Unformatted text preview: Exam 3 Review
Exam 3 Monday Apr. 28 Chaps. 9,10 & 11 Linear Momentum
n The linear momentum of a particle or an object that can be modeled as a particle of mass m moving with a velocity v is defined to be the product of the mass and velocity:
n p=mv
n The terms momentum and linear momentum will be used interchangeably in the text Newton and Momentum
n n Newton called the product mv the quantity of motion of the particle Newton's Second Law with constant mass (i.e. dm/dt = 0) can be used to relate the momentum of a particle to the resultant force acting on it This is a generalization of Newton's 2nd Law, useful
when the mass is not constant. Ballistic Cannon
A projectile (Mp) is fired from a cannon cart (Mcc). The projectile lands in the receiver cart (Mrc). In this DEMO Mcc= Mp + Mrc There are no horizontal external forces. Thus the total momentum remains constant. Ballistic Cannon
Initially the momentum of the system is zero. Since only internal forces are involved the total system momentum stays zero. p=0 After the cannon fires and before the projectile lands in the receiver cart: MccVcc + MpVp = 0 Vcc = (Mp/Mcc)Vp When the projectile lands in the receiver cart the momentum is still zero and the cannon cart is still going to the left. 0 = MccVcc + (Mp+Mrc)Vrc Since Mcc = Mp + Mrc, Vcc =  Vrc Impulse and Momentum
n n n From Newton's Second Law, F = dp/dt Solving for dp gives dp = Fdt Integrating to find the change in momentum over some time interval
tf Dp = p f  p i = Fdt = J
n The integral is called the impulse, I, of the force F acting on an object over Dt ti Problem Deformed Ball
A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in contact with the pavement, the lower side of the ball is temporarily flattened. Suppose that photographs show the maximum depth of the dent is 1 cm. Compute an estimate for the acceleration of the ball while it is in contact with the pavement and the time duration it is deformed. Solution Deformed Ball
To find the contact time we can use the Impulse equation using the average force. To find the average force we use kinematics to find the average acceleration, assuming constant a gives an approximate average a. J = Fave Dt = Pf  P
To find a, first find the velocity the ball has at the instant the bottom side hits the ground assuming the ball falls 1.50 m. v 2 = v 2 + 2a(y f  yi ) f i v 2 = 0 + 2(9.80m / s 2 )(1.50m  0) f v f = 5.42m / s Solution Deformed Ball
We can use the same equation to find the acceleration, assumed constant, as the ball moves from where it first touched the ground until it stopped. v 2f = v 2 + 2a(y f  yi ) i
0 = (5.42m / s) 2 + 2a(0  .01m) a = 1470m / s 2 Since this equation used is only valid for constant acceleration this is approximately the average acceleration. The momentum of the ball goes from Pi = m(5.42 m/s) to 0 when it is being deformed. Using the impulse theorem: Problem Deformed Ball
J = Fave Dt = p f  pi Dt = p f  pi Fave = p f  pi ma 0  m(5.42m / s) @ m1.47 10 3 m / s 2 Dt @ 3.69 10 3 s
Since it bounces back to nearly the same height the time to deform and undeform is approximately twice this time. Types of Collisions
n In an elastic collision, momentum is conserved and particles bounce off each other.
n In a perfectly elastic collision the ball bounces back at the same speed n In a perfectly inelastic collision, momentum is conserved and the objects stick together after the collision Forces in Elastic Collisions
http://www.physics.umd.edu/lecdem/services/demos/demosc7/c725.htm Show that a larger impulse is imparted by a perfectly elastic collision. Use a superball for the perfectly elastic collision and an ordinary ball for the nonperfect elastic collision. tf ti J = Fdt = p f  p i 1) p f = p i fi J = 2p i 2) p f << p f fi J @ p i 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 putting the f' s on one side and i' s on the other side 1 2 1 2 mv f + mgy f = mvi + mgyi 2 2 Energy of Falling Object
1 2 1 2 mv f + mgy f = mvi + mgyi 2 2 1 2 Define kinetic energy as K = mv 2 and gravitational potential energy as U = mgy The top equation says Kf + Uf = Ki + Ui 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 U = mgy
n n U is a form of potential energy U associated with position dimensions kg (m/s2) m = kg m2/s2
n For U 1 kgm2/s2 = 1 Joule 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. 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. Hookes Law
Fsp = kDs where k is the spring constant and Ds is the spring displacement from its equilibrium position 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 a dummy variable u = (s  se ) hence du = ds (se is constant)
2 1 1 2 Ds 1 2 =  kudu =  ku =  k (Ds f ) + k (Dsi ) Ds 2 2 2 Ds
f i i Ds f vf
i 1 1 mv s dv s = mv2  mv2 f i 2 2 v 2 1 2 1 2 1 2 1 mvi + k (Dsi ) = mvf + k (Ds f ) 2 2 2 2 Spring Potential Energy
n Define the spring potential energy as U sp 1 2 = k(Ds) 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. 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
1 2 Substitute for v1,f in the energy equation m2 m1 ( 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 ) 2 = = 1.8m / s m 2 (1+ m 2 m1 ) 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 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. BulletBlock Problem
Find (a) the speed at which the bullet emerges from the block and BulletBlock Problem
(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. 400 m/s 5.00 cm v mv i = MVbl + mv
1 2 1 2 MVbl = kx 2 2 BulletBlock 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 Work
W = F Dr cosq Work
n W = F Dr cos q
n n n The displacement Dr is along the point of application of the force A force does no work on the object if the force does not move through a displacement The work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application Work
n The normal force, n, and the gravitational force, mg, do no work on the object moving along the table
n cos q = cos 90 = 0 n The force F does do work on the object equal to FDrcosq Work Example w/friction
F=10.N q= 45, k = 0.1 m =4.0kg Dr = 0.1m f Work Example w/friction
x : F cos q  f = ma y : n + F sin q  mg = 0 n = mg  F sin q f = mn = m (mg  F sin q ) = .10[(4.0kg)(9.80m / s 2 ) 10.N sin 45 o ] f = 3.2N WF = 10N(0.5m) cos 45 o = 3.2J W f = 3.2N(0.5) cos180 o = 1.6J Kinetic Energy & Work
n Kinetic Energy is the energy of a particle due to its motion
n K = 1/2 mv2
n n n K is the kinetic energy m is the mass of the particle v is the speed of the particle n A change in kinetic energy is the result of doing work to transfer energy into a system Kinetic Energy, cont
n Calculating the work: dv W = Fdx = madx = m dx dt dv dx dv W = m dx = m vdx = mvdv dx dt dx 1 1 W = mv  mv 2 2
xf xf xf xi xi xi xf xf vf xi xi vi 2 2 f i WorkKinetic Energy Theorem
n n In the case in which work is done on a system and the only change in the system is in its speed, the net work done by the net force equals the change in kinetic energy of the system. W = DK = Kf  Ki always true even w/friction
n where K = 1/2 mv2 Summary
1) Only Conservative Forces: e.g. gravity, springs Wcon = DK or since Wcon= DU Mechanical Energy Conservation Kf + Uf = Ki + Ui 2) Non conservative (friction) + Conservative forces W = DK Wfr = DEMech Wfr = (Kf + Uf)  (Ki + Ui) Work Done with Friction A 6.0 kg block is pulled from rest by a force F = 12 N at an angle of 5 over a surface with a coefficient of kinetic friction 0.17. Find the speed of the block after it has been moved 3.0 m. Work Done by Friction
W fr = fDx cos180 =  fDx f = mk n y : Fy = n  mg + F sin q = 0 n = mg  F sin q W fr = m k (mg  F sin q )Dx
Note that n 0. If Fsinq mg n =0 and the block is lifted off the table. Total Work Done
Work Done by the Applied Force WF = FDx cos q
Work Kinetic Energy Theorem: W = Dk 1 2 1 2 W fr + WF = mv f  mvi 2 2 1 2 1 2 [m k (mg  F sin q ) + F cos q ]Dx = mv f  mvi 2 2 Work Done with Friction
m = 6.0kg F = 12.N q = 5 o m k = 0.17 Dx = 3.0m [m k (mg  F sin q ) + F cos q ]Dx vf = m /2 [.17(58.8 12 sin 5) +12 cos 5]3.0 vf = = 1.5m / s 3.0
w/o friction v = 3.5 m/s Power
n n The time rate of energy transfer is called power The average power is given by when the method of energy transfer is work Instantaneous Power
n The instantaneous power is the limiting value of the average power as Dt approaches zero n This can also be written for constant F as Power Generalized
n n Power can be related to any type of energy transfer In general, power can be expressed as n dE/dt is the rate rate at which energy is crossing the boundary of the system for a given transfer mechanism Units of Power
n The SI unit of power is called the watt
n 1 watt = 1 joule / second = 1 kg.m2 / s2 n A unit of power in the US Customary system is horsepower
n 1 hp = 746 W = 550 ftlbs/s n Units of power can also be used to express units of work or energy
n 1 kWh = (1000 W)(3600 s) = 3.6 x106 J Elevator Power
An elevator car has a mass of 1,600kg and is carrying passengers having a combined mass of 200kg. A constant frictional force of 4,000N retards its motion upward. What power must the motor deliver when the speed is 3.00m/s if the elevator is accelerating at 1.00m/s2. Elevator Power
Apply Newton's 2nd Law
T  f  Mg = Ma T = f + M (g + a) T = 4,000N + (1800kg)(9.80m / s 2 +1.00m / s 2 ) T = 23.4 10 3 N P = Tv = (23.4 10 3 N)(3.00m / s) P = 70.2kW P = (70.2kW) 1hp = 94.1hp .746kW ...
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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, Mass, Momentum

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