PY1052 Problem Set 7 – Autumn 2004 Solutions

PY1052 Problem Set 7 – Autumn 2004
(1) In a game of pool, the cue ball strikes another ball of the same mass
that is initially at rest. After the collision, the cue ball moves at 3.50 m/s
along a line making an angle of 22◦ with its original direction of motion
and the second ball has a speed of 2.00 m/s. Find (a) the angle between
the direction of motion of the second ball and the original direction of
motion of the cue ball and (b) the original speed of the cue ball. (c) Find
the vector impulse of each ball in terms of the unit vectors î and ĵ and
show that the sum of the two impulses is zero. (d) If the balls were in
contact for 1.20 ms, what was the average force acting on each ball?
22 deg
Conservation of momentum tells us that
p~bef ore = p~af ter
m~vcue,i + 0 = m~vcue,f + m~vball,f
where m is the mass of the two balls. Looking at the conservation of momentum in
the x and y directions separately, taking the x direction to coincide with the original
direction of motion of the cue:
mvcue,i = mvcue,f cos(22) + mvball,f cos θ
vcue,i = vcue,f cos(22) + vball,f cos θ
0 = mvcue,f sin(22) − mvball,f sin θ
0 = vcue,f sin(22) − vball,f sin θ
(a) We are given vcue,f = 3.50 m/s and vball,f = 2.00 m/s. Thus, we can use the y
equation to solve for θ:
vball,f sin θ = vcue,f sin(22)
sin(22) = 3.50
sin θ = vball,f
sin θ = 0.656 −→ θ = 41◦
(b) Now use the x equation to solve for the initial speed of the cue ball:
vcue,i = vcue,f cos(22) + vball,f cos(41) = (3.50 m/s) cos(22) + (2.00 m/s) cos(41) =
4.75 m/s
(c) We have from parts (a) and (b) the initial and final velocities of the cue ball and
the other ball:
vcue,i = 4.75 m/s î
vcue,f = (3.50 m/s) cos(22) î + (3.50 m/s) sin(22) ĵ
vball,i = 0
vball,f = (2.00 m/s) cos(41) î − (2.00 m/s) sin(41) ĵ
The initial and final momenta of the balls are obtained simply by multiplying these
by the mass of the balls, which we can call m:
pcue,i = 4.75m kg m/s î
pcue,f = 3.24m kg m/s î + 1.31m kg m/s ĵ
pball,i = 0
pball,f = 1.51m kg m/s î − 1.31m kg m/s ĵ
The impulses of the two balls are:
Jcue = pcue,f − pcue,i = (3.24m − 4.75m) kg m/s î + 1.31m kg m/s ĵ
Jcue = −1.51m kg m/s î + 1.31m kg m/s ĵ
Jball = pball,f − pball,i = 1.51m kg m/s î − 1.31m kg m/s ĵ
We can immediately see that the sum of the two impulses is zero:
Jcue + Jball = (−1.51m + 1.51m) kg m/s î + (1.31m − 1.31m) kg m/s ĵ = 0
(d) The magnitude of the impulse acting on each ball is given by
Jx2 + Jy2 =
(1.51m kg m/s)2 + (1.31m kg m/s)2 = 2.00m kg m/s
Therefore, the average force acting on the ball is
J = Favg ∆t −→ Favg = J/∆t
2.00m kg m/s
Favg = 1.20×10−3 s = 1667m N
(2) A ball having a mass of 150 g strikes a wall with a speed of 5.2 m/s
and rebounds with only 50% of its initial kinetic energy. (a) What is the
speed of the ball just after it leaves the wall? (b) What is the impulse on
the ball due to the wall? (c) What is the impulse on the wall due to the
ball? (d) If the ball is in contact with the wall for 7.6 ms, what was the
average force on the ball from the wall during the collision?
(a) KEf = 21 mvf2 = 21 KEi = 12 [ 12 mvi2 ]
vf2 = 21 vi2
vf = −vi / 2 = −5.2 m/s/ 2 = −3.68 m/s
where there is a negative sign in front of vf because the direction of vf is opposite to
that of vi .
(b) The impulse is therefore
J = pf − pi = mvf − mvi = m(− √vi2 − vi ) = −m( √vi2 + vi )
J = −(0.150 kg)(3.68 m/s + 5.20 m/s) = −1.34 kg m/s
The magnitude of the impulse of the ball due to the wall is 1.34 kg m/s, and the
impulse is directed in the negative direction, which is away from the wall.
(c) Because of Newton’s Third Law, the impulse of the wall due to the ball is equal
and opposite to the impulse of the ball due to the wall – +1.34 kg m/s.
(d) If the ball is in contact with the wall for 7.6 ms, the average force on the ball is
= −1.34 kg m/s0.0076 s = 1.76 × 103 N
Favg = ∆t
(3) A railroad car moves at a constant speed of 3.20 m/s under a grain
elevator. Grain drops into the car at a rate of 540 kg/min. (a) What is the
magnitude of the force that is needed to keep the car moving at constant
speed if friction is negligible? (b) What would be the initial acceleration
of the car when the grain began to drop in if this force were not applied
(a) Newton’s second law taking into account the possibility of varying mass is
F = dp
= m dv
+ v dm
is being added to the car), we see that dv
will be negative
Since dm
in the absence of other forces (F = 0) and the car will slow down. Since the speed is
constant, dv
= 0, and the required force is
F = v dm
= 540 kg/min = 9.00 kg/s:
Since the rate of change of the mass is dm
F = v dt = (3.20 m/s)(9.00 kg/s) = 28.8 N
(b) If this force is not applied,
+ m dv
F = 0 = v dm
v dm
m dt
3.20 m/s
= − m (9.0 kg/s) = 28.8
where m is the mass of the railroad car at the initial time.
(4) The speed with which cheetahs can run has been measured by observers driving along with the cheetahs in Jeeps. Imagine trying to measure a cheetah’s speed by keeping your Jeep even with the cheetah while
also glancing from time to time at your speedometer, which is registering
114 km/hr. You keep the Jeep at a constant 8.0 m from the cheetah, but
the cheetah keeps veering away from the Jeep and ends up moving in a
circular arc with a radius of 92 m. Since you maintain the Jeep at a distance of 8.0 m from the cheetah, the radius of the circular arc along which
you drive is 100 m. (a) What is the angular speed of you and the cheetah
around the circular arcs? (b) What is the linear speed of the cheetah along
its path?
(a) The angular speed of the jeep (and cheetah) is given by
v = rω −→ ω = vr
The jeep’s speed is
1 hr
× 1000
× 3600
= 31.7 m/s
114 km
1 km
ω = vr = 31.7
= 0.317 rad/s
100 m
(b) The linear speed of the cheetah is then given by
v = rω = (92 m)(0.317 rad/s) = 29.1 m/s = 105 km/hr.
When the running speeds of cheetah’s were first estimated in precisely this way, the
researchers involved forgot to take account of the fact that the circle in which the
cheetah was running was smaller than the one along which the jeep was driving, so
that these first cheetah speeds were overestimated (the claimed speeds were 114 km/hr
rather than 105 km/hr)!
(5) What is the moment of inertia of a metre stick with a mass of 0.56 kg
about an axis perpendicular to the stick going through its center (treat
the metre stick like a thin rod)? Using the parallel-axis theorem, calculate
the moment of inertia of the metre stick about an axis perpendicular to
the stick and located at the 20 cm mark.
The moment of inertia of the metre stick for rotation about an axis going through its
centre is given by (“thin rod about axis through centre perpendicular to length”):
M L2
Imid = 12
where M is the mass of the metre stick and L is its length (1 m):
Imid = 12
(0.56 kg)(1.0 m)2 = 0.047 kg m/s
The parallel axis theorem says:
I = Icom + M h2
where Icom is the moment of inertia for rotation about an axis that is parallel to the
one in which we are interested and passes through the centre of mass, and h is the
distance between the COM axis and our axis. In our case, Icom is just equal to Imid ,
since the axis for Imid is parallel to the one we are considering and passes through
the metre stick’s centre of mass. The distance h is just equal to the distance between
the middle of the metre stick (the 0.50 m mark) and our axis (the 0.20 m mark), or
0.30 m. Therefore:
I = Icom + M h2 = 0.047 kg m/s + (0.56 kg)(0.30 m)2 = 0.097 kg m/s
(6) A delivery truck operates by using the energy stored in a rotating
flywheel, which is a solid cylindre with a mass of 500 kg and a radius of
1.0 m. The truck is charged by using an electric motor to get the flywheel
up to its top speed of 200π rad/s. (a) What is the kinetic energy of the
flywheel after charging? (b) If the truck operates with an average power
requirement of 8.0 kW, for how many minutes can it operate between
chargings (1 W = 1 Watt = 1 J/s)?
(a) The kinetic energy of rotation of the flywheel is
KE = 21 Iω 2
We know that ω = 200π rad/s, and must determine the moment of inertia I of the
flywheel to find its kinetic energy. Your table indicates that the moment of inertia of
a solid cylinder or disk about its central axis is
I = 21 M R2 = 12 (500 kg)(1.0 m)2 = 250 kg m/s
Therefore, the kinetic energy is
KE = 21 (250 kg m/s)(200π rad/s)2 = 4.93 × 107 J
(b) The average power is given by
Pavg = Work
Here, the work done between when the wheel is fully charged (rotating at its maximum
angular speed) and uncharged (when it no longer rotates) is equal to the change in
the wheel’s kinetic energy. The time is what we are looking for. Therefore
4.93×107 J
= 8.0×10
time = PKE
3 W = 6.16 × 10 s = 103 min
(7) A thin spherical shell has a radius of 1.90 m. An applied torque
of 960 N m gives the shell an angular acceleration of 6.20 rad/s 2 about an
axis passing through the centre of the shell. (a) What is the moment of
inertia of the shell? (b) What is the mass of the shell?
(a) You can find from a table of moments of inertia that the moment of inertia of a
thin spherical shell about any diameter is
I = 23 M R2
but in this situation, we can’t use this directly, because we don’t know the mass of
the shell. Instead, use Newton’s second law for rotation:
τ = Iα −→ I = ατ
960 N m
I = 6.20
= 155 kg m/s
(b) Now, using the expression for I for a spherical shell and the answer to (a), we
can find the shell’s mass:
I = 32 M R2
3(155 kg m/s
2(1.90 m)2
= 64.4 kg
(1b) A 6100 kg rocket is set for vertical firing from the ground. If the
exhaust speed is 1200 m/s, how much gas must be ejected each second if
the thrust is to (a) equal the magnitude of the gravitation force on the
rocket and (b) give the rocket an initial upward acceleration of 21 m/s 2 ?
(a) The thrust is given by Rvrel , where R is the rate at which the fuel is consumed
and vrel is the velocity of the rocket relative to the ejected gases. In order for the
thrust to be equal to the gravitational force on the rocket,
Rvrel = M g −→ R = vMrelg
kg)(9.8 m/s
1200 m/s
= 50 kg/s
(b) In order for the rocket to have an upward acceleration of 21 m/s 2 ,
Thrust − M g = M a −→ Thrust = M (g + a)
Rvrel = M (g + a)
M (g+a)
kg)((9.8+21) m/s
1200 m/s
= 156 kg/s
(2b) A pulsar is a rapidly rotating neutron star that emits a radio beam
in much the same way that a lighthouse emits a light beam. We receive
a radio pulse once each time the star rotates, and we can measure the
period T of the rotation by measuring the time between the pulses. The
pulsar in the Crab Nebula has a period of rotation of T = 0.033 s, and this
period is increasing at a rate of 1.2 × 10−5 s/year. (a) What is the pulsar’s
angular acceleration? (b) The pulsar was born in a supernova explosion
seen in the year 1054. What was the initial period T of the pulsar if its
angular acceleration has been constant since the time it was born?
The relationship between the period and angular velocity is
T = 2π
= 2πω −1
Taking the time derivative of both sides:
= −2πω −2 dω
= −2πω −2 α
= −2πT −2 dT
α = − ω2π dT
We are given that T = 0.033 s and dT /dt = 1.2 × 10−5 s/year:
1 year
1 day
= 1.2 × 10−5 year
× 365.25
× 24
× 3600
= 3.8 × 10−13 s/s
1 hr
α = −2π(0.033 s)−2 (3.8 × 10−13 s/s) = −2.2 × 10−9 rad/s2
The negative sign indicates that the ω is decreasing, which makes sense because the
period is increasing.
(b) If the angular acceleration is constant, dT /dt is constant. There are 949 years
between 1054 and 2003, so
∆t −→ T1054 = T2003 − dT
T2003 = T1054 + dT
T1054 = 0.033 s − (1.2 × 10−5 s/year)(949 years) = 0.033 s − 0.011 s = 0.022 s