Transcript Momentum and Collisions

Momentum and Collisions
Momentum and Impulse

The momentum of an object is the product
of its mass and velocity:
p=mv
 Units of momentum: kg·m/s
Momentum and Impulse
change in momentum
p
F

time interval
t
p mv f - mv i m(v f - vi )
F


t
t
t

If the resultant force F is zero, the
momentum of the object does not change.
Impulse-Momentum Theory
I = FΔt = Δp = mvf – mvi
(I = Impulse)

*If we exert a force on an object for a
time interval Δt, the effect of this force is
to change the momentum of the object
from some initial value mvi to some final
value mvf.
Example: A golf ball of mass 50. g is struck with a club. Assume that the ball
leaves the club face with a velocity of +44m/s.
A) Estimate the impulse due to the collision.
B) Estimate the length of time of the collision and the average force on the ball.
pi = mvi = 0 (ball starts at rest)
pf = mvf = (0.05 kg)(44 m/s)
pf = 2.2 kg·m/s
Impulse = Δp = pf – pi
Δp = 2.2 kg·m/s - 0 kg·m/s
Δp = 2.2 kg·m/s
Example: A golf ball of mass 50. g is struck with a club. Assume that the ball leaves
the club face with a velocity of +44m/s.
A) Estimate the impulse due to the collision.
B) Estimate the length of time of the collision and the average force on the ball.

Estimate distance ball travels while in
contact with club = displacement of ball
while in contact ≈ 2.0 cm
x
v
t
x 0.02m
-4
t 
 4.5 10 s

vf
44m/s
Ft  p
p 2.2kg  m/s
3
F

 4  4.9  10 N

t
4
.
5

10
s

Applications of impulse momentum
theory: Ft = mvf - mvi
Follow through in golf swing, batting,
tennis.
 Catching a water balloon.
 Moving with the punch in boxing.
 Padding boxing gloves, goalie gloves in
hockey, baseball mitts, inside of helmets…

Example: In a crash test, a car of mass 1500 kg collides with a wall. The initial
velocity of the car is 15.0 m/s east and the final velocity of the car is 2.6m/s
west. The collision lasts for 0.150 s. find (A) How can the car’s initial velocity
be to the east and its final velocity to the west? Calculate (B) the impulse due
to the collision and (C) the average force exerted on the car.
Example: In a crash test, a car of mass 1500 kg collides with a
wall. The initial velocity of the car is 15.0 m/s east and the
final velocity of the car is 2.6m/s west. The collision lasts for
0.150 s. find (A) How can the car’s initial velocity be to the
east and its final velocity to the west? Calculate (B) the
impulse due to the collision and (C) the average force exerted
on the car.
pi = mvi
pi = (1500kg)(-15m/s) = -2.25x104kg·m/s
pf = mvf
pf = (1500kg)(2.6m/s) = +3.90x103kg·m/s
I = Δp = pf – pi
I = 3900 – (-22500)
I = 2.6x104kg·m/s
Example: In a crash test, a car of mass 1500 kg collides with a wall. The
initial velocity of the car is 15.0 m/s east and the final velocity of the car is
2.6m/s west. The collision lasts for 0.150 s. find (A) How can the car’s
initial velocity be to the east and its final velocity to the west? Calculate
(B) the impulse due to the collision and (C) the average force exerted
on the car.
p
F
t
4
2.64 10 kg  m/s
F
0.150s
F  1.8 10 N
5
Rebounding

Ft = mvf - mvi
Is it better to bounce or to hit and
stick/stop?
Conservation of Momentum
pi = pf
 Law
of conservation of momentum:
when no external forces act on a
system consisting of 2 objects, the
total momentum of the system
before the collision is equal to the
total momentum of the system after
the collision.
Example: A baseball player uses a pitching machine to help him improve his
batting average. He places the 50.-kg machine on a frozen pond. The machine
fires a 0.15-kg baseball with a speed of 36 m/s in the horizontal direction. What
is the recoil velocity of the machine?
Example: A baseball player uses a pitching machine to help him improve his
batting average. He places the 50.-kg machine on a frozen pond. The machine
fires a 0.15-kg baseball with a speed of 36 m/s in the horizontal direction. What
is the recoil velocity of the machine?
pbefore = pafter
m1v1i + m2v2i = m1v1f + m2v2f
(+50.kg 0.15kg)(0) = (50.kg)(v1f) + (.15kg)(36m/s)
0 = (50.kg)(v1f) + (.15kg)(36m/s)
(50.kg)(v1f) = -5.4kg·m/s
v1f = -0.11 m/s
(moves .11 m/s opposite to the motion of the ball)
Collisions
 For
any type of collision, the total
momentum is conserved.
 However,
the total kinetic energy is
generally NOT conserved.
Elastic Collisions


Elastic collisions: both momentum and kinetic
energy are conserved.
Example: two balls bouncing perfectly off each
other.
m1v1i + m2v2i = m1v1f + m2v2f
and
½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2v2f2
Inelastic Collisions

Inelastic collisions: momentum is
conserved, but kinetic energy is not.
 When 2 objects collide and stick
together, the collision is PERFECTLY
INELASTIC; in this case, their final
velocities are the same.
For perfectly inelastic collisions,
m1v1i + m2v2i = (m1 + m2)v2f
Example: A large luxury car with a mass of 1800. kg stopped at a traffic light is
struck from the rear by a compact car with a mass of 900. kg. The two cars become
entangled as a result of the collision. If the compact car was moving at 20.0 m/s
before the collision, what is the velocity of the entangled mass after the collision?
Example: A large luxury car with a mass of 1800. kg stopped at a traffic
light is struck from the rear by a compact car with a mass of 900. kg.
The two cars become entangled as a result of the collision. If the
compact car was moving at 20.0 m/s before the collision, what is the
velocity of the entangled mass after the collision?
pbefore = pafter
M1v1i + m2v2i = (m1+m2)vf
(1800 kg)(0) + (900.kg)(20.0m/s) = (1800+900)vf
1.80 x 104 kg·m/s = (2700)vf
1.80x104 = (2700)vf
v1f = 6.67 m/s
Example: Two kids are playing marbles. The
shooter rolls toward a stationary marble at 5.0 cm/s.
They two collide and the smaller marble speeds away
at 8.0 cm/s. What is the final velocity of the shooter?
Assume the shooter is three times as massive as the
smaller marble. (have faith with the masses…)
pi = pf
3mvsi + mvmi = 3mvsf + mvmf
3m(5.0cm/s) + 0 = 3mvsf + m(8.0cm/s)
15 cm/s + 0 = 3vsf + 8.0cm/s
7cm/s = 3vsf
Vsf = 2.3 cm/s