf2 - v02).
The magnitude of her change in momentum is
p = m v.
Now you just need to take the ratio of K / p.
2. Impulse [714507]
A ball experiences an impulse from the floor with a magnitude of 161 kg m/s. The impulse acts for a duration of 0.03 s. What is
the magnitude of the average force acting on the ball from the floor?
*02*d*It depends on the mass of the ball.
*10*a*5366.67 N
*02*e*It depends on the speed of the ball.
*01*c*0.07 N
*05*b*4.83 N
Solution or Explanation
From either the textbook, notes or website, we have that
J = Fav t.
You then solve for Fav.
3. Center of Mass [813712]
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A baseball bat of length L has a peculiar linear density (mass per unit length) given by
center of mass in terms of L.
= 0 x4/L4. Find the x coordinate of the
*10*a*0.83 L
*04*b*0.17 L2
*05*d*1.25 L
*00*e*1
*01*c*0.5 L
Solution or Explanation
Here we use the continuous form of the center-of-mass equation
rcm = [( 0L x
dx) i] / [( 0L
dx)].
rcm = [( 0L x 0(x4/L4) dx) i] / [( 0L 0(x4/L4) dx)].
Since 0 is a constant it can be pulled out of each of the integrals and cancels
rcm = [( 0L (x5/L4) dx) i] / [( 0L (x4/L4) dx)].
We have multiplied through by x in the numerator. Now we can solve each integral to get
rcm = [(x6/(6 L4))|0L i] / [ (x5/(5L4))|0L].
rcm = [(L6/(6 L4)) i] / [ (L5/(5L4))].
rcm = [((L2/6 ) i] / [(L/5)].
rcm = (0.167 L2) i / (0.200L).
rcm = 0.835 L i.
4. Impulse [716871]
What is the impulse exerted by the force shown in the graph below? Assume that the force starts acting at 0 s and ends at 10 s.
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*10*a*30 Ns
*03*c*1.20 Ns
*00*e*None of the other answers is correct.
*03*b*312 Ns
*01*d*60 Ns
Solution or Explanation
The impulse is given by the equation
J = 0 s10 s F dt.
But this is just the area under the curve, which is
A = 1/2 bh = 1/2 (10 s) (60 N).
5. Center of Mass [813716]
Find the center of mass rcm of the three masses in Figure 8-46, given that m1 = 1 kg, m2 = 2 kg, and m3 = 8 kg.
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Figure 8-46.
*10*a*3.36 m j
*02*d*2.33 m i
*03*e*3.67 m i
*03*c*2.33 m j
*06*b*3.36 m i
Solution or Explanation
The center of mass in this case is given by the discrete equation
(mi ri) / M = (m1 r1 + m2 r2 + m3 r3) / (m1 + m2 + m3) = [(1 kg) (1 m) j + (2 kg) (2 m) j + (8 kg) (4 m) j] / [(1 kg) + (2 kg)+ (8 kg)]
= (3.36 m) j.
6. Rotation [544774]
A racecar goes around a circular track at 2200 m/s. The track has a radius of 250 m. What is the angular speed of the car?
*01*e*0.00 rad/s
*10*a*8.80 rad/s
*01*d*2200.00 rad/s
*05*b*550.00 krad/s
*07*c*0.04 m/s
Solution or Explanation
The angular speed is simply v/R.
7. Rotation [813711]
A wheel starts from rest with constant angular acceleration of 2.8 rad/s2. After 6 s, through what angle has the wheel turned?
*10*a*50.4 rad
*01*d*84 rad
*03*b*16.8 rad
*01*e*117.6 rad
*05*c*100.8 rad
Solution or Explanation
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The angle that the wheel has turned is given by
= 1/2
t2 = 1/2 (2.8 rad/s2) (6 s)2 = 50.4 rad.
8. Rotation [921208]
Starting from rest, a disk takes 175 revolutions to reach an angular velocity 5 . At constant angular acceleration, how many
revolutions did it take from rest to reach an angular velocity of 2 ?
*02*d*0.05 rev
*10*a*28 rev
*06*e*0.14 rev
*01*c*5 rev
*05*b*16128 rev
Solution or Explanation
The easiest way to solve this problem is by looking at it as two separate problems. In each, we have
2
- 02 = 2
( - 0).
For the part of the problem where the angular velocity goes from 0 to 2 , we have
(2 )2 - 0 = 2
( - 0).
For the part where the angular velocity goes from 0 to 5 , we have
(5 )2 - 0 = 2
[(175 rev) (2
rad / 1 rev)].
You can now divide these two equations to get
4 / 25 = ( ) / (1100 rad).
= 175.93 rad = 28.00 rev.
9. Impulse [714508]
Two cars of equal mass (1500 kg) collide head-on at an intersection and stick together. One is initially traveling at a speed of 5
m/s and the other at 10 m/s. The time it takes them to come to a full stop is 50 s. What is the magnitude of the impulse on each
car from the time just before they collide until they come to a full stop?
*05*b*1.5 104 kg m/s for both cars
*05*c*4.5 104 kg m/s for both cars
*10*a*1.5 104 kg m/s for one car and 7.5 103 kg m/s for the other
*04*e*It is nearly zero because p = 0 for the system.
*01*d*3.8 105 kg m/s for both cars
Solution or Explanation
Here we just use the fact that
J = p = m v.
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10. Momentum and Energy [714510]
Two identical cars, one coming from the south and the other from the west, collide and stick together. After the collision, they
move directly northeast. If the speed of each car just before the collision is 52 km/hr, what is their speed just after the collision?
(Note: The momentum of both cars after the collision is the sum of momenta of both cars before the collision. In other words, the
total momentum of both cars is conserved.)
*00*d*52.00 km/hr
*00*e*I need to know the mass of the cars to determine this.
*07*b*73.54 km/hr
*10*a*36.77 km/hr
*05*c*18.38 km/hr
Solution or Explanation
The initial momentum of both cars (before the collision) is
p0 = m v01 + m v02.
If we set the x-axis as east then
p0 = - m (52 km/hr) i - m (52 km/hr) j.
After the collision, we have both cars moving together and their momentum is now
pf = 2 m vf.
In a collision momentum is always conserved. Thus,
pf = p0.
Thus,
vf = 1/2 (- 52 km/hr) i - m (52 km/hr) j
and
v=
[1/4 (- 52 km/hr)2 + 1/4 (- 52 km/hr)2] = 36.77 km/hr.
Assignment Details