Download Physics 130 - UND: University of North Dakota

10/25 Momentum in 2D
Text: Chapter7
HW: HW Handout “Pendulum and Block”
Due Wednesday, 10/30
HW Questions?
Examples
Impulse
Simply the name we give Fnett
Another way to find the change in momentum
when we know the force and the time
Can also find force if we know the change in
momentum and the time.
Two frictionless rocket sleds are fired from rest toward each other
and collide after they both burn out. They stick together after the
collision. What is the final velocity of the system?
vi = 0
1
m = 60kg
Thrust = 50N
t = 20s (burn time)
1
1
2
vSYS,f = ?
What you know
vi = 0
2
m = 30kg
Thrust = 25N
t = 10s (burn time)
2
What is the total momentum of the system after they collide?
The same. No net force so FSYS,net = 0 so pSYS = 0
What is the final velocity of the system after they collide?
pSYS = 750kg m/s = mSYSvSYS
vSYS = 750/90 = 8.3m/s right,
same direction as pSYS
Would the answer change if the rockets
NO!!!!! Once the collision is done and
were still burning and accelerating as
the rockets are burnt out, the end result
they collided?
is the same.
Now a third rocket sled is added as shown in the top view below. All
three stick together after the collision. What is the final velocity of
the system?
v
=?
SYS,f
vi = 0
1
m = 60kg
NGAS,R = 50N
t = 20s (burn time)
Fnett = p
Our new
friend!
1
2
3
vi = 0
3
vi = 0
2
m = 30kg
NGAS,R = 25N
t = 10s (burn time)
m = 45kg
NGAS,R = 15N
t = 30s (burn time)
Can we take advantage of “the system” and find the total impulse on the system?
Impulse = the sum of all the individual impulses.
= 1000kg m/s right + 250kg m/s left + 450kg m/s up
750kg m/s right
Now a third rocket sled is added as shown in the top view below. All
three stick together after the collision. What is the final velocity of
the system?
vSYS,f = ?
450
1

2
3
750
Impulse = the sum of all the individual impulses.
= 1000kg m/s right + 250kg m/s left + 450kg m/s up
750kg m/s right
pSYS = (750)2 + (450)2 = 875kg m/s
vSYS = 875/mTotal = 875/135 = 6.5m/s
pSYS,i = 0
in the same direction as pSYS,f
pSYS,f = 875kg m/s
May seem odd that v is less now but
 = Arctan (450/750) = 31° above
there is more mass in the system!
horizontal, to the right
Momentum in 2-D
A 0.5 kg object moving at constant velocity of 8 m/s east is struck
by a hammer. The average force is 300 N directed due north and
contact lasts for 0.01 seconds. What is the final velocity
(magnitude and direction) of the object?
Is the momentum of the object conserved?
vi = 8 m/s
0.5 kg
No, there is an impulse from an external force.
How big is the Impulse?
Impulse = Fnett = 3 kg m/s
What is the initial momentum?
pi = mv = 4 kg m/s
What is the final momentum?
Momentum in 2-D
A 0.5 kg object moving at constant velocity of 8 m/s east is struck
by a hammer. The average force is 300 N directed due north and
contact lasts for 0.01 seconds. What is the final velocity
(magnitude and direction) of the object?
y
Momentum must be added as vectors.
pf
vi = 8 m/s
x
The Impulse is in the y
0.5 kg
direction and does not
change the momentum in the
x direction.
pi = mv = 4 kg m/s east
pf = 5 kg m/s
at 37° above x
axis.
Impulse = Ft = 3 kg m/s north = py
v = 10 m/s
A car and a truck, velocities and masses shown, collide
and lock bumpers. Find the final velocity (magnitude
and direction) of the pair.
y
vf
vc = 15 m/s
370
Mc = 2000 kg
vt = 5 m/s
Mt = 8000 kg

x
pt,i,y = ?24,000 kg m/s
pc,i,y = 0?
psys,f,y = 24,000
?
kg m/s
pt,i,x = ?32,000 kg m/s
pc,i,x = 30,000
?
kg m/s
psys,f,x = ?62,000 kg m/s
A car and a truck, velocities and masses shown, collide
and lock bumpers. Find the final velocity (magnitude
and direction) of the pair.
y
vf
vc = 15 m/s
370
Mc = 2000 kg
vt = 5 m/s
Mt = 8000 kg

x
psys,f,y = 24,000 kg m/s
psys,f,x = 62,000 kg m/s
psys,f = ?66,500 kg m/s
vsys,f = ?6.65 m/s
 = tan
? -1 24,000/62,000 = 21°