2 Momentum is conserved in this rocket launch Momentum is conserved in this rocket launch. The velocity of the rocket and its payload is determined by the mass and velocity of the expelled gases. Photo: NASANASA
3 Objectives: After completing this module, you should be able to: State the law of conservation of momentum and apply it to the solution of problems.Distinguish by definition and example between elastic and inelastic collisions.Predict the velocities of two colliding bodies when given the coefficients of restitution, masses, and initial velocities.
4 A Collision of Two Masses When two masses m1 and m2 collide, we will use the symbol u to describe velocities before collision.m1u1m2u2BeforeThe symbol v will describe velocities after collision.Afterm1v1m2v2
5 A Collision of Two Blocks m1u1m2u2BeforeCollisionm1Bm2“u”= Before“v” = Afterm2v2m1v1After
6 Conservation of Energy m1m2u1u2The kinetic energy before colliding is equal to the kinetic energy after colliding plus the energy lost in the collision.
7 Example 1. A 2-kg mass moving at 4 m/s collides with a 1-kg mass initially at rest. After the collision, the 2-kg mass moves at 1 m/s and the 1-kg mass moves at 3 m/s. What energy was lost in the collision?It’s important to draw and label a sketch with appropriate symbols and given information.m2u2 = 0m1u1 = 4 m/sm1 = 2 kgm1 = 1 kgBEFOREm2v2 = 2 m/sm1v1 = 1 m/sm1 = 2 kgm1 = 1 kgAFTER
8 Energy Conservation: K(Before) = K(After) + Loss Example 1 (Continued). What energy was lost in the collision? Energy is conserved.m2u2 = 0m1u1 = 4 m/sm1 = 2 kgm1 = 1 kgv2 = 2 m/sv1 = 1 m/sBEFORE:AFTEREnergy Conservation: K(Before) = K(After) + LossLoss = 16 J – 3 JEnergy Loss = 15 J
10 Conservation of Momentum The total momentum AFTER a collision is equal to the total momentum BEFORE.mAvA + mBvB = mAuA + mBuBABuAuBvAvB-FADtFB DtRecall that the total energy is also conserved:Kinetic Energy: K = ½mv2KA0 + KB0 = KAf + KBf + Loss
11 mAvA + mBvB = mAuA + mBuB mBvB mA mAvA = - mBvB vA = - Example 2: A 2-kg block A and a 1-kg block B are pushed together against a spring and tied with a cord. When the cord breaks, the 1-kg block moves to the right at 8 m/s. What is the velocity of the 2 kg block?ABThe initial velocities are zero, so that the total momentum before release is zero.mAvA + mBvB = mAuA + mBuBvA = -mBvBmAmAvA = - mBvB
13 Example 2 (Cont.): Ignoring friction, how much energy was released by the spring? 2 kg1 kgAB8 m/s4 m/sCons. of E: ½kx2 = ½ mAvA + ½mBvB22½kx2 = ½(2 kg)(4 m/s)2 + ½(1 kg)(8 m/s)2½kx2 = 48 J½kx2 = 16 J + 32 J = 48 J
14 Elastic or Inelastic?An elastic collision loses no energy. The deform-ation on collision is fully restored.In an inelastic collision, energy is lost and the deformation may be permanent. (Click it.)
15 Completely Inelastic Collisions Collisions where two objects stick together and have a common velocity after impact.BeforeAfter
16 Example 3: A 60-kg football player stands on a frictionless lake of ice. He catches a 2-kg football and then moves at 40 cm/s. What was the initial velocity of the football?AGiven: uB= 0; mA= 2 kg; mB= 60 kg; vA= vB= vC vC = 0.4 m/sBMomentum:mAvA + mBvB = mAuA + mBuBInelastic collision:(mA + mB)vC = mAuA(2 kg + 60 kg)(0.4 m/s) = (2 kg)uAuA= 12.4 m/s
17 Example 3 (Cont.): How much energy was lost in catching the football? ½(2 kg)(12.4 m/s)2 = ½(62 kg)(0.4 m/s)2 + Loss154 J = 4.96 J + LossLoss = 149 J97% of the energy is lost in the collision!!
18 General: Completely Inelastic Collisions where two objects stick together and have a common velocity vC after impact.Conservation of Momentum:Conservation of Energy:
19 Common speed after colliding: 2.4 m/s. Example 4. An 87-kg skater B collides with a 22-kg skater A initially at rest on ice. They move together after the collision at 2.4 m/s. Find the velocity of the skater B before the collision.ABuB = ?uA = 0Common speed after colliding: 2.4 m/s.22 kg87 kgvB= vA = vC = 2.4 m/s(87 kg)uB = (87 kg + 22 kg)(2.4 m/s)(87 kg)uB =262 kg m/suB = 3.01 m/s
20 Example 5: A 50 g bullet strikes a 1-kg block, passes all the way through, then lodges into the 2 kg block. Afterward, the 1 kg block moves at 1 m/s and the 2 kg block moves at 2 m/s. What was the entrance velocity of the bullet?1 kg2 kguA= ?2 kg1 kg1 m/s2 m/s
21 mAuA + mBuB + mCuC = mBvB + (mA+mC) vAC 50 gFind entrance velocity of bullet: mA= 0.05 kg; uA= ?2 kg1 kg1 m/s2 m/sMomentum After = Momentum BeforemAuA + mBuB + mCuC = mBvB + (mA+mC) vAC(0.05 kg)uA =(1 kg)(1 m/s)+(2.05 kg)(2 m/s)(0.05 kg) uA =(5.1 kg m/s)uA= 102 m/s
22 Completely Elastic Collisions Collisions where two objects collide in such a way that zero energy is lost in the process.APPROXIMATIONS!
23 Velocity in Elastic Collisions uBuA1. Zero energy lost.AB2. Masses do not change.vAvBAB3. Momentum conserved.Equal but opposite impulses (F Dt) means that:(Relative Dv After) = - (Relative Dv Before)vA - vB = - (uA - uB)For elastic collisions:
24 From conservation of energy (relative v): Example 6: A 2-kg ball moving to the right at 1 m/s strikes a 4-kg ball moving left at 3 m/s. What are the velocities after impact, assuming complete elasticity?AB3 m/s1 m/svAvB1 kg2 kgvA - vB = - (uA - uB)vA - vB = uB - uAvA - vB = (-3 m/s) - (1 m/s)From conservation of energy (relative v):vA - vB = - 4 m/s
25 Example 6 (Continued) mAvA + mBvB = mAuA + mBuB vA + 2vB = -5 m/s 1 kg2 kgEnergy: vA - vB = - 4 m/sMomentum also conserved:mAvA + mBvB = mAuA + mBuB(1 kg)vA+(2 kg)vB=(1 kg)(1 m/s)+(2 kg)(-3 m/s)vA + 2vB = -5 m/sTwo independent equations to solve:vA - vB = - 4 m/s
26 Example 6 (Continued) vA + 2vB = -5 m/s vA - vB = - 4 m/s vA 1 kg2 kgvA + 2vB = -5 m/svA - vB = - 4 m/sSubtract:0 + 3vB2 = - 1 m/svB = m/svA2 - ( m/s) = - 4 m/sSubstitution:vA= m/svA - vB = - 4 m/s
27 (0.150 kg)vA+ (2 kg)(40 m/s) = (0.150 kg)(715 m/s) Example 7. A kg bullet is fired at 715 m/s into a 2-kg wooden block at rest. The velocity of block afterward is 40 m/s. The bullet passes through the block and emerges with what velocity?BAuB = 0(0.150 kg)vA+ (2 kg)(40 m/s) = (0.150 kg)(715 m/s)0.150vA+ (80 m/s) = (107 m/s)0.150vA = 27.2 m/s)vA = 181 m/s
28 Example 8a: Inelastic collision: Find vC. B5 kg7.5 kguB=02 m/sAfter hit: vB= vA= vC(5 kg)(2 m/s) = (5 kg kg)vCABCommon vC aftervC12.5 vC =10 m/svC = m/sIn an completely inelastic collision, the two balls stick together and move as one after colliding.
29 Example 8. (b) Elastic collision: Find vA2 and vB2 5 kg7.5 kgvB1=02 m/sConservation of Momentum:(5 kg)(2 m/s) = (5 kg)vA2 + (7.5 kg) vBABvAvB5 vA vB = 10 m/sFor Elastic Collisions:Continued . . .
30 Example 8b (Cont). Elastic collision: Find vA & vB Solve simultaneously:AB5 kg7.5 kgvB =02 m/svAvBx (-5)5 vA v B = 10 m/s5 vA vB = 10 m/s-5 vA vB = +10 m/svA m/s = -2 m/svA = m/s12.5 vB = 20 m/svB = 1.60 m/s
31 General: Completely Elastic Collisions where zero energy is lost during a collision (an ideal case).Conservation of Momentum:Conservation of Energy:
32 The ballistic pendulum! Example 9: A 50 g bullet lodges into a 2-kg block of clay hung by a string. The bullet and clay rise together to a height of 12 cm. What was the velocity of the 50-g mass just before entering?uABABA12 cmThe ballistic pendulum!
33 Example (Continued): mAuA+0= (mA+mB)vC vC = 2gh Collision and Momentum:BA12 cm50 guA2.05 kg2 kgmAuA+0= (mA+mB)vC(0.05 kg)uA = (2.05 kg)vCTo find vA we need vC .After collision, energy is conserved for masses.vC = 2gh
34 Example (Continued): mAuA+0= (mA+mB)vC uA = 62.9 m/s vC = 2gh = 2(9.8)(0.12)BA12 cm50 guA2.05 kg2 kgAfter Collision: vC = 1.53 m/sMomentum Also Conserved:mAuA+0= (mA+mB)vC(0.05 kg)uA = (2.05 kg)(1.53 m/s)uA = 62.9 m/s
35 Conservation of Momentum: Summary of Formulas:Conservation of Momentum:Conservation of Energy:For elastic only:
36 CONCLUSION: Chapter 9B Conservation of Momentum