Momentum Conservation

Conservation of Linear MomentumConservation of Linear MomentumThe net force acting on an object is the rate of change of its momentum:

If the net force is zero, the momentum does not change:

Conservation of Linear MomentumConservation of Linear MomentumInternal Versus External Forces:

Internal forces act between objects within the system.

As with all forces, they occur in action-reaction pairs. As all pairs act between objects in the system, the internal forces always sum to zero:

Therefore, the net force acting on a system is the sum of the external forces acting on it.

Conservation of Linear MomentumConservation of Linear MomentumFurthermore, internal forces cannot change the momentum of a system.

However, the momenta of components of the system may change.

Conservation of Linear MomentumConservation of Linear Momentum An example of internal forces moving components of a system:

Canoe ExampleTwo groups of canoeists meet in the middle of a lake. After a brief visit, a person in Canoe 1 (total mass 130 kg) pushes on Canoe 2 (total mass 250 kg) with a force of 46 N to separate the canoes.

Find the momentum of each canoe after 1.20 s of pushing.

1, 21,

1

46 N0.354 m/s

130 kgx

x

Fa

m

2, 22,

2

46 N0.184 m/s

250 kgx

x

Fa

m

21, 1, ( 0.354 m/s )(1.20 s) 0.425 m/sx xv a t

22, 2, (184 m/s )(1.20 s) 0.221 m/sx xv a t

1, 1 1, (130 kg)( 0.425 m/s) 55.3 kg m/sx xp m v

2, 2 2, (250 kg)(0.221 m/s) 55.3 kg m/sx xp m v Net momentum = 0

Example: Velocity of a BeeExample: Velocity of a Bee

A honeybee with a mass of 0.150 g lands on one end of a floating 4.75 g popsicle stick. After sitting at rest for a moment, it runs to the other end of the stick with a velocity vb relative to still water. The stick moves in the opposite direction with a velocity of 0.120 cm/s.

Find the velocity vb of the bee.s s sp m v

Conservation of x-axis momentum: 0s bp p b b b s s sp m v p m v

(4.75 g)(0.120 cm/s) 3.80 cm/s

(0.150 g)s

b sb

mv v

m

Inelastic CollisionsInelastic Collisions

Collision: two objects striking one another.

Time of collision is short enough that external forces may be ignored.

Inelastic collision: momentum is conserved but kinetic energy is not: pf = pi but Kf ≠ Ki.

Completely inelastic collision: objects stick together afterwards: pf = pi1 + pi2

Inelastic CollisionsInelastic CollisionsA completely inelastic collision:

1 1, 2 2, 1 2( )i i fm v m v m m v

Example: Goal-Line StandExample: Goal-Line StandOn a touchdown attempt, a 95.0 kgrunning back runs toward the endzone at 3.75 m/s. A 111 kg line backermoving at 4.10 m/s meets the runnerin a head-on collision and locks hisarms around the runner.

(a) Find their velocity immediatelyafter the collisions.

(b) Find the initial and final kinetic energies and the energy K lost in the collision.

1 1, 2 2,

1 2

(95.0 kg)(3.75 m/s) (111.0 kg)( 4.10 m/s)0.480 m/s

(95.0 kg) (111.0 kg)i i

f

m v m vv

m m

1 12 21 22 2

(95.0 kg) (111.0 kg) ( 0.480 m/s) 23.7 Jf fK m m v

1 1 1 12 2 2 21 1, 2 2,2 2 2 2

(95.0 kg)(3.75 m/s) (111.0 kg)( 4.10 m/s) 1600 Ji i iK m v m v

1576 JK

Inelastic CollisionsInelastic CollisionsBallistic pendulum: the height h can be found using conservation of mechanical energy after the object is embedded in the block.

Example: Ballistic PendulumExample: Ballistic PendulumA projectile of mass mis fired with an initialspeed v0 at the bob of apendulum. The bob hasmass M and is suspendedby a rod of negligible mass.After the collision the projectile and bob stick together and swing at speed vf through an arc reaching height h.

Find the height h. 0Momentum Conservation: fmv m M v

1 22

Energy Conservation: fE m M v m M gh

Inelastic CollisionsInelastic Collisions For collisions in two dimensions, conservation of momentum is applied separately along each axis:

Example: A Traffic AccidentExample: A Traffic Accident A car of mass m1 = 950 kg and a speed v1,i = 16 m/s approaches an intersection. A minivan of mass m2 = 1300 kg and speed v2,i = 21 m/s enters the same intersection. The cars collide and stick together.

Find the direction and final speed vf of the wrecked vehicles just after the collision.

1 1 1 2x-momentum: ( ) cosfm v m m v

2 2 1 2y-momentum: ( ) sinfm v m m v

1 22 2

1 1 1 2

( ) sin sintan

( ) cos cosf

f

m m vm v

m v m m v

2 2

1 1

(1300 kg)(21 m/s)arctan arctan 61

(950 kg)(16 m/s)

m v

m v

1 1

1 2

(950 kg)(16 m/s)14 m/s

( )cos (950 kg) (1300 kg) cos61f

m vv

m m

Explosions An explosion in which the particles of a system move apart from each other after a brief, intense interaction, is the opposite of a collision. The explosive forces, which could be from an expanding spring or from expanding hot gases, are internal forces. If the system is isolated, its total momentum will be conserved during the explosion, so the net momentum of the fragments equals the initial momentum.

Elastic Collision31. • A 722-kg car stopped at an intersection is rear-ended by a 1620-kg truck moving with a speed of 14.5 m/s. If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of both vehicles after the collision.

before afterP P

' 'c c t t c c t tm v m v m v m v

0

before afterK K