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Unformatted text preview: Chapter 10
Energy Introduction to Energy
n n n The concept of energy is one of the most important topics in science Every physical process that occurs in the Universe involves energy and energy transfers or transformations Energy is not easily defined Energy Approach to Problems
n n The energy approach to describing motion is particularly useful when the force is not constant and there is no friction An approach will involve Conservation of Energy
n This could be extended to biological organisms, technological systems and engineering situations Object Falling Due to Gravity Alone, i.e. no air drag
Recall the kinematic equation v 2 = v 2 + 2a y (y f  yi ) f i Multiply both sides by 1/2m and put a y = g (for projectile motion) 1 2 1 2 mv f = mvi  mg(y f  yi ) 2 2 Rearrange terms 1 2 1 2 mv f + mgy f = mvi + mgyi 2 2 Energy of Falling Object
1 2 1 2 mv f + mgy f = mv i + mgyi 2 2 1 2 Define kinetic energy as K = mv 2 and gravitational potential energy as U (or GPE) = mgy The top equation says K f + GPE f = K i + GPE i This is a form of energy conservation Units of Energy
n K = 1/2 mv2
n n Energy associated with motion dimensions kgm2/s2
n For KE 1 kgm2/s2 = 1 Joule n GPE = mgy
n n GPE is a form of potential energy U associated with position dimensions kg m/s2 m = kg m2/s2
n For GPE 1 kgm2/s2 = 1 Joule Potential Energy
n Potential energy is the energy associated with the configuration of a system of objects that exert forces on each other
n n This can be used only with conservative forces. More on conservative forces later. When conservative forces act within an isolated system, the kinetic energy gained (or lost) by the system as its members change their relative positions is balanced by an equal loss (or gain) in potential energy.
n This is Conservation of Mechanical Energy Types of Potential Energy
n There are many forms of potential energy, including:
n n n n Gravitational Electromagnetic Chemical Nuclear n One form of energy in a system can be converted into another Potential Energy
n n n The energy storage mechanism is called potential energyin general denoted by U. A potential energy can only be associated with specific types of forces Potential energy is always associated with a system of two or more interacting objects Total Mechanical Energy
n The sum of kinetic and potential energies is the mechanical energy (E) of the system.
n E = K + Ui n If there is no friction the total energy is conserved, i.e. constant. Gravitational Potential Energy
n Gravitational Potential Energy is associated with an object at a given distance above Earth's surface Gravitational System Example
n n n This system consists of Earth and a book Initially the book is at rest (K = 0) at y = yb (U = mgyb) When the book falls and is at ya the kinetic energy is:
n Ka + mgya = 0 +mgyb n Ka=mg(yb  ya) >0 n GPE is converted into K GPE
n Note that in the last slide the kinetic energy that was converted from potential energy depends only on the relative displacement [Ka=mg(yb  ya)]. n It does not depend on the absolute position, i.e. it doesn't matter where we take y = 0. Try yb = 100 m & ya = 50 m or yb = 125 m & ya = 75 m Gravitational Potential Energy
Before : K b = 1/2 mv = 6.25J, GPE b = 0 After : K a = 0, GPE a = mgy = 0.196y a K a + GPE a = K b + GPE b 0 + 0.196ya = 6.25 + 0 ya = 31.9m K was converted into GPE = 6.25J.
2 Conservation of Energy, Pendulum
n n n n As the pendulum swings, there is a continuous change between potential and kinetic energies At A, the energy is potential At B, all of the potential energy at A is transformed into kinetic energy n Let zero potential energy be at B At C, the kinetic energy has been transformed back into potential energy Path Independence of GPE
n n In the examples of the falling book and the ball thrown upward we considered 1D motion. We will now analyze a problem of something going up an incline.
n We will see that the energy conservation equation is the same as before Energy Conservation for a Frictionless Incline
A box is given a push and reaches the velocity vi = 3.00m/s and then coasts up the hill. What is its velocity L = 3.00m up the hill. We will transform Newton's 2nd Law into the energy equation. We will call the increment of distance up the incline ds. dv s Fs = mas = m dt dv s ds dv s mg sin q = m = mv s ds dt ds Energy Conservation for a Frictionless Incline
mg sin q = m dv s ds dv = mv s s ds dt ds multiply both sides by ds mg sin qds = mv s dv s but sin qds = dy Integrating from initial to final positions 1 2 1 mv f + mgy f = mv 2 + mgyi i 2 2 We get the same equation as for the thrown ball Indicating that the equation is independent of the path Downhill Sledding
Use energy conservation to find v.
1 2 mv 2 + mgy0 = 1 mv12 + mgy1 2 0 v1 = v 2 + 2g(y0  y1 ) 0 v1 = 10.1m / s
Notice:1) the mass cancelled out 2) Only the difference in y is needed. We could have taken the bottom of the hill as y = 10.0m and the top y = 15.0m and got the same answer. Other Forms of Potential Energy
n Elastic energy in springs
n Depends on the position of the end of the spring Hooke's Law n The force exerted by the spring is Fs =  kx
n n x is the position of the block with respect to the equilibrium position (x = 0) k is called the spring constant or force constant and measures the stiffness of the spring N/m n This is called Hooke's Law Hooke's Law, cont.
n n n When x is positive (spring is stretched), F is negative When x is 0 (at the equilibrium position), F is 0 When x is negative (spring is compressed), F is positive Hookes Law
Fsp = kDs where k is the spring constant and Ds is the spring displacement from its equilibrium position Hooke's Law
n n n The force exerted by the spring is always directed opposite to the displacement from equilibrium F is called the restoring force If the block is released it will oscillate back and forth between x and x if there is no friction Measuring the Force Constant F = F  mg = ma
y sp y When the motion stops ay = 0 kDs  mg = 0 mg k= N /m Ds Measure m, & Ds to determine k. Elastic Potential Energy
Take s as the arbitrary ball position, it will vary from si to sf. F = ma
s s dv s k(s  se ) = m dt Use the chain rule dv s ds k(s  se ) = m ds dt dv s k(s  se ) = mv s ds Energy of Springs
k(s  se )ds = mv s dv s Define u = (s  se ) = Ds hence du = ds
2 2 1 2 Ds 1 1  k(s  se )ds =  kudu =  ku Ds =  k (Ds f ) + k (Dsi ) s Ds 2 2 2 sf Ds f
f i i i vf 1 2 1 2 mv s dv s = mv f  mv i v 2 2
i 2 2 1 2 1 1 2 1 mv i + k (Dsi ) = mv f + k (Ds f ) 2 2 2 2 Energy of Springs
Note that in the total energy equation the elastic potential energy depends on the square of the displacement from the spring's equilibrium position.
2 2 1 2 1 1 2 1 mv i + k (Dsi ) = mv f + k (Ds f ) 2 2 2 2 The same amount of potential energy can be converted to K if the spring is extended or contracted by the same displacement. Spring Gun Problem
2 2 1 2 1 1 2 1 mv i + k (Dx i ) = mv f + k (Dx f ) 2 2 2 2 2 1 1 1 + k ( x1  x e ) = mv 2 + k(x e  x e ) 2 f 2 2 2 0 k(x  x e ) 2 (10N /m)(0.10  0) 2 vf = = = 3.16m / s m 0.010kg
1 Conservation of Energy, w/two U's
n yc = ? n n Choose point A as the initial point and C as the upper point where v = 0. What is yc? The spring is compressed by 0.120m = (ybya) and Ug= mgy = 0 at yA. EA = EC
n n n k = 935N/m, m = 35.0g KA + UgA + UsA = KC +UgC + UsC 1/2 kx2 = mgyc yc= (kx2/2mg) = 20.0 m ya = 0 yb = .120m Notice that this is the energy of the ball and that we didn't need to use the intermediate point b. Pushing Apart
Find the block velocities after they are pushed apart by the spring. Since we have two unknowns we need two equations. We will use energy and momentum conservation. Energy Conservation K i +U sp,i = K f +U sp , f
1 2 2 2 mv1,i + 1 mv 2 + 1 k(Dx i ) 2 = 1 mv1, f + 1 mv 2 f + 1 k(Dx f ) 2 2 2 2 2 2 2,i 2, 2 0 + 0 + 1 k(Dx i ) 2 = 1 mv1, f + 1 mv 2 f + 0 2 2 2 2, Pushing Apart
Momentum Conservation m1v1,i + m 2 v 2,i = m1v1, f + m 2 v 2, f 0 = m1v1, f + m 2 v 2, f m2 v1, f =  v 2, f m1
Substitute for v1,f in the energy equation
1 2 m1 ( m2 v 2, f ) 2 + 1 m 2 v 2 = 1 k(Dx1 ) 2 2 2 2,f m1 Pushing Apart m2 ^ 2 m 2 1+ ~v 2,f = k(Dx1 ) 2 m1 v 2, f = k(Dx1 ) = 1.8m / s m 2 (1+ m 2 m1 )
2 m2 v1, f =  v 2, f = 3.6m / s m1 Types of Collisions
n In an elastic collision, momentum and kinetic energy are conserved
n n Perfectly elastic collisions occur on a microscopic level In macroscopic collisions, only approximately elastic collisions actually occur n In an inelastic collision, kinetic energy is not conserved although momentum is still conserved
n If the objects stick together after the collision, it is a perfectly inelastic collision Collisions, cont
n n n In an inelastic collision, some kinetic energy is lost, but the objects do not necessarily stick together Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types Momentum is conserved in all collisions Elastic Collisions 1D
n Both momentum and kinetic energy are conserved (U=0) 1D Elastic Collisions
n n n Typically, there are two unknowns (v2f and v1f) to solve for so you need two equations Ki = Kf & pi = pf as well as v2i and v1i The kinetic energy equation can be difficult to use With some algebraic manipulation the two equations, can be used to solve for the two unknowns n The following can only be used with a onedimensional, elastic collision between two objects 1D Elastic Collisions, cont.
K f = Ki 1 1 1 1 2 2 m1 v1 f + m2 v 2 f = m1 v1i + m2 v 2i 2 2 2 2 2 2 1 1 1 1 2 2 m1 v1 f  m1 v1i = m2 v 2i  m2 v 2 f 2 2 2 2 2 2 m1 ( v1 f  v1i )( v1 f + v1i ) = m2 ( v 2 f  v 2i )( v 2 f + v 2i ) p f = pi m1 v1 f + m2 v 2 f = m1 v1i + m2 v 2i put all the i' s on one side & f' s on the other side m1 ( v1 f  v1i ) = m2 ( v 2 f  v 2i ) 1D Elastic Collisions, cont'd
KEi = KE f fi m1 (v1 f  v1i )(v1 f + v1i ) = m2 (v 2 f  v 2i )(v 2 f + v 2i ) pi = p f fi m1 (v1 f  v1i ) = m2 (v 2 f  v 2i )
Substitute m2 (v 2 f  v 2i ) for m1 (v1 f  v1i ) in the lhs of the KE equation m 2(v2f  v2i)(v1f + v1i) = m2(v2f  v2i)(v2f + v2i) dividing by the red terms leaves (v1f + v1i) = (v2f + v2i) 1D Elastic Collisions, cont'd
(v1 f + v1i ) = (v 2 f + v 2i )
Substituting v 2 f = (v1 f + v1i  v 2i ) & v1 f = (v 2 f + v 2i  v1i ) in the momentum equation : m1 (v1 f  v1i ) = m2 (v 2 f  v 2i ) results in two equations for v1 f & v 2 f 1D Elastic Collisions, cont.
Combining the manipulated energy and momentum equations it is found (see the text) that: v1 f v2 f m1  m2 ^ 2m2 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 2m1 ^ m2  m1 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 1D Elastic Collisions m1=m2
n m1 = m2. particles exchange velocity
v1 f m1  m2 ^ 2m2 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 v1 f = v 2i
v2 f 2m1 ^ m2  m1 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 v 2 f = v1i 1D Elastic Collisions m1<<m2 , v2i = 0 e.g. pingpong ball hitting a bowling ball
v1 f v1 f
v2 f m1  m2 ^ 2m2 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 0  m2 ^ 2m2 ^ ~v1i + ~0 = v1i 0 + m2 0 + m2 2m1 ^ m2  m1 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 0 ^ m2  0 ^ = ~v1i + ~0 = 0 0 + m2 0 + m2 v2 f 1D Elastic Collision, m1>> m2 v2=0
v1 f v1 f v2 f v2 f m1  m2 ^ 2m2 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 m1  0 ^ 0 ^ = ~v1i + ~0 = v1i m1 + 0 m1 + 0 2m1 ^ m2  m1 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 2m1 ^ 0  m1 ^ = ~v1i + ~0 = 2v1i m1 + 0 m1 + 0 Velocity Multiplier m1 = M, m2 =M/2, v1i=v,v2i=0
v1 f m1  m2 ^ 2m2 ^ = ~v + ~v 2i m1 + m2 m1 + m2 v1 f v2 f M  M /2 ^ v = ~v + 0 = M + M /2 3 2m1 ^ m2  m1 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 v2 f 2M ^ 4v = v+0 = = 1.33v ~ M + M /2 3 m1 = M/2, m2 =M/4, v1i=4v/3,v2i=0
v1 f m1  m2 ^ 2m2 ^ = ~v + ~v 2i m1 + m2 m1 + m2 v1 f v2 f M /2  M / 4 ^ 4v 4v = ~ +0 = M /2 + M / 4 3 9 2m1 ^ m2  m1 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 v2 f 2M /2 ^ 4v 16v = = 1.78v ~ +0 = M /2 + M / 4 3 9 m1 = M/4, m2 =M/8, v1i=16v/9,v2i=0
v1 f m1  m2 ^ 2m2 ^ = ~v + ~v 2i m1 + m2 m1 + m2 v1 f v2 f M / 4  M /8 ^ 16v 16v = +0 = ~ M / 4 + M /8 9 27 2m1 ^ 16v m2  m1 ^ = + ~ ~v 2i m1 + m2 9 m1 + m2 v2 f 2M / 4 ^ 16v 64v = +0 = = 2.37v ~ M / 4 + M /8 9 27 m1 = M/8, m2 =M/16, v1i=64v/27,v2i=0
v1 f m1  m2 ^ 2m2 ^ = ~v + ~v 2i m1 + m2 m1 + m2 v1 f v2 f M /8  M /16 ^ 64v 64v = +0 = ~ M /8 + M /16 27 81 2m1 ^ 16v m2  m1 ^ = + ~ ~v 2i m1 + m2 9 m1 + m2 v2 f 2M /8 ^ 64v 256v = +0 = = 3.16v ~ M /8 + M /16 27 81 Neutron Moderator
In a nuclear reactor neutrons are produced when an atom splits in a process called fission. The neutrons have a speed of about 107 m/s after fission. They must be slowed down to about 103 m/s to produce another fission neutron. They are slowed down by colliding them with heavy water D2O as the moderator and then allowed to strike a fissionable nucleus, e.g. U. The collision process is elastic collisions with the vm = 0. K=(1/2)mnvni2 After the collision: vnf = [(mnmm)/(mn+mm)]vni Neutron Moderator
Knf = (1/2) mnvni2 [(mnmm)/(mn+mm)]2 The fraction of Kn possessed after the collision is fn =Knf/Kn = [(mnmm)/(mn+mm)]2 If the neutron collides with D nucleus mm = 2mn and fn = 1/9 After many collisions the neutron velocity is reduced enough to make a fission collision with a fissile nucleus e.g. U Start 3/31 Conservation Laws
n If no external force momentum (p=mv) remains constant
n pbefore = pafter mvbefore = mvafter Conservation Laws
n If there is no friction then mechanical energy E is conserved (Note that friction is a special kind of force called a nonconservative force = more in this later.)
n E = K +U
n K=1/2mv2 U =mgy, 1/2kx2 etc Elastic Collisions 1D
n Both momentum and kinetic energy are conserved (U=0) 1D Elastic Collisions
n n n Typically, there are two unknowns (v2f and v1f) to solve for so you need two equations Ki = Kf & pi = pf as well as v2i and v1i The kinetic energy equation can be difficult to use With some algebraic manipulation the two equations, can be used to solve for the two unknowns n The following can only be used with a onedimensional, elastic collision between two objects 1D Elastic Collisions, cont.
Combining the manipulated energy and momentum equations it is found (see the text) that: v1 f v2 f m1  m2 ^ 2m2 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 2m1 ^ m2  m1 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 1D Elastic Collisions m1=m2
n m1 = m2. particles exchange velocity
v1 f m1  m2 ^ 2m2 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 v1 f = v 2i
v2 f 2m1 ^ m2  m1 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 v 2 f = v1i 1D Elastic Collisions m1=m2
v1 f = v 2i
& v 2 f = v1i Special case: v2i = 0, v1i 0 Then v1f 0 & v2f = v1i = ball 1 gives all i ts speed to 2 1D Elastic Collisions m1<<m2 , v2i = 0 e.g. pingpong ball hitting a stationary bowling ball
v1 f v1 f
v2 f m1  m2 ^ 2m2 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 0  m2 ^ 2m2 ^ ~v1i + ~0 = v1i 0 + m2 0 + m2 2m1 ^ m2  m1 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 0 ^ m2  0 ^ = ~v1i + ~0 = 0 0 + m2 0 + m2 v2 f Ball 1 reflects off ball 2 as it would reflect from a wall 1D Elastic Collision, m1>> m2 v1i= 0, v2i = v
v1 f v1 f v2 f v2 f m1  m2 ^ 2m2 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 m1  0 ^ 0 ^ ~0 + ~(v) = 0 m1 + 0 m1 + 0 2m1 ^ m2  m1 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 2m1 ^ 0  m1 ^ ~0 + ~(v) = +v m1 + 0 m1 + 0 Ball bouncing off a stationary wall, i.e. it reflects 1D Elastic Collision, m1>> m2 v2i=0, v1i 0
v1 f v1 f v2 f v2 f m1  m2 ^ 2m2 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 m1  0 ^ 0 ^ ~v1i + ~0 = v1i m1 + 0 m1 + 0 2m1 ^ m2  m1 ^ = ~v1i + ~v 2i m1 + m2 m1 + m2 2m1 ^ 0  m1 ^ ~v1i + ~0 = 2v1i m1 + 0 m1 + 0 This is like a moving wall hitting a stationary ball. TwoDimensional Elastic Collisions
n n The momentum is conserved in all directions Use subscripts for
n n n identifying the object indicating initial or final values the velocity components n If the collision is elastic, use conservation of kinetic energy as a second equation
n Remember, the simpler, previously derived equation can only be used for onedimensional situations ProblemSolving Strategies TwoDimensional Collisions
n Set up a coordinate system and define your velocities with respect to that system
n It is usually convenient to have the xaxis coincide with one of the initial velocities n In your sketch of the coordinate system, draw and label all velocity vectors and include all the given information ProblemSolving Strategies TwoDimensional Collisions, 2
n Write expressions for the x and ycomponents of the momentum of each object before and after the collision
n Remember to include the appropriate signs for the components of the velocity vectors n Write expressions for the total momentum of the system in the xdirection before and after the collision and equate the two. Repeat for the total momentum in the ydirection. ProblemSolving Strategies TwoDimensional Collisions, 3
n n If the collision is inelastic, kinetic energy of the system is not conserved, and additional information is probably needed If the collision is perfectly inelastic, the final velocities of the two objects are equal. Solve the momentum equations for the unknowns. ProblemSolving Strategies TwoDimensional Collisions, 4
n n If the collision is elastic, the kinetic energy of the system is conserved Equate the total kinetic energy before the collision to the total kinetic energy after the collision to obtain more information on the relationship between the velocities TwoDimensional Collision, example
n n n Particle 1 is moving at velocity v1i and particle 2 is at rest In the xdirection, the initial momentum is m1v1i In the ydirection, the initial momentum is 0 TwoDimensional Collision, example cont
n n After the collision, the momentum in the xdirection is m1v1f cos q + m2v2f cos f After the collision, the momentum in the ydirection is m1v1f sin q + m2v2f sin f TwoDimensional Collision, example cont
n Energy conservation:
n n Momentum Conservation:
n n 1/2m1v1i2 =1/2m1v1f2 + 1/2m2v2f2 x: m1v1i = m1v1f cosq+m2v2fcosf y: 0 = m1v1f sinq+m2v2fsinf n Note that we have three equation and 4 unknowns (q, f, v1f ,v2f). We cannot solve for the unknowns unless we have one final angle or final velocity given. Billiard Ball Example
A player wishes to sink the blue target ball in the corner pocket. The angle to the corner pocket is 35. At what angle is the white cue ball deflected? Billiard Ball Example, cont'd.
n Let 1 be the cue ball and 2 the target ball. 1 1 1 Energy conservation : 2 2 m1v1i = m1v1 f + m1v 2 f 2 2 2 2
Since m1 = m2
2 2 v1i = v1 f + v 2 f 2 Momentum Conservation: r r r m1v1i = m1v1 f + m2 v 2 f r r r v1i = v1 f + v 2 f Billiard Ball Example, cont'd.
r r r v1i = v1 f + v 2 f r r r r r r v1i v1i = (v1 f + v 2 f ) (v1 f + v 2 f )
2 v1i
n n Take the dot product of the momentum equation with itself. 2 = v1 f + v2 2f r r 2 + 2v1 f v 2 f = v1 f + v 2 f + 2v1 f v 2 f cos(q + 35) 2 From the energy equation we know that: v = v +v
2 2 1i 1f 2
2f Therefore: 2v1 f v 2 f cos(q + 35) = 0 Consequently:
n q + 35 = 90 or q = 55 If m1 =m2 it's always true that q+f =90 Start 4/2 Internal Energy
n n So far we have considered situations where there is no friction. In the absence of friction the mechanical energy is conserved, i.e. E = K + U is constant. Where friction is present some energy is converted to internal energy, usually heat.
n Here K+E constant Internal Energy ProblemBullet slows in passing through block due to friction
An m = 5.00 g bullet moving with an initial speed of 400 m/s is fired into and passes through a M = 1.00kg block. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 900 N/m. The block moves 5.00 cm to the right after impact and momentarily comes to rest. The block then is pushed back and comes to rest. Problem 967
Find (a) the speed at which the bullet emerges from the block and (b) the mechanical energy converted into internal energy in the collision. Internal Energy Problem
400 m/s (a) Find the speed at which the bullet emerges from the block Momentum conservation when the bullet leaves the block and the block has reached Vbl tells us: Energy conservation tells us that the block will compress the spring a distance x. 5.00 cm v mv i = MVbl + mv
1 2 1 2 MVbl = kx 2 2 Internal Energy Problem
400 m/s kx 2 (900N/m)(0.05m)2 Vbl = = = 1.50m/s M 1.00kg 5.00 cm v Using the momentum equation we can find v: mv i  MVbl (5.00 103 kg)(400m/s)  (1.00kg)(1.5m/s) v= = m 5.00 103 kg v = 100m/s Internal Energy Problem
(b) Find the mechanical energy lost in the collision when the spring is compressed and the block has stopped.
DE = DK + DU = 1 2 1 2 5.00 103 kg (100 m s )  5.00 103 kg (400 m s) 2 2 ( ) ) ( ) 1 + (900 N m) 5.00 10 2 m 2 ( 2 DE = 374 J , or there is an energy loss of 374 J . The lost energy is due to the friction between the bullet and block. The block heats up a little. Energy Loss in Perfectly Inelastic Collisions Energy Loss in Perfectly Inelastic Collisions In a perfectly inelastic collision momentum is conserved but energy is not conserved. What is the energy Loss? Energy Loss in Perfectly Inelastic Collisions
Take m1 = 11.0 kg; m2 = 1.00 kg; v1 = 2.00m/s; v2 = 4.00m/s Momentum conservation: (m1v1 + m2v2 =(m1+m2)v12
v12 = [(11.0kg)(2.00m/s)+(1.00kg)(4.00m/s)]/(11.0kg+1.00kg) v12 = 1.50m/s Energy Loss in Perfectly Inelastic Collisions
m1 = 11.0 kg; m2 = 1.00 kg; v1 = 2.00m/s; v2 = 4.00m/s Energy before: 1/2m1v12 +1.2m2v22 = (1/2) 11.0 kg(2.00m/s)2 +(1/2)1.00kg(4.00m/s)2 = 30.0 J Energy After: (1/2)(m1+m2) v122 = =1/2(11.0kg + 1.00kg)(1.50m/s)2 = 13.5J Lost energy (16.5J) went into internal energy (heat). Energy Diagrams
Energy of a particle in free fall: 1/2mv2+mgy = TE Four Pictures of a Particle's Motion w/Gravity SpringParticle Energy
TE = Mechanical Energy PE = Usp TE = K+Usp Usp =1/2k(xxe)2 @ x=xe+xR & x=xexL K=0 & PE =TE SpringParticle Energy
Think of the PE curve as a "potential well". At the bottom of the well K is a maximum. At the top where PE=TE: K=0. The particle is trapped in the well. The bottom is the equilibrium position. SpringParticle Energy SpringParticle Energy
n n Recall that for a spring Fsp = k(xxe) The spring forces are restoring forces. The spring force always point to xe,it tries to restore the spring to its equilibrium position.
n n n x > xe F<0 x < xe F>0 x = xe F=0 Generalized Energy Diagram
PE U Note there are two potential wells, separated by a potential hump. At the bottom of the wells and the hump (dU/dx)=0 Equilibria Equilibria are at points where dU/dx = 0. At equilibria F = 0 Away from equilibria dU/dx 0 and F 0. Forces and Potential Energy
n The force is related to the potential energy function through r dU ^ dU ^ dU ^ F =ijk dx dy dz n The x component of a conservative (no friction) force acting on an object within a system equals the negative derivative of the potential energy of the system wrt x, etc.. Forces and Potential Energy Check w/Spring
n Look at the case of a deformed spring
dU sp d 1 Fx = =  ( 2 k(x  x e ) 2 ) = k(x  x e ) dx dx
n This is Hooke's Law Forces and Potential Energy Check w/Gravity
n For gravity U = mgy dU Fg = = mg dy Energy Diagrams and Equilibrium n n n Motion in a system can be observed in terms of a graph of its position and energy In a springmass system example, the block oscillates between the turning points, x = xmax The block will always accelerate back toward x = 0 Energy Diagrams and Stable Equilibrium
n n n n At x = 0, F =0 called equilibrium The x = 0 position is one of stable equilibrium Configurations of stable equilibrium correspond to those for which U(x) is a minimum x=xmax and x=xmax are called the turning points Fsp Energy Diagrams and Unstable Equilibrium
n n n n Fx = 0 at x = 0, so the particle is in equilibrium For any other value of x, the particle moves away from the equilibrium position This is an example of unstable equilibrium Configurations of unstable equilibrium correspond to those for which U(x) is a maximum dU >0 dx F<0 dU <0 dx F>0 Neutral Equilibrium
n n Neutral equilibrium occurs in a configuration when U is constant over some region A small displacement from a position in this region will produce either restoring or disrupting forces Potential Energy in Molecules
n There is potential energy associated with the force between two neutral atoms in a molecule (e.g. N2) which can be modeled by the LennardJones function Potential Energy Curve of a Molecule n n Find the minimum of the function (take the derivative and set it equal to 0) to find the separation for stable equilibrium The graph of the LennardJones function shows the most likely separation between the atoms in the molecule (at minimum energy 2.9x1010m ) Force Acting in a Molecule n n n n The force is repulsive (positive) at small separations The force is zero at the point of stable equilibrium The force is attractive (negative) when the separation increases At great distances, the force approaches zero Some Energy Problems
Inelastic collision with v2i = 0. What is the energy After the collision? Energy in Inelastic Collisions
pi = m1vi pi = p f Ei = p f = (m1 + m2 )v f m1 fi vf = vi m1 + m2 Ef = Ef
1 m + m v2 1 2 f 2 1 m v2 2 1 i ( ) = 2 1 mv 2 1 i m1 ^ ~ m1 + m2 m1 ^ 1 Energy Ratio = ~= m Ei m1 + m2 1+ 2 m1 Energy in Inelastic Collisions
Energy Ratio
1.2 1 0.8 If m2 << m1 very little kinetic energy is lost. If m2 = m1 half the kinetic energy is lost. Note the kinetic energy gets converted into other forms of energy. e.g.heat Ef/Ei 0.6 0.4 0.2 0 0 2 4 6 8 10 m2/m1 Roller Coaster
A rollercoaster car is released from rest at the top of the first rise and then moves freely with negligible friction. The roller coaster has a circular loop of radius R in a vertical plane. (a) Suppose first that the car barely makes it around the loop: at the top of the loop the riders are upside down and feel weightless, i.e. n = 0. Find the height of the release point above the bottom of the loop, in terms of R so that n = 0. Roller Coaster
a) At the top of the loop the car and riders are in free fall, i.e. n=0: 2 mv mg down = down R v = Rg
Energy of the carridersEarth system is conserved between release and top of loop:
Ki + U gi = K f + U gf
1 2 0 + mgh = mv + mg(2 R) 2 1 gh = Rg + g(2 R) 2 If h < 2.5R the car falls, if h > 2.5R car goes in h = 2.50R a circle past the top. Roller Coaster
(b) Now assume that the release point is at or above the minimum required height. Show that the normal force on the car at the bottom of the loop exceeds the normal force at the top of the loop by six times the weight of the car. The normal force on each rider follows the same rule. Roller Coaster: at Bottom
Energy Cons. 1 2 mgh = mvbot or v 2 = 2gh bot 2 N' s 2 nd Law y is up toward center of circle Fy = ma y : mv2 bot nbot  mg = R nbot mv2 bot + mg = m 2gh + mg = mg(1+ 2h ) = R R R Roller Coaster: at Top
1 2 Energy Cons. At the top of the loop, mgh = mvtan + mg(2R) 2 v 2 = 2gh  4gR tan N' s 2 nd Law Fy = ma y mv2 tan ntop + mg = R ntop ntop m = mg + (2gh  4gR) R 2h = mg(5 + ) R Roller Coaster: ntop & nbot
ntop nbot 2h = mg(5 + ) R 2h = mg(1+ ) R nbot  ntop = 6mg Roller Coaster
Such a large normal force is dangerous and very uncomfortable for the riders. For a 150lb person the difference in normal force is 900lbs. Roller coasters are therefore not built with circular loops in vertical planes. They are helical so that some of the gravitational energy goes into axial motion. Axial Roller coaster
1 Energy : mgh = m(v 2 + v 2 ) t z 2 v 2 = 2gh  v 2 t z mv2 Fy = nbot  mg = t R nbot vz 2h = mg(1+ )  m R R
2 Start 4/4 Pulley w/weights
13. Two objects are connected by a light string passing over a light frictionless pulley as shown. The object of mass 5.00kg is released from rest. Using the principle of conservation of energy, (a) determine the speed of the 3.00kg object just before the 5.00kg object hits the ground. (b) Find the maximum height to which the 3.00kg object rises. Pulley w/weights
1) No friction Energy conserved. 2) Take the zero of gravitational potential to be at the bottom y = 0 of the three kg mass. (Note since we we are considering the masses to be point particles it is the location of the entire mass.) a) Determine the speed of the 3.00kg object just before the 5.00kg object hits the ground. . . Initial values: m2: K2i = 0, U2i = 0 m1: K1i = 0, U1i =(5.00kg)(9.80m/s2)(4.00m) = 196 J Final Values: m2: K2f = 1/2(3.00kg)v2, U2f=(3.00kg)(9.80m/s2)(4.00m)=118J m1: K1f = 1/2(5.00kg)(v)2, U1f = 0 Energy Conservation: Ki + Ui = Kf +Uf 196 J = (4.00)v2 +118 J v2 = 19.6 v = 4.43 m/s Pulley w/weights
(b) Find the maximum height (H) to which the 3.00kg object rises. Initial values: m2i: K2i = 1/2(3.00kg)(4.43m/s)2= 29.4J , U2i = (3.00kg)(9.80m/s2)(4.00m) = 118 J Final Values: m1: K1f = 0, U1f =m1gH = (3.00kg)(9.80m/s2)H = 29.4h Energy Conservation: Ef = Ei 29.4H = 117.6 J +29.4 J = 147 J H = 5.00 m Blocks, Spring & Pulley
n n n n The system consists of the two blocks, the spring, and Earth.The spring is in its equilibrium position. Gravitational and potential energies are involved The kinetic energy is zero if our initial and final configurations are at rest What is h, where the blocks come to rest and the spring is extended? k = 900. N/m m1 = 2.00 kg m2 = 18.0 kg Blocks, Spring & Pulley
K1i = K 2i = 0, U si = 0, U1ig = m1gH, U 2ig = m2 gh 1 K1 f = K 2 f = 0, U sf = kh 2 , U1 fg = m1gH, U 2 fg = 0 2 Ei = m1gH + m2 gh 1 2 E f = m1gH + kh 2 E f = Ei 1 2 kh = m2 gh 2 2m2 g 2(18.0kg)(9.80m / s 2 ) h= = = 0.392m k 900.N /m y = H y = 0 Problem 1069
A 10 kg box slides 4.0 m down the frictionless ramp shown in the Figure, then collides with a spring whose spring constant is 250 N/m. a. What is the maximum compression of the spring? b. At what compression of the spring does the box have its maximum velocity? Problem 1069
a. What is the maximum compression of the spring? Assume energy is conserved. Choose the origin of the coordinate system at point of max compression. Take the coordinate along the ramp to be s. Problem 1069
a. What is the maximum compression of the spring?
K 2 +U 2,sp +U 2,g = K1 +U1,sp +U1,g 0+ 1 k 2
1 2 Ds) + 0 = 0 + 0 + mg(4.0m + Ds) sin 30 o (
2 2 250N /m )(Ds) = (10kg) 9.8m / s 2 (4.0 + Ds)0.5 (
2 ( ) 125N /m )(Ds)  49kgm / s 2 Ds 196kgm / s 2 = 0 ( Ds = 1.46m,1.07m(unphysical) ( ) Problem 1069
b. At what compression of the spring does the box have its maximum velocity? For this part of the problem its easiest to take the origin at the point where the spring at its equilibrium position. Problem 1069
We equate the energy at the initial position to some arbitrary x and y position. We then take the derivative of v wrt Ds and set it equal to zero.
1 2 mv + mgy + k (Ds) = 0 + mgy1 + 0
2 1 2 2 1 2 mv 2 + mg(Ds sin 30 o ) + 1 k (Ds) = mg(4.0m sin 30 o ) 2 k (Ds)  (mg sin 30 o )(Ds ) + 1 mv 2  mg(4.0m sin 30 o ) = 0 2 dv =0 dDs
2 2 1 2 To find max v wrt Ds take the derivative of this equation and put
1 2 k2(Ds)  mg sin 30 o + 1 mv2 2 dv =0 dDs mg sin 30 o Ds = = 0.196m k Problem 1074
A sled starts from rest at the top of the frictionless, hemispherlcal, snowcovered hill shown in the figure. a. Find an expression for the sled's speed When it is at angle f. b. Use Newton's laws to find the maximum speed the sled can have at angle f without leaving the surface. c. At what angle f max does the sled "fly off" the hill? Problem 1074
a. Find an expression for the sled's speed when it is at angle f. K f +U f = K 0 +U 0
1 mv2 f 2 1 mv2 f 2 2 + mgyf = 1 mv0 + mgy0 2 2 + mgR cos f = 1 mv0 + mgR 2 vf = 2gR(1 cos f ) Problem 1074
b. Use Newton's laws to find the maximum speed the sled can have at angle f without leaving the surface. Fr = mar mv2 v2 ^ n + mg cos f = fi n = m g cos f + ~ R R n decreases as v increases. n can' t change sign so when n = 0 the sled leaves the hill n 0 fi vmax = gR cos fmax Problem 1074
c. At what angle f max does the sled "fly off" the hill? We have vf for an arbitrary angle and vmax. Equating the two we have 2gR(1 cos fmax ) = gR cos fmax 2gR(1 cos fmax ) = gR cos fmax cos fmax 2 1 2 = fi cos = 48.2 o 3 3 fmax does not depend on R ...
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This note was uploaded on 05/19/2008 for the course PHYS 161 taught by Professor Hammer during the Spring '07 term at Maryland.
 Spring '07
 Hammer
 mechanics, Energy

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