Chapter 6 Momentum and Impulse Holt Physics Chapter 6 Momentum and Impulse

6-1 Linear Momentum A vector quantity defined as the product of an object’s mass and velocity p = mv “the quantity of motion” Units = kg•m/s A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck? M = 2250 kg V – 25 m/s to the east P = ? P = mv = (2250 kg)(25 m/s) = 5.6 x 104 kg•m/s

Read more Sample Problems & do Practice Questions on momentum. Sample Problem 6A A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck? m = 2250 kg V = 25 m/s east p = mv = (2250kg)(25 m/s) = 5.6 x 104 kg m/s east Read more Sample Problems & do Practice Questions on momentum.

6-1 Impulse A change in momentum caused by a force acting over a period of time. p = F t A small force acting for a long time can produce the same change in momentum as a large force acting for a short time. (SciLinks: HF2061, Momentum)

6-1 Impulse-Momentum Theorem F t = mvf - mvi A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the magnitude of the force exerted on the car during the collision. M = 1400 kg Vi = 15 m/s to the west = - 15 m/s t = 0.30 s Vf = 0 m/s F = ? F t = mvf – mvi F = (mvf – mvi ) / (t ) = (1400 kg)(0 m/s) – (1400 kg)(-15 ms/) / 0.30 s = 7.0 x 104 N to the east

Sample Problem 6B A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the magnitude of the force exerted on the car during the collision.

Sample Problem 6B Solution m = 1400 kg vi = -15 m/s t = 0.30 s vf = 0 m/s F t = mvf – mvi F = mvf – mvi = 7.0 x 104 N t Read more Sample Problems & do Practice Questions on impulse.

6-1 Stopping Time/Distance A 2250 kg car traveling to the west slows down uniformly from 20.0 m/s to 5.00 m/s. How long does it take the car to decelerate if the force on the car is 8450 N to the east? How far does the car travel during the deceleration? M = 2250 kg Vi = 20.0 m/s to the west = - 20.0 m/s Vf = 5.00 m/s to the west = -5.00 m/s F = 8450 N to the east = +8450 N t = ? x = ? F t = mvf – mvi t = (mvf – mvi) / F = (2250kg)(-5.00 m/s) – (2250 kg)(-20.0 m/s) / 8450 kg m2/s2 = 4.00 s x = ½ (vi + vf) t = ½(-20.0 m/s – 5.00 m/s)(4.00 s) = -50.0 m = 50.0 m to the west

Sample Problem 6C A 2250 kg car traveling to the west slows down uniformly from 20.0 m/s to 5.00 m/s. How long does it take the car to decelerate if the force on the car is 8450 N to the east? How far does the car travel during the deceleration?

Sample Problem 6C Solution m = 2250 kg vi = -20.0 m/s vf = -5.00 m/s F = 8450 N Ft = mvf – mvi t = mvf – mvi = 4.00 s F x = ½ (vi + vf) t = -50.0 m Read more Sample Problems & do Practice Questions on stopping distance.

6-1 Time/Force A large force exerted over a short time. A. Causes the same change in the egg’s momentum as a small force exerted over a longer time (b)

6-1 Review | Assignment A pitcher claims he can throw a 0.145 kg baseball with as much momentum as a speeding bullet. Assume that a 3.00 g bullet moves at a speed of 1.50  103 m/s. What must the baseball’s speed be if the pitcher’s claim is valid? Which has greater kinetic energy, the ball or the bullet? The speed of a particle is doubled. By what factor is its momentum changed? What happens to its kinetic energy? When a force is exerted on an object, does a large force always produce a larger change in the object’s momentum than a smaller force does? Explain.

6-2 Conservation of Momentum The total momentum of all objects interacting with one another remains constant regardless of the nature of the forces between the objects. (SciLinks: HF2062, Rocketry) m1v1,i + m2v2,i = m1v1,f + m2v2,f (a) Before the collison, ball A has momentum pA and ball B has no momentum. (b) After the collision, ball B gains momentum pB.

Sample Problem 6D A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/ s to the right, what is the final velocity of the boat?

Sample Problem 6D Solution m1 = 76 kg m2 = 45 kg V1,I = 0 v2,I = 0 V1,f = 2.5 m/s m1v1,f + m2v2,f = 0 v2,f = (76 kg x 2.5 m s-1) = -4.2 m/s 45 kg Read more Sample Problems & do Practice Questions on conservation of momentum.

6-2 Forces in Real Collisions are Non-constant Change in momentum = area under curve on a Force vs. time graph. Newton’s third law leads to conservation of momentum. Check out Holt Physics Interactive Tutor CD-ROM Module 7: Conservation of Momentum

6-2 Review | Assignment A 44 kg student on in-line skates is playing with a 22 kg exercise ball. Disregarding friction, explain what happens during the following situations. The student is holding the ball, and both are at rest. The student then throws the ball horizontally, causing the student to glide back at 3.5 m/s. Explain what happens to the ball in part (a) in terms of the momentum of the student and the momentum of the ball. The student is initially at rest. The student then catches the ball, which is initially moving to the right at 4.6 m/s. Explain what happens in part © in terms of the momentum of the student and the momentum of the ball.

6-3 Perfectly Inelastic Collision A collision in which two objects stick together and move with a common velocity after colliding. m1v1,i +m2v2,i = (m1 + m2)vf A 1850 kg luxury sedan stopped at a traffic light is struck from the rear by a compact car with a mass of 975 kg. The two cars become entangled as a result of the collision. If the compact car was moving at a velocity of 22.0 m/s to the north before the collision, what is the velocity of the entangled mass after the collision? M1 = 1850 kg M2 = 975 kg Vi,I = 0 m/s V2,I = 22.0 m/s to the north Vf = ? m1v1,i +m2v2,i = (m1 + m2)vf Vf = (m1v1,i +m2v2,i ) / (m1 + m2) = (1850 kg)(0) + (975 kg)(22.0 m/s) / (1850 kg + 975 kg) = 7.59 m/s North

Sample Problem 6E A 1850 kg luxury sedan stopped at a traffic light is struck from the rear by a compact car with a mass of 975 kg. The two cars become entangled as a result of the collision. If the compact car was moving at a velocity of 22.0 m s-1 to the north before the collision, what is the velocity of the entangled mass after the collision?

Sample Problem 6E Solution m1 = 1850 kg m2 = 975 kg v2,I = 22.0 m s-1 v1,I = 0 vf = (m1 v1,I + m2 v2,I) / (m1 + m2) = 7.59 m s-1 Read more Sample Problems & do Practice Questions on momentum of perfectly inelastic collisions.

6-3 Kinetic Energy in Inelastic collisions Kinetic energy is converted into other forms of energy in an inelastic collision. Sound Energy Internal Energy Deformation Two clay balls collide head-on in a perfectly inelastic collision. The first ball has a mass of 0.500 kg and an initial velocity of 4.00 m/s to the right. The mass of the second ball is 0.250 kg, and it has an initial velocity of 3.00 m/s to the left. What is the final velocity of the composite ball of clay after the collision? What is the decrease in kinetic energy during the collision? M1 = 0.500 kg M2 = 0.250 kg Vi,I = 4.00 m/s to the right = + 4.00 m/s V2,I = 3.00 m/s to the left = - 3.00 m/s Vf = ? K = ? m1v1,i +m2v2,i = (m1 + m2)vf Vf = (m1v1,i +m2v2,i ) / (m1 + m2) = (0.500kg)(4.00m/s) + (0.250 kg)(-3.00 m/s)/ (0.500 kg + 0.250 kg) = 1.67 m/s to the right. Ki = K1,I + K2,I = ½ m1v1,i2 + ½ m2v2,i2 = ½(0.500kg)(4.00 m/s)2 + ½ (0.250 kg)(-3.00 m/s)2 = 5.12 J Kf = Ki,f + K2,f = ½(m1 + m2)vf2 = ½(0.750 kg)(1.67 m/s)2 = 1.05 J K = Kf – Ki = 1.05 J – 5.12 J = -4.07 J

Sample Problem 6F Two clay balls collide head-on in a perfectly inelastic collision. The first ball has a mass of 0.500 kg and an initial velocity of 4.00 m s-1 to the right. The mass of the second ball is 0.250 kg, and it has an initial velocity of 3.00 m s-1 to the left. What is the final velocity of the composite ball of clay after the collision? What is the decrease in kinetic energy during the collision?

Sample Problem 6F Solution m1 = 0.500 kg m2 = 0.250 kg v1,i = 4.00 m s-1 v2,i = -3.00 m s-1 vf = (m1 v1,i + m2 v2,i) / (m1 + m2) = 1.67 m s-1 Ki = ½ m1 v1,i2 + ½ m2 v2,i2 = 5.12 J Kf = ½ m1 v1,f2 + ½ m2 v2,f2 = 1.05 J K = Kf – Ki = -4.07 J Read more Sample Problems & do Practice Questions on kinetic energy of perfectly inelastic collisions.

6-3 Elastic Collisions A collision in which the total momentum and the total kinetic energy remain constant. m1v1,i + m2v2,i = m1v1,f + m2v2,f 1/2m1v1,i2 + 1/2m2v2,i2 = 1/2m1v1,f2 + 1/2m2v2,f2

Sample Problem 6G A 0.015 kg marble moving to the right at 0.225 m s-1 makes an elastic head-on collision with a 0.030 kg shooter marble moving to the left at 0.180 m s-1. After the collision, the smaller marble moves to the left at 0.315 m s-1. Assume that neither marble rotates before or after the collision and that both marbles are moving on a frictionless surface. What is the velocity of the 0.030 kg marble after collision?

Sample Problem 6G Solution m1 = 0.015 kg m2 = 0.030 kg v1,i = 0.225 m s-1 v2,i = =0.180 m s-1 v1,f = -0.315 m s-1 v2,f = [m1v1,I + m2 v2,I - m1 v1,f] / m1 = 9.0 x 10-2 m s-1 Read more Sample Problems & do Practice Questions on elastic collisions.

6-3 Types of Collisions

6-3 Review | Assignment A 90.0 kg fullback moving south with a speed of 5.0 m/s has a perfectly inelastic collision with a 95.0 kg opponent running north at 3.0 m/s. Calculate the velocity of the players just after the tackle. Calculate the decrease in total kinetic energy as a result of the collision. Two 0.40 kg soccer balls collide elastically in a head-on collision. The first ball starts at rest, and the second ball has a speed of 3.5 m/s. After the collision, the second ball is at rest. What is the final speed of the first ball? What is the kinetic energy of the first ball before the collision? What is the kinetic energy of the second ball after the collision?