2. The wind is blowing northward at 10 km/h,
carrying the rain with it. When you stand still, the rain
appears to be falling at an angle. If you begin to run northward
at 10 km/h, why does the rain appear to be falling straight
down?
This question is a matter of reference frame. When you are
running northward at 10 km/h, you are moving along with the wind
and so it is not moving with respect to you. In this frame of
reference, the only velocity and acceleration that the raindrops
will have is that due to gravity acting vertically downward.
This will cause the raindrops to move straight down with respect
to you. But, both you and the raindrops are moving at an
additional velocity of 10 km/h northward, so the velocity of the
raindrops with respect to the reference frame of the ground is
the vector sum of their velocity due to gravity and their
velocity due to the wind. One of these (gravity) is vertically
downward while the other (wind) is horizontally northward, giving
a net velocity that is at an angle to vertical.
4. Bumper cars are an amusement park ride in
which people drive small electric vehicles around a rink and
intentionally bump them into one another. All of the cars travel
at about the same speed. Why are head-on collisions more jarring
than other types of collisions?
Lets view a head-on collision from the reference frame of one of
the drivers. Our driver views the other car as approaching at
about twice the speed of the car relative to the ground, with our
car as being still. We know that a collision with a fast moving
object is more jarring than one with a slower moving
object.
Now, what does it look like to our driver when the other car
approaches from some other direction? Suppose the other car is
approaching from one side. Now, part of the other car's velocity
is along the same direction as our car's and part of it is at a
right angle to our direction (using the idea of vector addition
for velocities). The part along our direction of travel simply
adds with our car's speed - being careful of positive and
negative signs where needed, while the part at a right-angle
becomes the other component of the net approach velocity. Our
driver sees the approaching car moving with a speed given by the
Pythagorean theorem as the 'square root of the sum of the
squares' of the net velocity components. This value is always
less than what you get from simply adding the two speeds
together, as in the head-on case. (Try this for yourself. Pick
any arbitrary pair of speed values and relative directions.) So,
this collision is less jarring than a head-on collision.
This collision may also be viewed with the concept of linear
momentum conservation. If you try this, remember that momentum
is a vector and its components will 'add' via Pythagoras'
theorem. Now, how does a head-on collision differ from one that
is off-angle in terms of momentum transfer to our car? (Hint:
The linear momentum components parallel to our car's motion and
at a right-angle to it must be conserved separately!) Try
considering some extreme cases to see how this works.
6. During rehabilitation after hand surgery,
patients are often asked to squeeze and kneed putty to strengthen
their muscles. How does the energy transfer in squeezing putty
differ from that in squeezing a rubber ball?
A rubber ball is made of pretty elastic (or springy) material.
When squeezed, it deforms in proportion to the applied force and
then recovers as the force is released. During this time it
exerts an equal reaction force on your hand.
In contrast, putty does not exert an elastic restoring force and
does not require an ever increasing force to continue deforming
it. The work you do in kneading putty can be at a constant rate
as it deforms.
12. As you look out the window of a swiftly
moving train you see a car moving in the same direction along a
nearby highway. However, the car appears to be moving backward.
Explain.
When you are riding on a train, you view objects out the window
in relation to your reference frame inside the train. As you
continue, objects that are stationary with respect to the ground
will appear to be moving backwards at about the forward speed of
your train. A car that is traveling along the road in the same
direction as your train will have a relative speed given by the
difference between its speed relative to the ground and the
train's. If the car's forward speed is less than that of the
train, it will appear to be moving backward with respect to
you.
16. If you drop a ball onto the floor from a
height of 1 m, it will rebound to a height of less than 1 m. Why
can't it bounce higher than 1 m? Why doesn't it even reach 1
m?
Before you release the ball, it has a certain amount of
gravitational potential energy with respect to the ground, and
just about zero kinetic energy. As it falls, some of the
gravitational potential energy becomes the ball's kinetic energy.
If the collision of the ball with the floor was perfectly
elastic, it would regain all of its kinetic energy as the ball
rebounds from its deformation upon collision. This kinetic
energy would now change back into gravitational potential energy.
Since the ball had kinetic energy equivalent to the potential
energy of a 1 m fall, it would regain the 1 m height before all
of the available kinetic energy is once again gravitational
potential energy. It cannot go any higher since it has no more
kinetic energy available.
But the collision will be less than perfectly elastic since we
are dealing with real materials. Some fraction of the collision
energy will be lost to thermal energy, as given through the
materials' coefficients of restitution. The remaining, rebound
energy of the ball will be its kinetic energy as it begins its
upward travel. The maximum height it now reaches will occur when
this lesser amount of kinetic energy is completely changed to
gravitational potential energy.
20. Why does it hurt less to land on a soft
foam pad than on bare concrete after completing a high
jump?
Landing on a bare concrete floor after a high jump hurts because
the concrete floor exerts a sufficient force to slow and stop
your body very quickly - where the product of this force with the
stopping time equals the momentum change of your body in going
from impact speed to zero. Equivalently, this force multiplied
by the deformation distance at the impact is about equal to your
kinetic energy on impact, as this quantity is the work done in
stopping you from moving. Since the stopping distance and time
are short, the applied force must be large.
When you land on a soft foam pad, both the stopping distance and
time increase since the pad gives way and compresses, storing
some of the energy elastically and changing some into heat. You
still have the same impact momentum and energy to use up as when
you landed on the bare floor, but the applied force is smaller.
This occurs because the force gets multiplied by a larger
quantity in each case, whether you consider the momentum change
or the work done. The smaller stopping force hurts less.
22. The surface of a clay tennis court dents
when the ball hits it and doesn't return to its original shape
when the ball rebounds. Use the definition of work to show that
the clay surface extracts energy from the ball (which is why clay
courts are "slower" than asphalt courts).
When a tennis ball hits the surface of a clay court, the ball
deforms and stores part of the impact energy elastically, like a
spring. The clay surface also dents through some distance. In
doing so, work is done on the clay surface equal to the impact
force (keeping the tennis ball from displacing the clay surface
any further) multiplied by the dent distance. The energy
corresponding to this work is no longer available to the tennis
ball since the clay surface does not rebound and push back harder
on the tennis ball. The stored deformation energy of the tennis
ball now converts back into kinetic energy as the ball rebounds.
Since the energy available to the tennis ball is smaller now, it
moves slower after its bounce.
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