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Momentum and Collisions Dr. Robert MacKay Clark College, Physics Introduction Review Newtons laws of motion Define Momentum Define Impulse Conservation of Momentum Collisions Explosions Elastic Collisions Introduction Newtons 3 laws of motion 1. Law of inertia 2. Net Force = mass x acceleration (F=MA) 3. Action Reaction Law of interia (1st Law) Every object continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it. acceleration = 0.0 unless the objected is acted on by an unbalanced force Newton’s 2nd Law Net Force = Mass x Acceleration F = MA Newton’s Law of Action Reaction (3rd Law) You can not touch without being touched For every action force there is and equal and oppositely directed reaction force Newton’s Law of Action Reaction (3rd Law) Ball 1 F2,1 Ball 2 F1,2 F1,2 = - F2,1 For every action force there is and equal and oppositely directed reaction force Momentum , p Momentum = mass x velocity is a Vector has units of kg m/s Momentum , p (a vector) Momentum = mass x velocity p=m v p=? 8.0 kg 6.0 m/s Momentum , p Momentum = mass x velocity p=m v p = 160.0 kg m/s 8.0 kg V= ? Momentum , p Momentum is a Vector p=m v p1 = ? p2 = ? V= +8.0 m/s m1= 7.5 kg V= -6.0 m/s m2= 10.0 kg Momentum , p Momentum is a Vector p=m v p1 = +60 kg m/s p2 = - 60 kg m/s m1= 7.5 kg V= -6.0 m/s V= +8.0 m/s m2= 10.0 kg Momentum , p Momentum is a Vector p=m v p1 = +60 kg m/s p2 = - 60 kg m/s the system momentum is zero., m1= 7.5 kg V= -6.0 m/s V= +8.0 m/s m2= 10.0 kg Momentum , p Momentum is a Vector p=m v Total momentum of a system is a vector sum: p1+p2+p3+…….. p3 p2 ptotal p1 Newton’s 2nd Law Net Force = Mass x Acceleration F=M a F = M (∆V/∆t) F ∆t = M ∆V F ∆t = M (V1-V2) F ∆t = M V1 - M V2 F ∆t = ∆p Impulse= F∆t The Impulse = the change in momentum Newton’s 2nd Law Net Force = Mass x Acceleration F=M a or F = ∆p/ ∆t Newton’s 2nd Law Net Force = Mass x Acceleration F ∆t = ∆p Impulse= F ∆t The Impulse = the change in momentum If M=1500 kg and Dt=0.4 sec, Find Dp and Favg 30° 50° Impulse The Impulse = the change in momentum F ∆t = ∆p Impulse The Impulse = the change in momentum F ∆t = ∆p Newton’s Law of Action Reaction (3rd Law) Ball 1 F2,1 Ball 2 F1,2 F1,2 = - F2,1 For every action force there is and equal and oppositely directed reaction force Newton’s Law of Action Reaction (3rd Law) Ball 1 F2,1 Ball 2 F1,2 F1,2 = - F2,1 F1,2∆t = - F2,1 ∆t ∆p2 = - ∆p1 Conservation of momentum Ball 1 F2,1 Ball 2 F1,2 If there are no external forces acting on a system (i.e. only internal action reaction pairs), then the system’s total momentum is conserved. “Explosions” 2 objects initially at rest A 30 kg boy is standing on a stationary 100 kg raft in the middle of a lake. He then runs and jumps off the raft with a speed of 8.0 m/s. With what V=8.0 m/s speed does the raft recoil? after V=? M=100.0 kg before M=100.0 kg “Explosions” 2 objects initially at rest A 30 kg boy is standing on a stationary 100 kg raft in the middle of a lake. He then runs and jumps off the raft with a speed of 8.0 m/s. With what speed does the raft recoil? after V=8.0 m/s V=? M=100.0 kg before M=100.0 kg p before = p after 0 = 30kg(8.0 m/s) - 100 kg V 100 kg V = 240 kg m/s V = 2.4 m/s Explosions If Vred=8.0 m/s Vblue=? “Stick together” 2 objects have same speed after colliding A 30 kg boy runs and jumps onto a stationary 100 kg raft with a speed of 8.0 m/s. How fast does he and the raft move immediately after the collision? V=8.0 m/s V=? M=100.0 kg before M=100.0 kg after “Stick together” 2 objects have same speed after colliding A 30 kg boy runs and jumps onto a stationary 100 kg raft with a speed of 8.0 m/s. How fast does he and the raft move immediately after the collision? V=8.0 m/s before M=100.0 kg V=? M=100.0 kg p before = p after 30kg(8.0 m/s) = 130 kg V 240 kg m/s = 130 kg V V = 1.85 m/s after “Stick together” 2 objects have same speed after colliding This is a perfectly inelastic collision V=8.0 m/s before M=100.0 kg V=? M=100.0 kg after A 30 kg boy runs and jumps onto a stationary 100 kg raft with a speed of 8.0 m/s. How fast does he and the raft move immediately after the collision? A 20 g bullet lodges in a 300 g Pendulum. The pendulum and bullet then swing up to a maximum height of 14 cm. What is the initial speed of the bullet? mv = (m+M) V Before and After Collision 1/ 2(m+M)V 2=(m+M)gh After collision but Before and After moving up 2-D Collisions X axis m1V10 = m1v1cos(50) + m2v2cos(40) Y axis 0 = m1v1sin(50) - m2v2sin(40) 2-D Stick together (Inelastic) Momentum Before = Momentum After P before= P after For both the x & y components of P. A 2000 kg truck traveling 50 mi/hr East on McLoughlin Blvd collides and sticks to a 1000 kg car traveling 30 mi/hr North on Main St. What is the final velocity of the wreck? Give both the magnitude and direction OR X and Y components. 2-D Stick together (Inelastic) Momentum Before = Momentum After P before= P after For both the x & y components of P. A 2000 kg truck traveling 50 mi/hr East on McLoughlin Blvd collides and sticks to a 1000 kg car traveling 30 mi/hr North on Main St. What is the final velocity of the wreck? Give both the magnitude and direction OR X and Y components. 2-D Stick together (Inelastic) A 2000 kg truck traveling 50 mi/hr East (V1) on McLoughlin Blvd collides and sticks to a 1000 kg car traveling 30 mi/hr North (V2) on Main St. What is the final velocity (V) of the wreck? Give both magnitude and direction OR X and Y components. P =P before PBx=PAy 2000Kg(50 mi/hr)=3000KgVx Vx=33.3 mi/h after & PBy=PAy & 1000kg(30mi/hr)=3000kgVy & Vy=10 mi/hr Or V= 34.8mi/hr = (sqrt(Vx2+Vx2) & q =16.7° = tan-1(Vy/Vx) V V1 2000Kg 3000Kg V2 1000Kg Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after m1 v1 m2 m1 v1,f m2 v2,f Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after m1v1 m1v1,f m 2 v2,f m1 v1 v1, f m2 v 2,f 1 1 1 2 m1v12 m1v1,f m 2 v22,f 2 2 2 2 m1v12 m1v1,f m 2 v22,f 2 m1v12 m1v1,f m 2 v22, f m1v1 v1,f v1 v1,f m 2 v2,f v2, f v1 v1,f v 2,f 2 m1 v12 v1,f m2 v 22,f v1 v2,f v1,f Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after m1 v1 v1, f m2 v 2,f v1 v2,f v1,f m1v1 m1v 2,f m1v1,f m1v1 m1v1,f m 2 v2,f + m1v1 m1v1,f m 2 v2,f m1v1 m1v 2,f m1v1,f 2m1v1 m1 m 2 v 2,f or v 2,f 2m1v1 m1 m 2 Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after m1 v1 v1, f m2 v 2,f v1 v2,f v1,f m 2 v1 m 2 v2,f m 2 v1,f m1v1 m1v1,f m 2 v2,f or m1v1 m1v1,f m 2 v2,f m 2 v1 m 2 v2,f m 2 v1,f m1 m 2 v1 m1 m 2 v1,f v1,f m1 m 2 v1 m1 m 2 Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after v1,f m1 v1 m1 m 2 v1 m1 m 2 m2 & v 2,f m1 2m1v1 m1 m 2 v1,f m2 v2,f Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after v1,f m m 2 v1 1 m1 m 2 & v 2,f 2m1v1 m1 m 2 if m1 = m2 = m, then v1,f = 0.0 & v2,f = v1 m m v1 m m v1,f= 0.0 v2,f = v1 Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after v1,f m m 2 v1 1 m1 m 2 & v 2,f if m1 <<< m2 , then m1+m2 ≈m2 v1,f = - v1 & v2,f ≈ 0.0 M m 2m1v1 m1 m 2 & m1-m2 ≈ -m2 m v1 v1,f=- v1 M v2,f ≈ 0.0 Elastic Collisions Bounce off without loss of energy if m1 <<< m2 and v 2 is NOT 0.0 v1 v2,f v1,f Speed of Approach = Speed of separation (True of all elastic collisions) M m v1 v2 m v1,f=- (v1 +v2 +v2) M v2,f ≈ v2 Elastic Collisions if m1 <<< m2 and v 2 is NOT 0.0 Speed of Approach = Speed of separation (True of all elastic collisions) A space ship of mass 10,000 kg swings by Jupiter in a psuedo elastic head-on collision. If the incoming speed of the ship is 40 km/sec and that of Jupiter is 20 km/sec, with what speed does the space ship exit the gravitational field of Jupiter? v2=20 km/s m v1 = 40 km/s M v1,f=? m M v2,f ≈ ? if m1 <<<Elastic m2 and vCollisions 2 is NOT 0.0 Speed of Approach = Speed of seperation (True of all elastic collisions) A little boy throws a ball straight at an oncoming truck with a speed of 20 m/s. If truck’s speed is 40 m/s and the collision is an elastic head on collision, with what speed does the ball bounce off the truck? v2=40 m/s m M v1,f=? m v1 = 20 m/s v2,f ≈ ? M Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after if m1 = m2 = m, then v1,f = 0.0 & v2,f = v1 1 m 2 v2 p2 2 KE mv 2 2m 2m m m m v1 m 90° Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after if m1 = m2 = m, then v1,f = 0.0 & v2,f = v1 p2f m m 1 m 2 v2 p2 2 KE mv 2 2m 2m 2 p12 p1f p22f 2m 2m 2m 90° p1 v1 p1 =p1f +p2f p1f p12 p1f2 p22f m m 90° Center of Mass The average position of the mass When we use F=ma We really mean F = m acm The motion of an object is the combination of – The translational motion of the CM – Rotation about the CM Center of mass m1x1 m 2 x2 m 3x3 ....... m n xn X cm m1 m 2 m 3 .......m n The average position of the mass CM=? M1=6.0 kg and M2=8.0 kg Ycm= Xcm= CM Center of Gravity 8 kg at (0,3) 4 kg at (-2,0) Where must a 10 kg mass be placed so the center of mass of the three mass system is at (0,0) ? (M+Dm)v=M(v+Dv)+Dm(v-ve) M Dv= Dmve Thrust=ve(dM/dt) t(sec) M(kg) y(m) V(m/s) 0 360 0 0 1 357.5 5.2083 10.417 2 355 20.87 20.906 4000 3500 3000 2500 2000 1500 dM=2.5 kg/s and dt =1.0 sec 1000 500 0 A 5 6 7 8 9 t(sec) 0 =A5+dt =A6+dt =A7+dt =A8+dt B M(kg) 360 =B5-dM*dt =B6-dM*dt =B7-dM*dt =B8-dM*dt C y(m) 0 =C5+(D5+D6)/2*dt =C6+(D6+D7)/2*dt =C7+(D7+D8)/2*dt =C8+(D8+D9)/2*dt D 0 50 100 V(m/s) 0 =D5+dM*dt/B5*1500 =D6+dM*dt/B6*1500 =D7+dM*dt/B7*1500 =D8+dM*dt/B8*1500 The End