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Text Elements
Momentum
p
The quantity of motion at a given frame of reference at a given point of time
momentum of the system before collision would be the same as the momentum of the system after collision
If 2 objects were to collide,
p
before
=
p
after
*for the system, not each individual object
Before
p
system
= 0
After
p = 20[Forward]
p = 20[Backward]
p
system
= 20 + -20 = 0
2 gals on frictionless ice.
Amy
who is 55kg, pushes off
Bamy
who is 50kg
Afterwards, is propelled at 1.5m/s [S]
Amy
Before
v
B
m
B
= 50kg
= 0
v
A
m
A
= 55kg
= 0
After
v
B
m
B
= 50kg
= ?
v
A
m
A
= 55kg
= 1.5[S]
β
β
p + p = p + p
A
B
B
A
β
β
0 = 50(v ) + (55)(-1.5)
B
β
v
B
β
= 1.55 [UP]
A 2000kg going at 4m/s [Forward] it collides and couples with a with a 1500kg load How fast would they be moving after the collision?
identical train car
train car
Before
2000kg
3500kg
4m/s
m
1
= 2000kg
v
1
= 4m/s
m
2
= 3500kg
v
2
= 0m/s
After
5500kg
m
f
= 5500kg
v m/s
f
v
f
= ?m/s
p + p = p + p
2
1
1
2
β
β
8000 = 5500(v )
f
f
v = 1.4545m/s
1500kg cannon
shoots a
10kg shot
velocity of shot is
200m/s [Forward]
whats the velocity of the
relative to
cannon
cannon relative to ground
?
Before
m
c
= 1500kg
v
c
= 0m/s
m
b
= 10kg
v
b
= 0m/s
After
m
c
= 1500kg
v
c
= ?
m
b
= 10kg
v
b
=
v
b
g
=
v
b
c
v
c
g
equivalent to
v
b
β
equivalent to
v
c
β
v
b
β
= 200 +
v
c
β
p + p = p + p
b
c
c
b
β
β
0 = 1500(v ) + (10)(200+v )
c
c
2000 = 1510(v )
c
v = -1.3245 v = 1.3245 [Back]
c
c
curling stone slide on ice at velocity of
20kg
1.5m/s [N]
it glances off an
identical stone at rest
after the collision, the curling stone moves at whats the velocity of the
second stone
o
1.1m/s [N15 E]
Before
v
1
= 1.5m/s
v
2
= 0m/s
m
2
= 20kg
m
1
= 20kg
After
v
1
= 1.1m/s [N15 E]
v
2
= ?m/s
m
2
= 20kg
m
1
= 20kg
15
o
o
ΞΈ
o
You must split the systemβs momentum into components
p + p = p + p
1
1
2
2
β
β
4500kg sailboat Moving through water at Experiences a force of Force is applied for 1 minute What is the final velocity of the boat?
o
12.5
o
8.33m/s
1000N
4500kg
80
o
- Find speed from force.
J = F(Ξt) = m(Ξv) = 1000(60) = (4500)(Ξv)
Ξv = 13.333
- Find final speed
v
x
β
1000N[N10 E]
8.33m/s[E12.5 N]
o
=
+ = 10.4473 ^0ePwx0jW
v
y
β
=
+ = 14.93 ^9G1bYltm
13.33sin(80)
8.33sin(12.5)
13.33cos(80)
8.33cos(12.5)
14.93
10.4473
55
18.22
o
m = 1kg v = v
1
1
45
o
30
o
m = 0.5kg v = ?
2
2
m = 0.5kg v = ?
3
3
A missile splits into 2 smaller missiles of identical mass. What is the speed of each of the 2 smaller missiles compared to the speed of the original?
p
= p - p
1y
3y
2y
= 0.5(v )(sin45) - 0.5(v )(sin30)
= 0
3
2
= 0.3535(v ) - 0.25(v )
3
2
0
0
p
= p + p
1x
3x
2x
= v
v
= (0.5)(v )(cos45) + (0.5)(v )(cos30)
3
2
= 0.3535(v ) + 0.433(v )
3
2
v
= 0.3535(v ) - 0.25(v )
3
2
0
= 0.3535(v ) + 0.433(v )
3
2
v
Elimination
-v = -0.25(v ) - 0.433(v )
2
2
v
2
= 1.464v
v
3
= 1.035v
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