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Unformatted text preview: P221l82000/Problem 2A. A deuteron (hydrogen2 nucleus) of mass m moving at speed 1), along the positive a:
direction collides elastically with an alpha particle (helium4 nucleus) of mass 2m
initially at rest. The deuteron is scattered through an angle of 90° from its initial
direction of motion, so it is traveling along the positive y direction with a speed of (which will be determined later). (a) [2 points] Sketch a diagram describing the motion of these two particles before
and after the collision. Your diagram should be qualitatively but not necessarily
quantitatively Correct. g?
@ev; @ ‘M
{gag—”Q M (b) [2 points] Determine the x component of the ﬁnal velocity of the alpha particle in terms of 121 and 'Uf.
f The initial momentum was mug in the positive a: direction, so the ﬁnal momentum
must have the same value. Since only the alpha particle has a com onent of its ﬁnal velocity along as, its momentum must have an x component of mug. Since its mass is 2m, the a: component of its ﬁnal velocity must be ﬁt; (c) [2 points] Determine the y component of the ﬁnal velocity of the alpha particle in
terms ofvi and of. The 3; component of its ﬁnal momentum of the deuteron is is mvf. Since there was no y component of the initial momentum, there is no y component of the ﬁnal momentum,
so the y component of the ﬁnal momentum of the alpha particle must be equal to — muf. That makes the y component of the ﬁnal velocity of the alpha particle (whose mass is 2m) 1
— Evf' (d) [4 points] Determine ’Uf in terms of 2), assuming that the collision is elastic.
To ﬁnd of in terms of U), we use the fact that the collision is elastic. The initial kinetic energy of the deuteron was §mvg The ﬁnal kinetic energy of
the two particles is "W? 3 _
4 + 1m(vf)°. imam? + étzmxevir + — tun? = Since the collision is elastic, the initial and ﬁnal kinetic energies must be equal, so mu2 3 my? = 4 z + %m(1}f)2 and imv? :7. Im(vf)2 SO Uf = ﬁvi. lob—I (e) [2 bonus points] Find the percentage of the deuteron's initial kinetic energy that
was transferred to the alpha particle. ‘
The initial kinetic energy was Ki = £77m? The ﬁnal kinetic energy of the deuteron was
de =émv§ = %m(71§vi)2 = %mv§ = %Ki. Since the deuteron cam’es off 1/3 the initial kinetic energy, and the collision was
elastic, the alpha particle has 2/3 the initial kinetic energy. Thus 2/3 of the deuteron’s
initial kinetic energy was transferred to the alpha particle. >. ‘ .45 53
. u u H ‘6
“ﬂ C
WW” I P221/SZOOO/Problem 28. A small particle in the vicinity of a
larger object has the potential energy
function U(x) shown in the ﬁgure tow
the right. n AMI: U(x) in joules ‘_ aaaango‘ x in meters §
E
S
.E (a) (4 points)0n the diagram to the
§ left, sketch the corresponding force
component Fz(:z:) acting on the
0  5 10 particle. Include scale for the force.
x in meters L____——————————————————' (b) [2 points] Explain in what direction the particle would move when it is placed,
initially at rest, at any point between B and C. It will move to the right, towards C
(c) If the total mechanical energy E of the system is — 2.0 J, find:
(i) [2 points] The potential energy when the particle is at point C.
U(z = 4 m) = — 5 J
(ii) [2 points] The kinetic energy of the particle when it is at point C.
K=E~U= —2J—(—SJ)= +31. (iii) [2 bonus points] If the mass of the particle is 0.40 kg, what is its speed at
= 5.00 m?
Atz=5.00m, = ——5JsoK=3Jasinpart(ii). l
2 v = \/ 2(3 J)/0.40 k = V 15 m/s z 4 m/s mm2 = 3 J means that t P221/SZOOO/Problem 2C. A uniform diskshaped pulley
(Icm = §mRz) has mass M = 4.0 kg and radius R = 0.50 m. It rotates freely about an axis perpendicular to the plane of this page, as shown in the ﬁgure to the
right. A block of mass m = 2.0 kg hangs by a massless, inextensible string that is tightly wrapped around the pulley. Assume the system starts at rest at t = 0.00 s. (a) [2 points] What is the magnitude of the moment of inertia of the pulley? For a uniform disk of mass m and radius R, the moment of inertia about an axis
perpendicular to the disk, through its center of mass, is I = 15ng, so I = (0.5)(40 kg)(0.50 m)2 = 050 kg  m2 (b) [4 points] Show that a, the magnitude of the acceleration of the block, is equal
to S m/s2. Let’s call the tension in the rope T. Using Newton's second law for the block,
mg—T=ma so T=m(g—a). For the pulley, we have force T acting on its edge, so the torque due to the rope
has magnitude 7" = TR, which must equal I a. However, since the rope is inextensible,
a = a/R. Putting this all together gives: T = TR = Ia = Ia/R so T = Ia/R2. Setting out two expressions for T equal to one another, and noting that M = 2m, m(g — a) = Ia/R2 so 9 ~ a = Ia/mRz = éMRza/ng = %(M/m)a = a.
This leads to g = 2a or a = (l/2)g = 5 m/sg. (c) [2 points] Determine the angular speed of the pulley 4.0 s after the block is
released. Since the block has acceleration of magnitude 5 m/sg, the linear acceleration of a point on the edge of the pulley has the same value, which means its angular acceleration
has magnitude a = a/R = (5 n1/52)/(O.50 m) = 10 rad/52. Then after 4 seconds, the
angular speed of the pulley is w = too + at = O + (10 rad/s2) (40 s) = 40 rad/s.
(d) [2 points] Find the speed of the block after it has fallen 3.00 m. The time it takes the block to fall 1.00 m is found by recognizing that the distance
d dropped is related to the acceleration and time by d = %at2 so t = \/ 2d/a . Its speed is then 12 = at = \/ Zad = i/2(5.0 ni/s2)(3.00 m) = \/30 m/s z 5.5 m/s. , ".l A MULTIPLECHOICE QUESTIOﬁS‘
Notice that these questions are numbered 1120. These questions are worth 2 points each. .
101. A. Begin by shading in the "A" bubble for 101 on the back of your machine
graded sheet. This is important! If it is not shaded in you willweceiveno credit for any multiplechoice answer. 11. A block of mass m slides up a frictionless surface to a height h above the
ground. What is the minimum initial 3 n  d it must have had? (A) 1/2x/gh (B) x/gh/Z (D)2\/29h (E) Zx/gh 12. A small object of mass m slides slides ‘ ‘
the frictionless looptheloop track of radius R shown in the diagram to the right. What is the smallest value of y such that the object will slide without losing contact with the track? 13. The (1:, y) coordinates (in meters)
of the center of mass of the 3particle
system shown to thehright are: (A) (0, 0)
(B (2.) ( (C): (1.4,1.9) >
(D) (1.9, 2.5)
(E) (1.4, 2.5) 14. A 5.0kg object can move along
the ccaxis. It is subjected to a net force F in the + a: direction. A
graph of F; as a function of time is
shown. The change in the velocity
component 11;. of the object is: (A) 4.0 m/s
(B) 2.4 m/s
(C) 1.6 m/s 0.8 m/s (E) cannot be determined without knowing the initial velocity 15. The mass ofa particle is m. In order for its total energy to be twice its mass
energy, its linear momentum must have ma ' (A) mc/z (B) mc/ﬁ (C) mc (E) 2mc 16. Which of the following is one of the basic postulates of Einstein‘s special theory
of relativity? (A) Moving clocks run more slowly than when they are at rest.
(B) Moving rods are shorter than when they are at rest. (C) Light has both wave and particle ro e '
e aws 0 physics must be the same in all inertial reference frames. ‘
' e o t e mear momentum o a par 1 s a ways greater than m1}. 17. The work that must be done to increase the speed of an elementary particle of
0.90 c to 0.95 c is approximately (B) 1.8 me2 (C) 2.7 ch (D) 3.6 mc2 (E) 4.5 'rnc2 18. A person of mass m = 50 kg is standing at the edge of a platform which has a
radius of 2.0 m and a moment of inertia of 100 kg  m2 and is rotating at 2.0 rad/s.
If the person moves to the center, the angular speed of the platform will then be: (A) 1.0 rad/s (3)2.0 rad/s (C) 3.0 rad/s (D) 4.0 rad/s (E) 6.0 rad/s 19. A beam 2.0 m long of mass 8.0 kg is resting at its ends on two'chairs. A block of
mass 4.0 kg is placed 0.50 m from one of the ends. What is the magnitude of the
normal force exerted on the beam by the chair nearest th :10 9 (A) 30 N (B) 40 N (C) 50 N (D) 60 N 20. Suppose F = (3 N)’i~ — (6 N) l: and F = (2m)i — (3 m) lg. What is the cross
product 5‘ x F? ' (A) —(21Nm>§ (B) —(3Nm)§ ((3)0
(E) +(21Nmﬁ
0 n A
2 ‘1}, A :1 A ...
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 Spring '08
 HerreraSiklody

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