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Chapter 4 Solutions
Exercises
5.
9.
14.
18.
20.
23.
25.
A boxing glove has padding which increases the time and distance over which the
momentum of the fist is changed. For the same given impulse (change in
momentum) a longer time implies a smaller force. Similarly, for a given energy,
more distance implies less force.
The bus and the bug interact with each other. By action reaction and/or
conservation of momentum, the force on the bug equals the force on the bus and
the change in momentum of the bug equals the change in momentum of the bus.
However, this change in momentum results in an extreme change in velocity for
the bug and a negligible change in velocity for the bus.
The collision would be more damaging if the cars bounced off of each other. This
is a greater change in velocity and momentum and therefore a greater danger to
the passengers.
A long barreled gun or cannon exerts a force on the cannonball for longer than a
short barreled one. Therefore, a longer barrel delivers a higher muzzle speed with
the exact same amount of gunpowder.
An object can have potential energy without having any momentum. For example
a heavy boulder on the top of a cliff has a lot of potential energy but no
momentum. However, any object with kinetic energy (K.E. = ½ mv2) will have
some momentum (p=mv). Any object with momentum will have some kinetic
energy. However, depending on the size of the object the potential energy can be
relatively big or small compared to the momentum. K.E. = p2/(2m) so a large
mass has lots of momentum and little energy. A small mass has lots of energy but
only a little momentum.
K.E. = ½ mv2. Since the ball has 5 times as much velocity it has 52 = 25 times as
much kinetic energy.
The total energy of the swinging pendulum is constant. It will have the most
potential energy when it is at it’s highest point. At these heights the kinetic
energy is zero because the pendulum comes to a stop. The potential energy is
least at the bottom. At these points the kinetic energy is a maximum. This is at
the bottom of the arc when the pendulum is swinging the fastest. IF the kinetic
energy is half of its maximum value, then the potential energy is half way
between its maximum and minimum values. (If we choose to set potential energy
to 0 at the bottom than the potential energy is half of its value too. However,
potential energy has no absolute zero point.)
Problems
1.
p = mv = 1000 kg * 20 m/s = 20,000 kg m/s
∆p=0-20,000 kg m/s = -20,000 kg m/s
F∆t = ∆p
so F=∆p/∆t = 20,000 kg m/s / 10 s = 2,000 kg m/s2 = 2000 N
3.
pi = pf
mbvb + ms0 = mbvf + msvf = (mb + ms) vf
vf = mb/(mb + ms) vb = 5 kg/( 5kg + 1 kg) * 1 m/s = 5/6 m/s
pi = pf
mbvb + msvs = mbvf + msvf = (mb + ms) vf
vf = (mbvb + msvs)/(mb + ms) = (5 kg * 1 m/s – 1 kg * 4 m/s)/(5 kg + 1 kg) = 1 m/s
5.
Consider my answer to question 20. If we combine K.E. = ½ mv2 and p=mv then
we get K.E. = p2/2m. Therefore, if we have twice the mass with the same
momentum we will have half the kinetic energy. (This can also be considered
with twice the mass and half the velocity.) This energy has gone into work done
coupling the cars, damage to the cars, and random thermal energy in the cars and
surrounding air.
6.
The energy in equals the energy out. Therefore, the work = Fd is the same.
Findin = Foutdout
Fout=(din/dout) Fin = (10 cm / 1 cm) * 10 N = 100 N
9.
e=Wout / Win = 100 W / 1000 W = .1 = 100%