AP Physics B - Impulse and Momentum PDF
AP Physics B - Impulse and Momentum PDF
AP Physics B - Impulse and Momentum PDF
AP Physics B
Impulse = Momentum
Consider Newtons 2nd Law and
the definition of acceleration
Units of Impulse: Ns
Units of Momentum: Kg x m/s
Example
A 100 g ball is dropped from a height of h = 2.00 m above the floor. It
rebounds vertically to a height of h'= 1.50 m after colliding with the
floor. (a) Find the momentum of the ball immediately before it collides
with the floor and immediately after it rebounds, (b) Determine the
average force exerted by the floor on the ball. Assume that the time
interval of the collision is 0.01 seconds.
EB = E A
Uo = K
mgho = 1 mv 2
2
v = 2 gho = 2 * 9.8 * 2 = 6.26 m / s
EB = E A
Ko = U
v = 2 gh = 2 * 9.8 *1.5 = 5.4 m / s
r
r
p = mv
pbefore = 0.100(6.26) = 0.626 kg * m / s
F = 116.6 N
Ft = mv = m(v vo )
F1 = F2
t1 = t 2
( Ft )1 = ( Ft ) 2
J1 = J 2
J1 = J 2
p1 = p2
m1v1 = m2 v2
m1 (v1 vo1 ) = m2 (v2 vo 2 )
m1v1 m1vo1 = m2 v2 + m2 vo 2
before
= p after
m1vo1 + m2 vo 2 = m1v1 + m2 v2
Momentum is conserved!
The Law of Conservation of Momentum: In the absence of
an external force (gravity, friction), the total
momentum before the collision is equal to the total
momentum after the collision.
po ( truck ) = mvo = (500)(5) = 2500kg * m / s
po ( car ) = (400)(2) = 800kg * m / s
po ( total ) = 3300kg * m / s
ptruck = 500 * 3 = 1500kg * m / s
pcar = 400 * 4.5 = 1800kg * m / s
ptotal = 3300kg * m / s
before
= p after
Example
EB = E A
pB = p A
K o ( swing ) = U swing
mT voT = m1v1 + m2 v2
1 mvo2 = mgh
2
vo2
(0.68) 2
=h=
= 0.024 m
2g
19.6
-0.680 m/s
Example
2
How many objects do I have after the collision?
pb = pa
m1vo1 + m2 vo 2 = mT vT
(80)(6) + (40)(0) = 120vT
vT = 4 m/s
Collisions in 2 Dimensions
The figure to the left shows a
collision between two pucks
on an air hockey table. Puck A
has a mass of 0.025-kg and is
vA
moving along the x-axis with a
vAsin
velocity of +5.5 m/s. It makes
a collision with puck B, which
has a mass of 0.050-kg and is
vAcos
initially at rest. The collision is
NOT head on. After the
vBcos
vBsin collision, the two pucks fly
vB
apart with angles shown in the
drawing. Calculate the speeds
of the pucks after the collision.
Collisions in 2 dimensions
p
ox
= px
vA
vAsin
vAcos
vBcos
vB
oy
= py
0 = m Av yA + mB v yB
vBsin
0.0300vB = 0.0227v A
vB = 0.757v A
Collisions in 2 dimensions
0.1375 = 0.0106v A + 0.040vB
vB = 0.757v A
0.1375 = 0.0106v A + (0.050)(0.757v A )
0.1375 = 0.0106v A + 0.03785v A
0.1375 = 0.04845v A
v A = 2.84m / s
vB = 0.757(2.84) = 2.15m / s