![CALCULATINg THE CHANgE Of
MOMENTUM
[ ]0 ( )p m v mv∆ = − − =
[ ]( ) 2p m v v mv∆ = − − =
November 29, 2017
( )
after before
after before
after before
p p p
mv mv
m v v
∆ = −
= −
= −
For the teddy bear
For the bouncing ball](https://image.slidesharecdn.com/linearmomentumandcollison-171129072833/85/Linear-momentum-and-collison-10-320.jpg)
Linear momentum and collison
- 1. LINEAR MOMENTUM AND COLLISION SUBMITTED TO- MR.DC BHATT
- 2. INDEX • Conservation of Energy • Momentum • Impulse • Conservation of Momentum • 1-D Collisions • 2-D Collisions November 29, 2017
- 3. CONSERvATION Of ENERgy • ∆ E = ∆ K + ∆ U = 0 if conservative forces are the only forces that do work on the system. • The total amount of energy in the system is constant. • ∆ E = ∆ K + ∆ U = -fkd if friction forces are doing work on the system. • The total amount of energy in the system is still constant, but the change in mechanical energy goes into “internal energy” or heat. 2222 2 1 2 1 2 1 2 1 iiifff kxmgymvkxmgymv ++=++ November 29, 2017 ++− ++=− 2222 2 1 2 1 2 1 2 1 iiifffk kxmgymvkxmgymvdf
- 4. LINEAR MOMENTUM • This is a new fundamental quantity, like force, energy. It is a vector quantity (points in same direction as velocity). • The linear momentum p of an object of mass m moving with a velocity v is defined to be the product of the mass and velocity: • The terms momentum and linear momentum will be used interchangeably in the text • Momentum depend on an object’s mass and velocity vmp = November 29, 2017
- 5. MOMENTUM AND ENERgy • Two objects with masses m1 and m2 have equal kinetic energy. How do the magnitudes of their momenta compare? (A) p1 < p2 (B) p1 = p2 (C) p1 > p2 (D) Not enough information is given November 29, 2017
- 6. LINEAR MOMENTUM, CONT’D • Linear momentum is a vector quantity – Its direction is the same as the direction of the velocity • The dimensions of momentum are ML/T • The SI units of momentum are kg m / s • Momentum can be expressed in component form: px = mvx py = mvy pz = mvz m=p v November 29, 2017
- 7. NEwTON’S LAw AND MOMENTUM • Newton’s Second Law can be used to relate the momentum of an object to the resultant force acting on it • The change in an object’s momentum divided by the elapsed time equals the constant net force acting on the object t vm t v mamFnet ∆ ∆ = ∆ ∆ == )( netF t p == ∆ ∆ intervaltime momentuminchange November 29, 2017
- 8. IMpULSE • When a single, constant force acts on the object, there is an impulse delivered to the object – – is defined as the impulse – The equality is true even if the force is not constant – Vector quantity, the direction is the same as the direction of the force tFI ∆= November 29, 2017 I netF t p == ∆ ∆ intervaltime momentuminchange
- 9. IMpULSE-MOMENTUM THEOREM • The theorem states that the impulse acting on a system is equal to the change in momentum of the system if vmvmpI −=∆= November 29, 2017 ItFp net =∆=∆
- 10. CALCULATINg THE CHANgE Of MOMENTUM [ ]0 ( )p m v mv∆ = − − = [ ]( ) 2p m v v mv∆ = − − = November 29, 2017 ( ) after before after before after before p p p mv mv m v v ∆ = − = − = − For the teddy bear For the bouncing ball
- 11. November 29, 2017 How Good Are the Bumpers? In a crash test, a car of mass 1.5103 kg collides with a wall and rebounds as in figure. The initial and final velocities of the car are vi=-15 m/s and vf = 2.6 m/s, respectively. If the collision lasts for 0.15 s, find (a) the impulse delivered to the car due to the collision (b) the size and direction of the average force exerted on the car
- 12. November 29, 2017 How Good Are the Bumpers? In a crash test, a car of mass 1.5103 kg collides with a wall and rebounds as in figure. The initial and final velocities of the car are vi=-15 m/s and vf = 2.6 m/s, respectively. If the collision lasts for 0.15 s, find (a) the impulse delivered to the car due to the collision (b) the size and direction of the average force exerted on the car smkgsmkgmvp ii /1025.2)/15)(105.1( 43 ⋅×−=−×== N s smkg t I t p Fav 5 4 1076.1 15.0 /1064.2 ×= ⋅× = ∆ = ∆ ∆ = smkgsmkgmvp ff /1039.0)/6.2)(105.1( 43 ⋅×+=+×== smkg smkgsmkg mvmvppI ifif /1064.2 )/1025.2()/1039.0( 4 44 ⋅×= ⋅×−−⋅×= −=−=
- 13. Impulse-momentum theorem • A child bounces a 100 g superball on the sidewalk. The velocity of the superball changes from 10 m/s downward to 10 m/s upward. If the contact time with the sidewalk is 0.1s, what is the magnitude of the impulse imparted to the superball? (A) 0 (B) 2 kg-m/s (C) 20 kg-m/s (D) 200 kg-m/s (E) 2000 kg-m/s if vmvmpI −=∆= November 29, 2017
- 14. Impulse-momentum theorem 2 • A child bounces a 100 g superball on the sidewalk. The velocity of the superball changes from 10 m/s downward to 10 m/s upward. If the contact time with the sidewalk is 0.1s, what is the magnitude of the force between the sidewalk and the superball? (A) 0 (B) 2 N (C) 20 N (D) 200 N (E) 2000 N t vmvm t p t I F if ∆ − = ∆ ∆ = ∆ = November 29, 2017
- 15. ConservatIon of momentum • In an isolated and closed system, the total momentum of the system remains constant in time. – Isolated system: no external forces – Closed system: no mass enters or leaves – The linear momentum of each colliding body may change – The total momentum P of the system cannot change.. November 29, 2017
- 16. ConservatIon of momentum • Start from impulse-momentum theorem • Since • Then • So if vmvmtF 222212 −=∆ if vmvmtF 111121 −=∆ November 29, 2017 tFtF ∆−=∆ 1221 )( 22221111 ifif vmvmvmvm −−=− ffii vmvmvmvm 22112211 +=+
- 17. Conservation of Momentum • When no external forces act on a system consisting of two objects that collide with each other, the total momentum of the system remains constant in time • When then • For an isolated system • Specifically, the total momentum before the collision will equal the total momentum after the collision ffii vmvmvmvm 22112211 +=+ November 29, 2017 ifnet ppptF −=∆=∆ 0=∆p 0=netF if pp =
- 18. November 29, 2017 The Archer An archer stands at rest on frictionless ice and fires a 0.5-kg arrow horizontally at 50.0 m/s. The combined mass of the archer and bow is 60.0 kg. With what velocity does the archer move across the ice after firing the arrow? ffii vmvmvmvm 22112211 +=+ fi pp = ?,/50,0,5.0,0.60 122121 ====== ffii vsmvvvkgmkgm ff vmvm 22110 += smsm kg kg v m m v ff /417.0)/0.50( 0.60 5.0 2 1 2 1 −=−=−=
- 19. ConservatIon of momentum • A 100 kg man and 50 kg woman on ice skates stand facing each other. If the woman pushes the man backwards so that his final speed is 1 m/s, at what speed does she recoil? (A) 0 (B) 0.5 m/s (C) 1 m/s (D) 1.414 m/s (E) 2 m/s November 29, 2017
- 20. types of CollIsIons • Momentum is conserved in any collision • Inelastic collisions: rubber ball and hard ball – Kinetic energy is not conserved – Perfectly inelastic collisions occur when the objects stick together • Elastic collisions: billiard ball – both momentum and kinetic energy are conserved • Actual collisions – Most collisions fall between elastic and perfectly inelastic collisions November 29, 2017
- 21. Collisions summary • In an elastic collision, both momentum and kinetic energy are conserved • In a non-perfect inelastic collision, momentum is conserved but kinetic energy is not. Moreover, the objects do not stick together • In a perfectly inelastic collision, momentum is conserved, kinetic energy is not, and the two objects stick together after the collision, so their final velocities are the same • Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types • Momentum is conserved in all collisions November 29, 2017
- 22. more about PerfeCtly inelastiC Collisions • When two objects stick together after the collision, they have undergone a perfectly inelastic collision • Conservation of momentum • Kinetic energy is NOT conserved 21 2211 mm vmvm v ii f + + = November 29, 2017 fii vmmvmvm )( 212211 +=+
- 23. more about elastiC Collisions • Both momentum and kinetic energy are conserved • Typically have two unknowns • Momentum is a vector quantity – Direction is important – Be sure to have the correct signs • Solve the equations simultaneously November 29, 2017 2 22 2 11 2 22 2 11 22112211 2 1 2 1 2 1 2 1 ffii ffii vmvmvmvm vmvmvmvm +=+ +=+
- 24. elastiC Collisions • A simpler equation can be used in place of the KE equation iffi vvvv 2211 +=+ ffii vmvmvmvm 22112211 +=+ November 29, 2017 )vv(vv f2f1i2i1 −−=− 2 22 2 11 2 22 2 11 2 1 2 1 2 1 2 1 ffii vmvmvmvm +=+ ))(())(( 2222211111 ififfifi vvvvmvvvvm +−=+− )()( 222111 iffi vvmvvm −=− )()( 2 2 2 22 2 1 2 11 iffi vvmvvm −=− ffii vmvmvmvm 22112211 +=+
- 25. summary of tyPes of Collisions• In an elastic collision, both momentum and kinetic energy are conserved • In an inelastic collision, momentum is conserved but kinetic energy is not • In a perfectly inelastic collision, momentum is conserved, kinetic energy is not, and the two objects stick together after the collision, so their final velocities are the same November 29, 2017 iffi vvvv 2211 +=+ ffii vmvmvmvm 22112211 +=+ ffii vmvmvmvm 22112211 +=+ fii vmmvmvm )( 212211 +=+
- 26. Conservation of momentum • An object of mass m moves to the right with a speed v. It collides head-on with an object of mass 3m moving with speed v/3 in the opposite direction. If the two objects stick together, what is the speed of the combined object, of mass 4m, after the collision? (A) 0 (B) v/2 (C) v (D) 2v (E) 4v November 29, 2017
- 27. Problem Solving for 1D Collisions, 1 • Coordinates: Set up a coordinate axis and define the velocities with respect to this axis – It is convenient to make your axis coincide with one of the initial velocities • Diagram: In your sketch, draw all the velocity vectors and label the velocities and the masses November 29, 2017
- 28. Problem Solving for 1D Collisions, 2 • Conservation of Momentum: Write a general expression for the total momentum of the system before and after the collision – Equate the two total momentum expressions – Fill in the known values November 29, 2017 ffii vmvmvmvm 22112211 +=+
- 29. Problem Solving for 1D Collisions, 3 • Conservation of Energy: If the collision is elastic, write a second equation for conservation of KE, or the alternative equation – This only applies to perfectly elastic collisions • Solve: the resulting equations simultaneously November 29, 2017 iffi vvvv 2211 +=+
- 30. One-Dimension vs Two-Dimension November 29, 2017
- 31. Two-Dimensional Collisions • For a general collision of two objects in two-dimensional space, the conservation of momentum principle implies that the total momentum of the system in each direction is conserved November 29, 2017 fyfyiyiy fxfxixix vmvmvmvm vmvmvmvm 22112211 22112211 +=+ +=+
- 32. Two-Dimensional Collisions • The momentum is conserved in all directions • Use subscripts for – Identifying the object – Indicating initial or final values – The velocity components • If the collision is elastic, use conservation of kinetic energy as a second equation – Remember, the simpler equation can only be used for one-dimensional situations fyfyiyiy fxfxixix vmvmvmvm vmvmvmvm 22112211 22112211 +=+ +=+ November 29, 2017 iffi vvvv 2211 +=+
- 33. 2-D Collision, example • Particle 1 is moving at velocity and particle 2 is at rest • In the x-direction, the initial momentum is m1v1i • In the y-direction, the initial momentum is 0 November 29, 2017 1iv r
- 34. 2-D Collision, example cont • After the collision, the momentum in the x-direction is m1v1fcos θ+ m2v2f cos φ • After the collision, the momentum in the y-direction is m1v1fsin θ+ m2v2f sin φ • If the collision is elastic, apply the kinetic energy equation φθ φθ sinsin00 coscos0 2211 221111 ff ffi vmvm vmvmvm −=+ +=+ November 29, 2017 2 22 2 11 2 11 2 1 2 1 2 1 ffi vmvmvm +=
- 35. November 29, 2017 Collision at an Intersection A car with mass 1.5×103 kg traveling east at a speed of 25 m/s collides at an intersection with a 2.5×103 kg van traveling north at a speed of 20 m/s. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision and assuming that friction between the vehicles and the road can be neglected. ??,/20,/25 105.2,105.1 33 ==== ×=×= θfviycix vc vsmvsmv kgmkgm
- 36. November 29, 2017 Collision at an Intersection ??,m/s20,m/s25 kg105.2,kg105.1 33 ==== ×=×= θfviycix vc vvv mm ∑ ⋅×==+= m/skg1075.3 4 cixcvixvcixcxi vmvmvmp ∑ +=+= θcos)( fvcvfxvcfxcxf vmmvmvmp θcos)kg1000.4(m/skg1075.3 34 fv×=⋅× ∑ ⋅×==+= m/skg1000.5 4 viyvviyvciycyi vmvmvmp ∑ +=+= θsin)( fvcvfyvcfycyf vmmvmvmp θsin)kg1000.4(m/skg1000.5 34 fv×=⋅×
- 37. November 29, 2017 Collision at an Intersection ??,/20,/25 105.2,105.1 33 ==== ×=×= θfviycix vc vsmvsmv kgmkgm 33.1 /1075.3 /1000.5 tan 4 4 = ⋅× ⋅× = smkg smkg θ 1.53)33.1(tan 1 == − θ m/s6.15 1.53sin)kg1000.4( m/skg1000.5 3 4 = × ⋅× = fv θcos)kg1000.4(m/skg1075.3 34 fv×=⋅× θsin)kg1000.4(m/skg1000.5 34 fv×=⋅×