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Wed
3 Nov:
Key
Concept(s) Today:
(Conservation of Ene
Journal:
L #17:
1- 16
Due
Today Wed 3Nov
(Sliding Blocks)
Thurs 4 Nov: Special Speaker
Lab: Acceleration of Carts Friday 5 Nov
Review: Monday 8 Nov
Test 9, 12, 13, 15, 16,
17 Tuesday 9 Nov
Momentum is a
conserved
quantity, meaning that the total momentum of any
closed system
(one not affected by external forces) cannot change. For a collision
occurring between object 1 and object 2 in an
isolated system,
the total momentum of the two objects before the collision is equal to
the total momentum of the two objects after the collision. That is, the
momentum lost by object 1 is equal to the momentum gained by object 2.
The amount of
momentum that an object has depends on two physical
quantities: the
mass and the
velocity of the moving object
in the
frame of reference. In
physics, the symbol for momentum is usually denoted by a
r
(it is a
vector); so this can be
written:
r1i +
r2i =
r1f +
r2f
(or)
- m1V1i
+ m2V2i = m1V1f
+ m2V2f
where:
- r
is the momentum
- m
is the mass
- v
the velocity
Examples:
http://video.google.com/videoplay?docid=5537533644685549811
1. A 120-kg lineman moving west at 2
m/s tackles an 80-kg football fullback moving east at 8
m/s. After the collision, both players move east at 2
m/s. Draw a vector diagram in which the before- and
after-collision momentum of each player is represented
by a momentum vector. Label the magnitude of each
momentum vector.
Answer to Question #1
m1V1i + m2V2i
= m1V1f + m2V2f
(80)(8) - (120)(2) =
640 - 240 = 400-kgm/s
2. A 3.0-kg ball is moving at 12 m/s to the right. It
hits a 3.0-kg ball moving to the left at 4.0 m/s in a totally elastic
collision.
a. What is the velocity of the two ball immediately
after the collision?
b. What is the total KE of the of the two balls
before & after the collision.
Answer to Question #2
a.) m1V1i
+ m2V2i = m1V1f
+ m2V2f
(3)(12) - (3)(4) =
36 - 12 = 24-kgm/s (at the start)
Balls have the same mass, so both
momentums are transferred.
left ball will be rebound to the
left at - 4.0m/s and right ball will rebound to the right at 12 m/s
b.) KEi
= KEf
= 1/2 mV2 +
1/2 mV2
= 1/2(3)(122)
+ 1/2(3)(42)
= 216 + 24 = 240 J
3. What is the final velocity of an inelastic
collision when a 2.0-kg cart is moving at 60m/s and a 4-kg brick (V =
0.0 m/s) is dropped on top of the car?
Answer to Question #3
a.) m1V1i
+ m2V2i = m1V1f
+ m2V2f
(2)(60)
- (4)(0) = 120 - 0 = 120-kgm/s
(at the start)
(at the finish)
120 = (6)(V2f
) -->
V2f
= 20m/s
4. Two objects (4-kg on the left and
6-kg on the right) are tied together with a compressed spring in between
them. Together, they are moving to the right at 10m/s. What is the final
velocity of the 4-kg object after the spring is released when the 6.0-kg
object final velocity is moving to the right at 20m/s?
Answer to Question #4
a.) m1V1i
+ m2V2i = m1V1f
+ m2V2f
(4)(10) + (6)(10) =
100-kgm/s (at the start)
(at the finish)
100-kgm/s
= m1V1f + m2V2f
100
= (4)V1f + (6)(20)
100 -
120 = (4)V1f
V1f
=
- 5.0m/s
(the 4-kg object is moving to the left at -5m/s)
Problems L#30: 2 -
9, 12 - 16, 18, 19 |