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**Momentum, Impulse, and Collisions**

Chapter 8 Momentum, Impulse, and Collisions

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Goals for Chapter 8 To learn the meaning of the momentum of a particle and how an impulse causes it to change To learn how to use the principle of conservation of momentum To learn how to solve problems involving collisions

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Goals for Chapter 8 To learn the definition of the center of mass of a system and what determines how it moves To analyze situations, such as rocket propulsion, in which the mass of a moving body changes

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Introduction In many situations, such as a bullet hitting a carrot, we cannot use Newton’s second law to solve problems because we know very little about the complicated forces involved. In this chapter, we shall introduce momentum and impulse, and the conservation of momentum, to solve such problems.

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**Momentum and Newton’s second law**

The momentum of a particle is the product of its mass and its velocity: Newton’s second law can be written in terms of momentum as What is the momentum of a 1000kg car going 25 m/s west?

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**Momentum and Newton’s second law**

The momentum of a particle is the product of its mass and its velocity: Newton’s second law can be written in terms of momentum as What is the momentum of a 1000kg car going 25 m/s west? p = mv = (1000 kg)(25 m/s) = 25,000 kgm/s west

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Impulse and momentum Impulse of a force is product of force & time interval during which it acts. Impulse is a vector! On a graph of Fx versus time, impulse equals area under curve.

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Impulse and momentum Impulse-momentum theorem: Impulse = Change in momentum J of particle during time interval equals net force acting on particle during interval J = Dp = pfinal – pinitial (note all are VECTORS!) J = Net Force x time = (SF) x (Dt) so… Dp = (SF) x (Dt) SF = Dp/(Dt)

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**Compare momentum and kinetic energy**

Changes in momentum depend on time over which net force acts But… Changes in kinetic energy depend on the distance over which net force acts.

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**Ice boats again (example 8.1)**

Two iceboats race on a frictionless lake; one with mass m & one with mass 2m. Wind exerts same force on both. Both boats start from rest, both travel same distance to finish line. Questions! Which crosses finish line with more KE? Which crosses with more Momentum?

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**Ice boats again (example 8.1)**

Two iceboats race on a frictionless lake; one with mass m and one with mass 2m. The wind exerts the same force on both. Both boats start from rest, and both travel the same distance to the finish line? Which crosses the finish line with more KE? In terms of work done by wind? W = DKE = Force x distance = same; So ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22

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**Ice boats again (example 8.1)**

Two iceboats race on a frictionless lake; one with mass m and one with mass 2m. The wind exerts the same force on both. Both boats start from rest, and both travel the same distance to the finish line? Which crosses the finish line with more KE? In terms of work done by wind? W = DKE = Force x distance = same; So ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22 Since m2> m1, v2 < v1 so more massive boat loses

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**Ice boats again (example 8.1)**

Two iceboats race on a frictionless lake; one with mass m and one with mass 2m. The wind exerts the same force on both. Both boats start from rest, and both travel the same distance to the finish line? Which crosses the finish line with more p? In terms of momentum? Force on the boats is the same for each, but TIME that force acts is different. The second boat accelerates slower, and takes a longer time. Since t2> t1, p2 > p1 so second boat has more momentum

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**Ice boats again (example 8.1)**

Check with equations! Force of wind = same; distance = same Work done on boats is the same, gain in KE same. ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22 And m1v1 = p1; m2v2 = p2 ½ m1v12 = ½ (m1v1) v1 = ½ p1 v1 & same for ½ p2 v2 ½ p1 v1 = ½ p2 v2 Since V1 > V2, P2 must be greater than P1! p2 > p1 √

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**A ball hits a wall (Example 8.2)**

A 0.40 kg ball moves at 30 m/s to the left, then rebounds at 20 m/s to the right from a wall. What is the impulse of net force during the collision, and if it is in contact for 0.01 s, what is the average force acting from the wall on the ball?

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**Kicking a soccer ball – Example 8.3**

Soccer ball 0.40 kg moving left at 20 m/s, then kicked up & to the right at 30 m/s at 45 degrees. If collision time is 0.01 seconds, what is impulse?

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An isolated system The total momentum of a system of particles is the vector sum of the momenta of the individual particles. No external forces act on the isolated system consisting of the two astronauts shown below, so the total momentum of this system is conserved.

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**Conservation of momentum**

External forces (the normal force and gravity) act on the skaters, but their vector sum is zero. Therefore the total momentum of the skaters is conserved. Conservation of momentum: If the vector sum of the external forces on a system is zero, the total momentum of the system is constant.

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**Remember that momentum is a vector!**

When applying conservation of momentum, remember that momentum is a vector quantity! Use vector addition to add momenta in COMPONENTS!

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**Recoil of a rifle – 1D conservation of momentum**

A rifle fires a bullet, causing the rifle to recoil.

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**Objects colliding along a straight line (example 8.5)**

Two gliders collide on an frictionless track. What are changes in velocity and momenta?

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**A two-dimensional collision**

Two robots collide and go off at different angles. You must break momenta into x & y components and deal with each direction separately

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Elastic collisions In an elastic collision, the total momentum of the system in a direction is the same after the collision as before if no external forces act in that direction: Pix = Pfx mavai +mbvbi =mavaf +mbvbf But wait – there’s more!

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Elastic collisions In an elastic collision, the total kinetic energy of the system is the same after the collision as before. KEi = KEf ½ mavai2 + ½ mbvbi2 = ½ mavaf2 + ½ mbvbf2 =

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**Elastic collisions In an elastic collision,**

Difference in velocities initially = (-) difference in velocities finally (vai - vbi) = - (vaf – vbf)

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**Elastic collisions In an elastic collision,**

Difference in velocities initially = (-) difference in velocities finally (vai - vbi) = - (vaf – vbf) Solve generally for final velocities: vaf = (ma-mb)/(ma+mb)vai + 2mb/(ma+mb)vbi vbf = (mb-ma)/(ma+mb)vbi + 2ma/(ma+mb)vai

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Elastic collisions

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Elastic collisions Behavior of colliding objects is greatly affected by relative masses.

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Elastic collisions

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Elastic collisions Behavior of colliding objects is greatly affected by relative masses.

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Elastic collisions

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Elastic collisions Behavior of colliding objects is greatly affected by relative masses. Stationary Bowling Ball!

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Inelastic collisions In any collision in which the external forces can be neglected, the total momentum is conserved. A collision in which the bodies stick together is called a completely inelastic collision In an inelastic collision, the total kinetic energy after the collision is less than before the collision.

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**Some inelastic collisions**

Cars are intended to have inelastic collisions so the car absorbs as much energy as possible. Example 8.7

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**The ballistic pendulum**

Ballistic pendulums are used to measure bullet speeds

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**A 2-dimenstional automobile collision (Example 8.9)**

Two cars traveling at right angles collide.

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**An elastic straight-line collision**

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**Neutron collisions in a nuclear reactor**

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**A two-dimensional elastic collision (page 258)**

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**Center of mass of symmetrical objects**

It is easy to find the center of mass of a homogeneous symmetric object

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**Motion of the center of mass**

The total momentum of a system is equal to the total mass times the velocity of the center of mass. The center of mass of the wrench at the right moves as though all the mass were concentrated there.

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**Tug-of-war on the ice Start 20 meters away; each 10 m from coffee!**

Pull light rope; when James moved 6 m, how far is Ramon?

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**External forces and center-of-mass motion**

When a body or collection of particles is acted upon by external forces, the center of mass moves as though all the mass were concentrated there.

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Rocket propulsion Conservation of momentum holds for rockets, too! F = dp/dt = d(mv)/dt = m(dv/dt) + v(dm/dt) As a rocket burns fuel, its mass decreases (dm< 0!)

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**Rocket propulsion Initial Values m = initial mass of rocket**

v = initial velocity of rocket in x direction mv = initial x-momentum of rocket vexh = exhaust velocity from rocket motor (This will be constant!) (v - vexh) = relative velocity of exhaust gases

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**Rocket propulsion Changing values**

dm = decrease in mass of rocket from fuel dv = increase in velocity of rocket in x-direction dt = time interval over which dm and dt change

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**Rocket propulsion Final values**

m + dm = final mass of rocket less fuel ejected v + dv = increase in velocity of rocket (m + dm) (v + dv) = final x-momentum of rocket (+x direction) [- dm] (v - vexh) = final x-momentum of fuel (+x direction)

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**final momentum of the ejected mass of gas**

Rocket propulsion Final values mv = [(m + dm) (v + dv)] + [- dm] (v - vexh) initial momentum final momentum of the lighter rocket final momentum of the ejected mass of gas

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**Rocket propulsion Final values**

mv = [(m + dm) (v + dv)] + [- dm] (v - vexh) mv = mv + mdv + vdm + dmdv –dmv +dmvexh

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**Rocket propulsion Final values**

mv = [(m + dm) (v + dv)] + [- dm] (v - vexh) mv = mv + mdv + vdm + dmdv –dmv +dmvexh

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Rocket propulsion Final values mv = mv + mdv + vdm + dmdv –dmv +dmvexh

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**Rocket propulsion Final values mv = mv + mdv + vdm + dmdv –dmv +dmvexh**

0 = mdv + dmdv + dmvexh

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**Rocket propulsion Final values mv = mv + mdv + vdm + dmdv –dmv +dmvexh**

0 = mdv + dmdv + dmvexh Neglect the assumed small term: m(dv) = -dmdv – (dm)vexh

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**Rocket propulsion Final values m(dv) = – (dm)vexh**

Gain in momentum of original rocket is related to rate of mass loss and exhaust velocity of gas!

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**Rocket propulsion Final values m(dv) = – (dm)vexh**

Gain in momentum of original rocket is related to rate of mass loss and exhaust velocity of gas! Differentiate both sides with respect to time!

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**Rocket propulsion Final values m(dv) = – (dm)vexh**

m(dv/dt) = – [(dm)/dt] vexh

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**Rocket propulsion Final values m(dv) = – (dm)vexh**

m(dv/dt) = – [(dm)/dt] vexh ma = Force (Thrust!) = – [(dm)/dt] (vexh)

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**Rocket propulsion Final values m(dv) = – (dm)vexh**

m(dv/dt) = – [(dm)/dt] vexh ma = Force (Thrust!) = – [(dm)/dt] (vexh) rate of change of mass Velocity of gas exhausted

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**Rocket propulsion Final values Thrust = – [(dm)/dt] (vexh)**

Example: vexhaust = 1600 m/s Mass loss rate = 50 grams/second Thrust =?

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**Rocket propulsion Final values Thrust = – [(dm)/dt] (vexh)**

Example Problem 8.62 vexhaust = 1600 m/s Mass loss rate = 50 grams/second Thrust = m/s (-0.05 kg/1 sec) = +80N

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**Rocket propulsion Gain in speed? m(dv) = – (dm)vexh**

dv = – [(dm)/m] vexh integrate both sides

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**Rocket propulsion Gain in speed? m(dv) = – (dm)vexh**

dv = – [(dm)/m] vexh (integrate both sides) vf - vi = (vexh) ln(m0/m) m0 = initial mass m0 > m, so ln > 1 “mass ratio” gain in velocity Velocity of gas exhausted

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**Rocket propulsion Gain in speed? vf - vi = (vexh) ln(m0/m)**

m0 = initial mass m0 > m, so ln > 1 “mass ratio” gain in velocity Velocity of gas exhausted Faster exhaust, and greater difference in mass, means a greater increase in speed!

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**Rocket propulsion Gain in speed? vf - vi = (vexh) ln(m0/m) Example:**

Rocket ejects gas at relative 2000 m/s. What fraction of initial mass is not fuel if final speed is 3000 m/s?

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**Rocket propulsion Gain in speed? vf - vi = (vexh) ln(m0/m) Example:**

Rocket ejects gas at relative 2000 m/s. What fraction of initial mass is not fuel if final speed is 3000 m/s? ln(mo/m) = (3000/2000) = 1.5 so m/mo = e-1.5 = .223

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**Rocket propulsion Gain in speed? vf - vi = (vexh) ln(m0/m)**

m0 = initial mass m0 > m, so ln > 1 “mass ratio” gain in velocity Velocity of gas exhausted STAGE rockets to throw away mass as they use up fuel, so that m0/m is even higher!

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