(and Collisions explosions)
The conservation of energy is a very powerful law. It is easy to use, mainly because it eliminates time and directions. But because it eliminates time and directions, it does have its limitations. In collisions, the directions of the pieces (both before and after) are important. To attack this problem, we will go back to Newton's Second Law.
Newton's Second Law
We have worked with Newton's Second Law in the following form:
x x y y
F = ma and F = ma , where acceleration is defined as:
x x y y
a = v / t and a = v / t . If mass doesn't change with motion, which
Momentum
x x y y
F = (mv )/ t and F = (mv )/ t . We can re-write this as:
x x y
(F )* t = (mv ) and (F )* t =
y
(mv )
Impulse
x x y
(F )* t = (mv ) and (F )* t =
y
(mv ) The quantity on the left side of the above equations also has a name:
x x y x
impulse = (F )* t and impulse = (F )* t .
Collisions
However, since Newton's Second Law relates the impulse to the change in momentum, we can determine the impulse by determining the change in momentum. Further, by determining the time during which the collision took place (t), we can obtain information about the (average) forces of collision!
Collisions
There is also another use of Newton's Second Law for collisions when we use it in combination with Newton's Third Law
by 2on 1 by 1on 2
(F
x on 1
=-F
exton 1
)
by 2on 1 1 x1
F
x on 2
=F
exton 2
+F
by 1on 2
= (m v )/ t
2 x2
F
=F
+F
= (m v )/ t.
Conservation of Momentum
xext on 1 xext on 2 x1 x2
F +F = (p + p ) / t . If the external forces are small, or if the time of the collision, t, is small, then we have: (px1 + px2) = 0. This can be re-written as: (px1 + px2)i = (px1 + px2)f . This is called Conservation of Momentum. This is a vector law, so a similar equation
1-D examples
In one dimensional collision cases, we can apply two laws: Conservation of Momentum and Conservation of Energy
(here we assume there are no PE's that change):
1 1i2 2 2i2
(1/2)m v + (1/2)m v
1 1f2
=
2 2f2 lost
(1/2)m v
1 1i 2 2i
+ (1/2)m v
1 1f
+E
2 2f
mv + mv
= mv + mv
1-D Collisions
We can divide the collisions into three cases: 1. The two objects can bounce off of each other without any Elost. This is called an
lost
elastic collision. This means E = 0. 2. The two objects can stick to each other. This means v1f = v2f . 3. The objects can be deformed but still not stick to one another. No special information is available in these cases.
1-D Collisions
In the first two cases, we can predict the final motion (solve for v1f , v2f and Elost) if we know the initial motion (m1, m2, v1i, v2i). In the third case, we would have to know something besides the initial motion to solve for the final. When you think about it, this does make sense, since in the third case the results of the collision should depend on the material of the colliding objects!
1-D Example
A lead bullet of mass 5 grams collides and sticks inside a block of wood of mass 400 grams. After the collision, the block of wood (with the bullet embedded) moves at a speed of 4 m/s. How fast was the initial speed of the bullet? How much energy was "lost" to denting the block?
1-D example, cont.
This is a collision problem, so we have
1 1i2 2 2i2
(1/2)m v + (1/2)m v
1 1f2
=
2 2f2 lost
(1/2)m v
1 1i 2 2i
+ (1/2)m v
1 1f
+E
2 2f
mv + mv
= mv + mv
We know: m1 = 5 grams, m2 = 400 grams,
1-D Example, cont.
Using the second equation, we can solve for
1 1i 2 2i 1 1f 2 2f
v1i: m v + m v
1i
= mv + mv
Since v2i = 0, and v1f = v2f = 4 m/s, we have v = (5 grams + 400 grams)*(4 m/s) / 5 grams = 324 m/s . Now we can use the first equation to solve for E :
1-D Example, cont.
1 1i2 2 2i2
(1/2)m v + (1/2)m v
1 1f2
=
2 2f2 lost
(1/2)m v
+ (1/2)m v
+E
E
= (1/2)*(.005 kg)*(324 m/s)2 -
1-D Example modified
What would be the speed of the block if the bullet were rubber and bounced off the block instead of denting it and sticking? Would the block be going faster or slower after the collision with the "rubber" bullet as opposed to the "lead" bullet?
1-D Example Modified
In this case, we still have a collision, so we have the same equations. But in this case, we know Elost = 0, and v1i = 324 m/s. We do not know v1f and v2f. Thus we still have two equations and need to solve them for two unknowns! Here, however, both unknowns are in both equations. We need to solve them simultaneously.
1-D Example Modified
We use the simpler second equation first:
1 1i 2 2i 1 1f 2 2f
mv + mv
1f
= mv + mv
to get v = (m1v1i - m2v )/m1 (since v2i = 0). We use this value for v1f in the second equation:
1 1i2 2 2i2
(1/2)m v + (1/2)m v
1 1f2
=
2 2f2 lost
1-D Example Modified
(1/2)m1v1i2 = (1/2)m1[(m1v1i - m2v )/m1]2 + (1/2)m2v
2
.
Comparisons
Notice that the rubber bullet gave about twice the kick to the wood block that the lead bullet did! If we think about it, the rubber bullet caused the wood block to catch it, and then throw it back! The lead bullet only was caught by the block - it was not thrown back. Hence the block was kicked more by the rubber bullet.
2-D Collisions
In two dimensions, we have the scalar Conservation of Energy equation plus the two component equations of Conservation of Momentum. This gives three equations, one more equation than we had with 1-D.. However, we have four more quantities: v1yi, v2yi, v1yf and v2yf. Usually we know the initial quantities, but this still leaves two more unknowns with only one more equation.
2-D Collisions
This does make sense, since there are lots of ways for two objects to collide in 2-D - a head on collision or a variety of glancing collisions! We should need to know something about how the collision happened! In this course we will not do any problems involving 2-D collisions due to the messy algebra involved.
Explosions
Explosions can be viewed as collisions run backwards! Instead of Elost , we will need Esupplied by the explosion. If you shoot a gun, there is a kick. If you hit a ball, there is a "kick" - if you swing and miss, you are out of balance since you expected a "kick" that didn't come! From Conservation of Momentum, we can see the reason for the "kick":
Explosions
Normally, in an explosion the initial object is in one piece and at rest. After the explosion, one piece goes forward. Conservation of Momentum says the other piece must then go backwards. (If we brace ourselves, we can compensate for that backwards push and not fall over.)
1 1f 2 2f 2f 1 1f 2
0 + 0 = m v + m v or v = - m v / m
Rockets
A special case of explosions is when the explosions are controlled and continuous: a rocket and a jet are examples. By throwing the exhaust gases out the back, the engine pushes the body of the plane or rocket forward - by conservation momentum. of But since this is a continuous process (rather than a series of discrete shots), we need the calculus to derive a nice formula.
Rockets
That nice formula from Conservation of Momentum using the calculus is: vf = vo + vexhln(mi/mf) where vexh is the speed with which the rocket or jet engine throws the gases out the back. As the rocket burns its fuel, its mass (including the remaining fuel) goes down.
Rockets
The thrust, or force, that the engine delivers to the rocket or plane depends on how fast the gases are exhausted and how much fuel is
on rocket exh
used: Thrust = F
=v
m/ t .
The Trolley-Sled computer homework program and the Moonlanding computer homework program are examples that show the result of using rockets to cause changes in motion.
Rotational Dynamics
In Part 2 we looked at uniform circular motion: r = constant (def. of circular motion) changes with time at uniform rate vr = 0 (radius doesn't change) v = r (where = /t) ar = 2r (v changes direction toward center) a = 0 (angular speed doesn't change)
Angular Acceleration
Here we generalize circular motion to include the case where the angular speed can change. We define angular acceleration as: = / t .
[Recall arclength = s, and =s/r, so s=r, v =r ]
Since a = v /t = ( r)/t = r( /t) = r.
Constant Angular Acceleration
In the special case of constant angular acceleration ( = constant), we have equations analogous to those we had for constant (regular) acceleration: = + t +(1/2)t and
Constant Angular Acceleration
If a car rolls on wheels that do not slip, we can relate the motion of the car to the spinning of the wheels: s = distance of car = arclength of wheels, v = speed of car = v of wheels, a = acceleration of car = a of wheels; = angle of wheel, = angular speed of wheel, and = angular acceleration of wheel:
Constant Angular Acceleration
s = r , v = r = v
r = constant
r
v =0
r
2
a = r = a
a = r (due to turning)
For(1/2)at acceleration: at constant s=s +vt+ v=v +
Torque and Rotations
Forces cause changes in motion. Torques cause changes in rotational motion. Forces are related to acceleration by Newton's Second Law: F = ma. How are and related?
Rotational Dynamics
rF
Consider: F = ma . = r F sin( ) .
Moment of Inertia
Note that the moment of inertia relates the torque to the angular acceleration just like the mass relates the force to the regular acceleration. Note that the moment of inertia depends on the mass of the object AND also on the shape of the object.
Moment of Inertia
Things that have most of their mass in the center have smaller moments of inertia than things that have most of their mass on the outside. Using the calculus, we can derive nice formulas for the moments of inertia of certain shapes: I = MR
Sphere versus Cylinder
The blue object in the figures is the axis side view of a cylinder with the same radius and volume (mass) R as the orange object (a sphere). Note that most of the mass is in the same place, but the top and bottom of the sphere must be moved away to provide the corners for the cylinder. Hence, the Moment of Inertia of the cylinder is slighly larger than that of the sphere: Isphere = (2/5)MR2 ; Icylinder = (1/2)MR2 .
Rotational Equations
In the same way, we can derive nice formulas for the kinetic energy and (angular) momentum due to the spinning of an object; basically we replace F with , m with I, v with , a with and p with L (where L is the angular momentum): F = ma = I
rotation 2
KE = (1/2)mv2 p = mv
KE = (1/2)I L = I
Rotational Equations
The rotational kinetic energy formula is used as another energy in the Conservation of Energy Law. The Angular Momentum (L) relation is used in the Law of Conservation of Angular Momentum, just as regular (linear) momentum is used in the Conservation of Momentum.
Rotational Energy Example: rolling ball
How fast will a ball be going at the base of a ramp if it is released at the top of the ramp, where the length of the ramp is 2 meters and the high end is 50 cm above the floor (base of the ramp)? We assume the ball rolls ...