Physics H7A, Fall 2011
Homework 7 Solutions
1. (Three pool balls) A pool ball with initial speed v is aimed right between two other
pool balls, as shown in the figure below. If the two right balls leave the (elastic) collision
with equal speeds, find the final velocities of all three balls.
At the instant the first ball collides with the other two, the three balls form an equilateral
triangle. Since the force that ball number 1 exerts on the other two must be normal to the
surface of contact, we know that ball 2 will come off with velocity vector v2 pointing 30◦
above the horizontal, ball 3 will have v3 pointing 30◦ below the horizontal, and ball 1 will
rebound (or carry forward, we don’t know which yet) in a purely horizontal direction with
velocity v1 = v1 x̂. By symmetry v3 = v2 (equal speed).
Putting these arguments into equations for energy and momentum conservation, we have
√
mv = mv1 + 2mv2 cos 30◦ = mv1 + 3mv2
(conservation of momentum in x direction),
1
1
mv 2 = mv12 + mv22
(conservation of energy).
2
2
Solving for v1 and v2 , after some simple algebra we find
√
1
2 3
v1 = − v, v2 =
v.
5
5
So the final velocity vectors are
1
v1 = − vx̂,
5
√
2 3
v2 =
v
5
√
!
3
1
x̂ + ŷ ,
2
2
1
√
2 3
v3 =
v
5
!
√
3
1
x̂ − ŷ .
2
2
2. *(Maximum scattering angle) A ball of mass m1 with initial speed v0 collides
elastically with a second ball of mass m2 that is initially at rest. The collision isn’t
head on, so after the collision the masses come off at some angles, as shown below. Let
their speeds be v1 and v2 , respectively, and let θ be the opening angle between the two
final velocity vectors.
(a) Show that for m1 ≥ m2 , the maximum opening angle is given by sin θmax =
m2 /m1 .
(b) Verify (as was discussed in lecture) that for equal mass balls, the opening angle
is always 90◦ .
Hint: First transform to the center-of-mass frame, which moves to the right with speed
1
v0 . Analyze the collision in this frame, then transform back to the lab
Vcm = m1m+m
2
frame.
2
(a) This problem is worked out in K&K as Example 4.19 (starting on page 193). I refer the
interested reader there.
(b) This problem was worked out in lecture using a geometric argument.
algebraic one.
Here is an
By conservation of momentum,
mv0 = mv1 + mv2
=⇒
v 0 = v1 + v2 .
Squaring the previous equation gives
v02 = v0 · v0 = (v1 + v2 ) · (v1 + v2 ) = v12 + v22 + 2v1 · v2 .
On the other hand, conservation of energy implies
1
1
1
mv 2 = mv12 + mv22
2 0
2
2
=⇒
v02 = v12 + v22 .
Comparing these equations we deduce v1 · v2 = 0, i.e. the final velocities are perpendicular.
3. (K&K Problem 3.4) An instrument-carrying projectile accidentally explodes at the
top of its trajectory. The horizontal distance between the launch point and the point
of explosion is L. The projectile breaks into two pieces which fly apart horizontally.
The larger piece has three times the mass of the smaller piece. To the surprise of the
scientist in charge, the smaller piece returns to earth at the launching station. How far
away does the larger piece land? Neglect air resistance and effects due to the earth’s
curvature.
Let t be the time between the launch of the projectile and the instant when it explodes. The
horizontal component of the projectile’s velocity must then be v = L/t. Analyze the motion
of the projectile in a frame that moves horizontally along the ground with this speed. In
this frame, the projectile shoots straight up in the air and then explodes at its highest point,
sending one piece flying forwards and the other backwards, but such that both pieces land
on the ground at the same time (which is 2t, i.e. the pieces take the same amount of time to
come down as they did to go up).
The smaller piece (of mass m/4) lands at horizontal position x0 = −2L, while the larger
(mass 3m/4) lands at horizontal position x1 . Since no external forces act on the system in the
horizontal direction, the horizontal component of the center of mass must move at constant
velocity, and hence for our choice of frame the center of mass must remain stationary at
xCM = 0. At the time when the pieces land this implies
xCM =
3m
m
x0 +
x1 = 0
4
4
3
2
=⇒ x1 = L.
3
The total distance traveled by the larger piece in the lab frame is then given by the separation
between the two pieces at the time that they land,
2
8
d = x1 − x0 = L + 2L = L.
3
3
4. (K&K Problem 3.14) N men, each with mass m, stand on a railway flatcar of mass
M . They jump off one end of the flatcar with velocity u relative to the car. The car
rolls in the opposite direction without friction.
(a) What is the final velocity of the flatcar if all the men jump at the same time?
(b) What is the final velocity of the flatcar if they jump off one at a time? (The
answer can be left in the form of a sum of terms.)
(c) Does case (a) or case (b) yield the largest final velocity of the flat car? Can you
give a simple physical explanation for your answer?
4
(a) Let v be the final velocity of the flatcar after the men jump off. Since we are told the
men jump off with relative velocity u, their final velocity relative to the ground will be
v − u. Expressing conservation of momentum in the horizontal direction (everything
starts at rest, and there are no external forces acting in this direction) we have
M v + N m(v − u) = 0
=⇒
v=
Nm
u.
M + Nm
(b) Let vk be the velocity of the flatcar after k men have jumped off (so v0 = 0, and vN is
the final flatcar velocity). After the kth man has jumped, the flatcar has velocity vk ,
and there are N − k men still on board, so the total momentum is
Pk = [M + (N − k)m]vk .
After the next man jumps, the flatcar (plus the N − k − 1 men on board) will have
velocity vk+1 , while the man who jumped will have velocity vk+1 − u (i.e. velocity u
relative to the flatcar), giving total momentum
Pk+1 = [M + (N − k − 1)m]vk+1 + m(vk+1 − u) = [M + (N − k)m]vk+1 − mu.
Since momentum in the horizontal direction is conserved, Pk = Pk+1 , which gives us the
relation
m
u.
vk+1 − vk =
M + (N − k)m
Summing the above expression from k = 0 to k = N − 1 gives
vN =
N
−1
X
N
−1
X
k=0
k=0
(vk+1 − vk ) =
N
X
m
m
u=
u,
M + (N − k)m
M + jm
j=1
where in the last equality we changed summation indices to j = N − k.
m
(c) Look carefully again at summand M +jm
u for part (b). Note that each of these terms is
m
greater than or equal to M +N m u, since j ≤ N . Thus
N
X
j=1
N
X
m
m
Nm
u≥
u=
u.
M + jm
M + Nm
M + Nm
j=1
Thus the final velocity in case (b) is greater than in case (a) for N ≥ 2.
5. (K&K Problem 3.19) A bowl full of water is sitting out in a pouring rainstorm.
Its surface area is 500 cm2 . The rain is coming straight down at 5 m/s at a rate of
10−3 g/cm2 ·s. If the excess water drips out of the bowl with negligible velocity, find
the force on the bowl due to the falling rain. What is the force if the bowl is moving
uniformly upward at 2 m/s?
5
A = 500 cm2 = 5 × 10−2 m2
v = 5 m/s
−3
g/cm2 ·s = 10−2 kg/m2 ·s
R
ρrain water =
= 2 × 10−3 kg/m3
v
R = 10
Stationary bowl:
dm/dt = ρAv = RA
F = dp/dt = (dm/dt)v = RAv = 2.5 × 10−3 N
Bowl moving upwards at speed u = 2 m/s:
dm/dt = ρA(v + u) = RA(v + u)/v
F = dp/dt = (dm/dt)(v + u) = RA(v + u)2 /v = 4.9 × 10−3 N
6. (Center of mass) Compute the center of mass rcm = (xcm , ycm ) for each of the
following objects. In each case assume the mass is distributed uniformly, and take the
origin of your coordinate system as indicated.
(a) An ‘L’ shaped plate with side lengths a.
(b) A uniform bar originally of length L, bent in the shape of an asymmetric ‘Z’.
(c) A circle of radius R with a cutout in the shape of a circular hole of radius r, a
distance d from the origin.
6
(a)
5
rCM = (x̂ + ŷ)
6
(b)
rCM =
13
5
Lx̂ + Lŷ
8
8
(c)
rCM = −
r2
d
R2 − r 2
7. (Moment of inertia) Compute the moment of inertia about the origin O for each of
the objects in the previous problem. (K&K discuss moments of inertia starting on pg.
249. Apply the parallel axis theorem.)
7
(a)
1 m 2
1
(a + a2 ) = ma2
12 3
18
m
2
2
I1 = Isquare,CM + [(a/2) + (a/2)2 ] = ma2
3
9
m
8
I2 = Isquare,CM + [(3a/2)2 + (a/2)2 ] = ma2
3
9
8
m
2
2
I3 = Isquare,CM + [(a/2) + (3a/2) ] = ma2
3
9
2
Itotal = I1 + I2 + I3 = 2ma
Isquare,CM =
(b)
M 2
1 M
M 2
7
a1 =
(2L)2 +
(L + L2 ) = M L2
2
12 2
2
6
M 2
1 M
M
13
I2 = I2,CM +
a2 =
L2 +
[(2L)2 + (L/2)2 ] = M L2
4
12 4
4
12
M 2
1 M
M
19
I2 = I3,CM +
a =
L2 +
(5L/2)2 = M L2
4 3 12 4
4
12
23
Itotal = I1 + I2 + I3 = M L2
6
I1 = I1,CM +
(c)
1
Isolid = Msolid R2
2
1
Icutout = Mcutout r2 + Mcutout d2
2
r2
Mcutout = − 2 Msolid
R
r2
M = Msolid + Mcutout = 1 − 2 Msolid
R
M
I = Isolid + Icutout =
(R4 − r4 − 2r2 d2 )
2
2(R − r2 )
8