PHY 101 Lecture 7 7 1 The ImpulseMomentum

PHY 101: Lecture 7 • 7. 1 The Impulse-Momentum Theorem • 7. 2 The Principle of Conservation of Linear Momentum • 7. 3 Collision in One Dimension • 7. 4 Collisions in Two Dimensions • 7. 5 Center of Mass

PHY 101: Lecture 7 Impulse and Momentum 7. 1 The Impulse-Momentum Theorem

Force and Collision Time

Definition of Impulse • Vector quantity • Impulse J of a force is product of average force Favg and time interval Dt during which the force acts ØJ = Favg. Dt • Direction of Impulse is same as direction of average force • SI Units of Impulse: Newton second (Ns)

Definition of Linear Momentum • Linear momentum p of an object is the product of the object’s mass m and velocity v Øp = mv • Linear momentum is a vector that has the same direction as velocity • Symbol: p • SI Units of Linear Momentum: kilogram meter/second (kg m/s)

Linear Momentum – Example 1 • What is magnitude of linear momentum of a 7. 1 -kg bowling ball traveling at 12 m/s? Øp = mv = 7. 1(12) = 85. 2 kg(m/s)

Linear Momentum – Example 2 • Linear momentum of a runner in a 100 -m dash is 7. 5 x 102 kg(m/s) • The runner’s speed is 10 m/s • What is his mass? Øp = mv Ø 750 = m(10) Øm = 750/10 = 75 kg

Momentum vs. Kinetic Energy • Momentum and Kinetic Energy are two different properties of matter • Massless radiation has Momentum but not Kinetic Energy, but that is PHY 102 • Formula relationship is ØKE = p 2 / 2 m

Impulse and Momentum Relationship Øp = mv • Dp = D(mv) = m(Dv) • Dp/Dt = m(Dv/Dt) = ma = F (Newton’s 2 nd Law) • Dp = F(Dt) = J • Note: Dp = pf – pi, Dt = tf – ti • Note: We assumed that m is constant, but m could also change. We will see examples where this is the case

Impulse – Momentum Theorem • When a net force acts on an object, the impulse of this force is equal to the change in momentum of the object (SF)Dt = mvf - mvi Impulse = Change in Momentum

Impulse-Momentum Theorem Example 1 • A 0. 20 -kg softball is tossed upward and hit horizontally by a batter • Softball receives an impulse of 3. 0 Ns • With what horizontal speed does the ball move away from the bat? ØJ = pf – pi = mvf - mvi ØInitial momentum is 0 ØJ = mvf Øvf = J/m = 3/0. 20 = 15 m/s

Impulse-Momentum Theorem Example 2 • Automobile with linear momentum of 3. 0 x 104 kg(m/s) in +x-direction is brought to a stop in 5. 0 s • What is the average braking force? ØJ = Ft = pf – pi ØFinal momentum is zero ØF = – pi /t = -(+3. 0 x 104) / 5 = - 6000 N

Impulse-Momentum Theorem Example 3 • A pool player imparts an impulse of 3. 2 Ns to a stationary 0. 25 -kg cue ball with a cue stick • What is the speed of the ball just after impact? ØJ = FDt = pf – pi = mvf - mvi ØInitial momentum is zero ØJ = mvf Øvf = J/m = 3. 2/0. 25 = 12. 8 m/s

Impulse-Momentum Theorem Example 4 • A car of 1500 kg is traveling in +x-direction at 20 m/s • Force acts on car in the +x-direction for 30 seconds • Final velocity of car is 40 m/sec in the +x-direction • What is the force? ØJ = FDt = pf – pi = mvf - mvi ØJ = 1500(+40) – 1500(+20) ØJ = 60000 – 30000 = 30000 ØF = 30000/Dt = 30000/30 = 1000 N

Impulse-Momentum Theorem Example 5 • Machine gun fires 15 bullets (mass 0. 020 kg) per second • Velocity of bullets is 800 m/sec in the +x-direction • What is the force on the machine gun? ØChange in momentum of 1 bullet = mvf – mvi = (0. 020)800 = 16 ØChange in momentum for all bullets in second = 16(15) = 240 N ØThis is average force on the bullets ØForce on gun is equal and opposite, -240 N

PHY 101: Lecture 7 Impulse and Momentum 7. 2 The Principle of Conservation of Linear Momentum

Conservation of Momentum Internal / External Forces • Two types of forces act on the system ØInternal forces – Forces that the objects within the system exert on each other ØExternal forces – Forces exerted on the objects by agents external to the system

Conservation of Momentum Collision between Two Balls 1 • System is two balls that are colliding • F 12 is the force exerted on ball 1 by ball 2 • F 21 is the force exerted on ball 2 by ball 1 • Forces are internal forces • Forces are equal in magnitude but opposite in direction ØF 12 = - F 21 • Force of gravity also acts on the balls • The weights are external forces • The weights are W 1 and W 2 • Impulse-Momentum Theorem is applied to the two balls

Conservation of Momentum Collision between Two Balls 2 • (W 1 + F 12)Dt = m 1 vf 1 – m 1 vi 1 • (W 2 + F 21)Dt = m 2 vf 2 – m 2 vi 2 • Add these equations • (W 1+W 2+F 12+F 21)Dt = (m 1 vf 1+m 2 vf 2) – (m 1 vi 1+m 2 vi 2) • (sum external forces+sum internal forces)Dt = pf - pi • Sum of internal forces is zero (Newton’s Third Law) • (sum of external forces)Dt = pf – pi • Suppose that the sum of external forces is zero • pf = p i • This is conservation of momentum

Conservation of Momentum • Total linear momentum of an isolated system remains constant • An isolated system is one for which the vector sum of the external forces acting on the system is zero • Note: It is important to realize that the total linear momentum may be conserved even when the kinetic energy is not conserved

Conservation of Momentum Example 1 • A 60 -kg astronaut floats at rest in space • He throws his 0. 50 -kg hammer such that it moves with a speed of 10 m/s in the -x-direction • What happens to the astronaut? ØAstronaut is object 1 and hammer is object 2 ØMomentum is conserved, pf = pi ØInitial momentum is zero Øm 1 v 1 f + m 2 v 2 f = m 1 v 1 i + m 2 v 2 i = 0 Ø 60 v 1 f + 0. 5(-10) = 0 Øv 1 f = 5/60 = +0. 0833 m/s

Conservation of Momentum Example 2 • An 800 -kg car travelling in the +x-direction with a velocity of 30 m/sec • It hits a stationary truck of 3200 kg • The car and truck stick together • What is speed of combined masses after the collision? ØThe car is object 1 and the truck object 2 Øpf = p i Ø(m 1 + m 2)vf = m 1 v 1 i + m 2 v 2 i Ø 4000 vf = 800(+30) + 3200(0) = 24000 Øvf = 24000 / 4000 = +6 m/s

Conservation of Momentum Example 3 • A car of mass 1200 kg is traveling 30 m/s in the +xdirection • The car collides with a truck of 3200 kg traveling in the –x-direction at 20 m/s • After the collision, the truck comes to a stop • What is the velocity of the car? ØThe car is object 1 and the truck object 2 Øpf = p i Øm 1 v 1 f + m 2 v 2 f = m 1 v 1 i + m 2 v 2 i Ø(1200)v 1 f = 1200(30) + (3200)(-20) = -28000 Øv 1 f = -28000/1200 = -23. 33 m/s

PHY 101: Lecture 7 Impulse and Momentum 7. 3 Collisions in One Dimension

1 -Dimension Collisions • Total linear momentum is conserved when two objects collide • If they are an isolated system (no external forces) • Collisions are classified according to whether the total kinetic energy changes during the collision: ØElastic collision – Total kinetic energy of system after collision is equal to total kinetic energy before collision ØInelastic collision – Total kinetic energy of system is not the same before and after collision ØIf the objects stick together after colliding, the collision is said to be completely inelastic • When a collision is completely inelastic, the greatest amount of kinetic energy is lost

Elastic Collision 1 • Ball 1 initially is moving in the +x-direction • Ball 2 initially is not moving • The two balls collide and bounce off each other with no loss of total kinetic energy • Each ball moves off with a separate velocity • What are the velocities of the two balls after the collision?

Elastic Collision 2 • m 1 = m 2 • v 1 = 0 • m 1 >>> m 2 • v 1 = u 1 • m 1 <<< m 2 • v 1 = -u 1 v 2 = 2 u 1 v 2 = 0

PHY 101: Lecture 7 Impulse and Momentum 7. 4 Collisions in Two Dimension Skipped

PHY 101: Lecture 7 Impulse and Momentum 7. 5 Center of Mass Skipped