Now that we have looked at motion problems (in Chapter 2) we will turn our attention
to force and momentum calculations. Some of these can give us a rather sobering
explanation of why it can be dangerous to drive a car recklessly. Let's consider such
a collision problem.
1. If a car strikes a tree when it is traveling at a velocity of 25 m/s and
comes to a complete stop in 0.20 seconds, how much force must be exerted on a
60-kg person by his seat belt to keep him from striking the windshield?
First we need to know the acceleration (actually deceleration) of the car.
a =
v / t = ( vF - vo ) / t = (0 m/s - 25 m/s) / 0.20 s
a = -125 m/s2 (the "-" means the car is slowing down or decelerating)
Since the mass of the person is given,
m = 60 kg, we can use Newton's second
law to find the force.
F = m a = 60 kg (-125 m/s2) = -7500 N
Can you imagine a 7500 N weight (about 3400 lb) sitting on your chest, even
for a few tenths of a second? Fortunately, most situations allow time for
the brakes to be applied gradually so that the change in velocity is not so sudden.
Most objects that you hit will also move somewhat, thus absorbing some of the impact.
It is still evident, however, that driving carefully is a very good idea.
Now let's try a problem involving momentum.
2. A boy standing on the flat roof of a building strikes a 0.65-kg ball with
a bat. The bat exerts an average horizontal force of 60 newtons on the ball for
0.48 seconds. What will be the magnitude of the velocity of the ball at the time
it leaves the bat?
Here we must first use Newton's Second Law of Motion to solve for the acceleration.
F = m a,
ora = F / m = 60 N /0.65 kg = 92 m/s2
Now we can use the motion equations from Chapter 2 to solve for the final velocity.
vF = vo + a t = 0 + 92 m/s2 (0.48 s) = 44 m/s
From this information, we can find some interesting things about the motion of the ball.
3. If the ball is in flight for 3.2 seconds, how tall is the building and how far from
the building does the ball strike the ground?
The ball travels in the horizontal direction at a constant velocity
(vH) after the impact with the bat described above.
If we neglect air friction, this constant horizontal velocity is, therefore,
equal to vF as calculated above. The acceleration due to
gravity only acts on the vertical motion. The ball will, therefore, strike the
ground at a distance:
dH = vH t = 44 m/s (3.2 s) = 140.8 m = 140 m or 1.4 x
102 m (to 2 sig. fig.)
The height of the building will be the distance that the ball would fall under the
influence of gravity in 3.2 seconds starting from rest. (See Figure 2.15 in the
previous chapter of the textbook.)
dV = 1/2 at2 = 1/2
(9.8 m/s2)(3.2 s)2 = 50 m downward
The distance dropped will be the same as the height of the building, so the building is 50 m tall.
4. With how much average force does the ball push against the bat?
This requires the application of Newton's third law. Since we already know
how much force the bat exerts on the ball, we have:
Fball on bat = -Fbat on ball = -60 N
The forces have the same magnitudes but are applied in
opposite directions.
Remember that force is a vector quantity and that the direction is just as important
as the magnitude.
There has been a great amount of discussion in the last few years about
establishing a manned space colony on Mars. When this occurs, the living
conditions will be quite different on Mars than they are on Earth. One of
the biggest differences will be in the pull of gravity.
5. Calculate how much a 60-kg person would weigh if he lived on the surface
of Mars instead of on the surface of Earth.
Both the mass and radius of Mars are different than those of Earth, so we need
to determine how much force would be exerted on the human being using Newton's
universal law of gravitation. The values for the mass and radius of Mars can be
found in Chapter 15 (The Solar System) so the calculation can be set up as:
weight = Fgrav = Ggrav Mman
MMars / RMars2
Remember
G is the universal gravitational constant and is
not the same
thing as g, the acceleration due to gravity near Earth's surface. As a matter
of fact, we will be calculating a value for "
g" on Mars in a moment, and
we will see that
g is quite different on Mars than it is on Earth. The value
for
G can be found in Table 3.1 in the textbook.
weight = (6.67 x 10-11 N m2/kg2)
(60 kg)(6.6 x 1023 kg) / (3.4 x 106 m)2
weight = 228.48 N = 230 N (downward)
The person would weigh only 230 newtons on Mars, whereas he would weigh
590 newtons on Earth; [
Mman g = 60 kg x 9.8 m/s
2].
Why couldn't you use some value for "
g" on Mars like we did for Earth?
We could, but this is not a number most science students know. As soon as we
establish a colony on Mars, we will probably all learn to use it like the value of
g on Earth is used today, but for now we will resort to calculating it.
gMars = G MMars / RMars2
gMars = (6.67 x 10-11 N m2/kg2)
(6.6 x 1023 kg) / (3.4 x 106 m)2 = 3.8 m/s2
Now the weight of any object on the surface of Mars can be easily calculated.
Our new colony member with a mass of 60 kg would find his weight to be:
weight = Fgrav = Mman gMars =
(60 kg)(3.8 m/s2) = 230 N (just as we found above.)
Notice that as predicted, the acceleration due to gravity on Mars
(3.8 m/s
2) is quite different from the acceleration due
to gravity near Earth's surface (9.8 m/s
2).
6. Find the new weight of a car that weighs 9200 N (about 3200 pounds)
on Earth, if that car were transported to Mars.
The weight will change, but the mass will remain the same, so first we must
find the mass of the car. On Earth we can find this using gEarth
= 9.80 m/s2.
m = weight / gEarth = 9200 N / 9.80 m/s2 = 940 kg
Now using the value for
gMars that we have calculated for objects
near the surface of Mars:
weightMars = m gMars = 940 kg (3.8 m/s2) = 3600 N
7. If the car is traveling with a velocity of 15 m/s, what would be the
momentum of the car on Mars and on Earth?
Momentum is defined as mass times velocity, so:
p = m v = 940 kg (15 m/s) = 14000 kg m/s = 1.4 x 104 kg m/s
Because the velocity is a given quantity in the problem, it will be the same in
both locations and, because the mass of an object is the same throughout the universe,
the
momentum will be the same for the car whether it is on Earth or on Mars.
It is interesting that some quantities remain the same while others change
as we travel from one location in the universe to another. Careful consideration
must be given to this situation when discussing such topics in physical science.
8. What will be the new velocity of a 4.5-kg brick that is being swung
at a velocity of 2.6 m/s at the end of a 3.0 m rope, if the rope is suddenly
shortened to a length of 1.5 m?
Conservation of angular momentum is the key to this calculation.
m v1 r1 = m v2 r2
Solving for
v2 we find:
v2 = m v1 r1 / m r2
Because
m is the same, it cancels out and the result will not depend
on the value of the mass at all.
v2 = v1 r1 / r2
= 2.6 m/s (3.0 m) / 1.5 m = 5.2 m/s
Angular momentum is one of the special quantities in physical science that
is conserved from one instant in time to the next. This is why we can deal
with values like v
1 and v
2—the velocity at
instant one and the velocity at instant two. Notice that the velocity itself
is not conserved; it can obviously change. It is the angular momentum that is
conserved, and this fact can be used in our calculations. In the next chapter,
we will look at
energy as a conserved quantity and see how
work
is related to energy conservation.
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