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Monday 15 Nov:
Target:
Understanding Momentum
Journal:
1. A
firecracker explodes in midair. Considering all the fragments upon
explosion
a)
the total
KE remains constant
b)
the total
momentum decreases
c)
the total
KE decreases
d)
the total
momentum remains constant
e)
None of these
2.
Two equal-mass bullets traveling with the same speed strike a target.
One of the bullets is rubber and bounces off. The other is metal and
penetrates, coming to rest in the target. Which exerts the greater
impulse on the target?
a)
the rubber
bullet
b)
the metal
bullet
c)
both exert
the same
d)
not enough information
3. A
tank car coasting frictionless horizontally along the rails has a leak
in its bottom and dribbles several thousand gallons of water onto the
roadbed. In the process it
a)
speeds up
b)
slows down
c)
gains
momentum
d)
loses momentum
4.
What happens to the momentum of a body of constant mass if while it’s
traveling its kinetic energy is doubled?
a)
It doubles
b)
It remains
the same
c)
It
increases by a multiplicative factor of square root of 2
d)
It decreases by a multiplicative factor of square root of
2
1. D, momentum
is conserved
2. A, a greater
change of (D) Velocity
3. D, maintains the
same speed
4. C
L# 21: 1,2, 4
- 6, 9-16 Due Mon 15 Nov
(Momentum)
Lab: Specific Heat
L# 22: 1- 5,
7- 9, 12, 14 - 20 Due Wed 17 Nov
(Calculus: Derivative & Integration )
L# 23: 1-4, 8,
13, 15, 16, 18, 20 Due Fri 19 Nov
(Center of Mass)
Review Monday 22 Nov
Test L# 19 - 23 Tues 23 Nov
Simple Machine Project:
Requirements:
1.
Individual
2.
Build and its
due Wed 24 Nov
(must be made and
not just a store example)
2.
Any size & any of the 6 types you want.
(as long as
you can carry it by your self through the doorway)
3.
One page paper explaining your simple
machine and its use in History
.
Bridge Contest:
Tower Contest:
L#22
Derivative is
slope at an instant in time.
Derivative:
The old exponent is
brought down as a coefficient and the new exponent is the old one
reduced by 1. Note: constants are = 0
Find Derivative:
Ex: 5t4
+ 4t2 + 7 where t = 3
--> 4(5t3)
+ 2(4t1) + 0
4(5)(27) + (2)(12) = 540 +24 =
564
2t3
+ 4t2 + 7t + 5 where t =
2
-->
3(2t2)
+ 2(4t1) + 7t0
3(8) + (2)(8) + 7 =
47
Integral is area
under the curve;
Increase the
exponent by one and divide by new exponent. Note: Constants are
considered as t0
Find Integral: Integral: (t =
3)
-2t2
+ 3t + 20
-2t3/3 + 3t2/2 + 20t1/1
-2(27)/3 +
3(9)/2 + 60
= -18 + 13.5 + 60
=
55.5
Integral: (t =
3)
Increase the
exponent by one and divide by new exponent.
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