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# Test 3 Additional Review-sols.pdf

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University of Texas at Austin

Physics

PHY 303K

Rodenborn

Spring

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Version 001 – Test 3 Additional Review – ﬂorin – (57850) 1
This print-out should have 13 questions. From the setup, the ﬁnal velocity of the col-
Multiple-choice questions may continue on lision is the initial velocity of the subsequent
the next column or page – ﬁnd all choices swing. During the swinging process the total
before answering. energy is conserved
Ballistic Pendulum 02 Ei= E f
1
001 (part 1 of 2) 10.0 points (m 1 m )v2=( m + m 1gh 2
2 f
Auntrsainm
experiment using an apparatus similar to that ▯
shown in the ﬁgure. vf= 2gh
Initially the bullet is ﬁred at the block while ▯
= 2(9 .8m /s )(0 .047 m)
the block is at rest (at its lowest swing point).
After the bullet hits the block, the block rises
to its highest position, see dashed block in the = 0.959792 m/s .
ﬁgure, and continues swinging back and forth.
The following data is obtained:
the maximum height the pendulum rises is 002 (part 2 of 2) 10.0 points
Find v1,theinitialspeedof m 1
4.7cm,
the mass of the bullet is 79 g, and
the mass of the pendulum bob is 945 kg. Correct answer: 11482 m/s.
Explanation:
The linear momentum is conserved in a
collision
47
▯ pi= p f
m v =( m + m )v
1 1 1 2 f
m 1 m 2
v1= m vf
1
(0.079 kg) + (945 kg)
= (0.079 kg)
4.7cm ▯ (0.959792 m/s)
v 945 kg v = 11482 m/s .
79 g i f
Find the ﬁnal velocity of the system (1 +
m 2immediatelyafterthecollisionandbefore CanoeCenterofMass
the pendulum starts to swing upward. The 003 (part 1 of 2) 10.0 points
2 Am nwhsmsi5kgndawmn
acceleration of gravity is 9.8m /s .
whose mass is 60 kg sit at opposite ends of a
Correct answer: 0.959792 m/s. canoe 4 m long, whose mass is 25 kg.
Assume the man is seated at x =0a d
Explanation:
the boat extends along the positive x axis
with the woman at the other end. Where is
the center of mass of the system consisting of
▯
Let : ▯ =46 .6332 , man, woman, and canoe?
m 179g=0 .079 kg ,
m 2945kg , and Correct answer: 1.70588 m.
h =4 .7cm=0 .047 m. Explanation: Version 001 – Test 3 Additional Review – ﬂorin – (57850) 2
With the man (m )1ttheorigiehe
SkaterRealAndPP
m 1 1 m x 2 2 x 3 3 005 10.0 points
x CM = Aserhstayrmaw ll
m 1 m + 2 3 She pushes on the wall with a force whose
0+(60kg)(4m)+(25kg)(2m)
= magnitude is F,ostelsnhr
85 kg + 60 kg + 25 kg with a force F (in the direction of her motion).
=1 .70588 m. As she moves away from the wall, her center
of mass moves a dista htedis.C
004 (part 2 of 2) 10.0 points following statements regarding energy.
Suppose that the man moves quickly to the
center of the canoe and sits down there, while I▯ K trans+▯ E internal Fd
the woman moves quickly 1/3 the length of II. ▯K trans▯ E internal ▯Fd
the canoe towards the man. How far does III. ▯K trans▯ E internal0
the canoe move in the water? Assume force IV. ▯K trans Fd
of friction between water and the canoe is V. ▯K trans= ▯Fd
negligible.
What is the correct form of the energy prin-
Correct answer: 0.529412 m. ciple for the skater as a real system and as a
point particle (PP) system?
Explanation:
Since friction is negligible, the net horizon-
tal force on the system (the man, woman and
canoe) is zero. Therefore, its center of mass 1. Real: IV, PP: III
momentum (and velocity) remain constant
(which is zero in this case). So the center of 2. Real: III,PP: IV correct
mass of the system remains stationary.
3. Real: II, PP: V
For this problem, let us take the origin to
be at the left end of the canoe (note that it
is not a stationary point). The new center of 4. Real: IV, PP: IV
mass is at 5. Real: V, PP: I
xCM = m 1 1 m x2 2m x 3 3 6. Real: I, PP: V
m 1 m +2m 3
(85 kg)(2 m) + (60 kg)(2.66667 m)
= 7. Real: I, PP: IV
85 kg + 60 kg + 25 kg
(25 kg)(2 m) 8. Real: II, PP: IV
+
85 kg + 60 kg + 25 kg
=2 .23529 m. 9. Real: III,PP: V
10. Real: V, PP: IV
At ﬁrst, the center of mass was .70588 m
from the left end of the canoe. Now the center
of mass is 2.23529 m from the left end of the
Explanation:
canoe. This means the boat moved As a real system, the wall does no work
on the skater since the contact point of the
2.23529 m ▯ 1.70588 m = 0.529412 m
skater’s hand and the wall does not move.
to the left as the man walked 0.529412 m to Also, the wall does not transfer energy to the
the right and the woman walked 1.33333 m to skater, i.e. by cooling down. The right hand
the left. side of the energy principle must therefore Version 001 – Test 3 Additional Review – ﬂorin – (57850) 3
▯ ▯
be zero for the real systemFurthermore, 1+ k 2
the left hand side must contain the skater’s 4. K = mv .
▯ 2 ▯
change in internal energy for a real system. k 2
III is correct. 5. K = 1+ mv .
As a point particle system, the skater’s cen- 2
6. K =[1+2 k]mv .
ter of mass moves a distance d while a force
F is applied. Thus, the work done is Fd.
Translational kinetic energy is the only type7. K =[1+ k]mv .
of energy a point particle system can have. IV ▯ ▯
k
is correct. 8. K = 1+ mv.
2
Wheel and a Weight 01
9. K =[1+2 k]mv.
006 (part 1 of 2) 10.0 points ▯ ▯
Consider a circular wheel with a mass 2 m, 10. K = 1+2 k mv . correct
and a radius R (as seen in the ﬁgure below), 2
mounted on a horizontal frictionless axle paExplanation:
ing through its center (not shown in ﬁgure). Basic Concepts: Rotational kinetic en-
The moment of inertia about the center of thergy is
wheel is I =2 kmR 2, where 0.5

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