Physics 113 –Morning Session, Summer 2012
Participation Points #2
To be turned in at the end
each discussion session, stapled on top of Participation Points #1
Student’s Name:
__________________________ Write previous # points accumulated:
|
|
|
|
|
|
|
|
|
|
|
3.4- Velocity
and acceleration. An object is moving in the
x-direction at 1m/s when it is subjected to an acceleration given by a
=0.5 j m/s2. What is its velocity vector after 4s?
Fill in the blank: After 4s, given that the acceleration is along the
___-direction, vx=_____m/s, and vy=______m/s. So, in term
of the unit vectors, the velocity vector after 4s is v=_______i +_______j (m/s).
3.5- Projectile
Motion. You toss dime horizontally
at 8m/s from a height of 5m. At the same time you drop a quarter from the same
height. How long does each take to reach the ground?
Diagram:
Fill in the blank: In the __________ direction, both coins suffer a constant
downward acceleration of (value and units) __________, so the equation relating
h (the height) to g and t is h =_______________________(this
is so because the initial velocity in this direction is _______). This leads to
t=_________, for the quarter, and t=_________, for the dime. This result makes
sense since motions along the vertical and horizontal directions are
________________.
3.6- Circular
Motion. A runner rounds the
semicircular end of a track whose curvature radius is 16m. She runs at constant
speed, with an acceleration of 1m/s2. How long would it take for her
to complete the turn?
Diagram:
Fill in the blank: Since a= v2/r,
the linear speed that leads to the given centripetal acceleration is _________.
Since the semicircular arc measures ____________, it would take ________π
s for the runner to complete the turn.
4.1- Newton’s
Second Law. The maximum breaking force
of a 2000 kg car is 8kN. Estimate the stopping distance when the car is
traveling at (a) 36 km/h; (b) 72 km/h. Note that this distance is NOT doubled
when the speed is doubled!!
Diagram:
Fill in the blank: According to
Newton
a = - ______________(formula) = -
____________(numbers and unit). Recall a useful formula, for constant
acceleration, relating initial and final speeds, and acceleration and distance,
which is 
. In our problem, v =
______ , consequently, x = -v02
/____________ , then (a)
x=__________________________ (b)
x=____________________________
4.2- Adding Forces. An elevator accelerates downward at 1.81 m/s2.
What force does the floor exert on a 50 kg passenger?
Diagram:
Fill
in the blank:
The passenger also accelerates downward at ay=_______ . So if the
floor exerts a force N upward on the passenger, mg-N=may, or
N=__________________________(formula)=______________(numbers and unit)
4.3- Newton’s Third Law. A 80-kg astronaut pushes
off a 400-kg satellite, exerting a force of 100N during 0.8s while in contact.
(a) Find the speed of the astronaut after they have separated (b) what is the speed of the satellite?
Diagram:
Fill
in the blank:
The satellite exerts on the astronaut a force of _________ during the same time
interval. So
(a)
the speed of the astronaut is va = aat = (______/______)
t = (______/______) (0.8s) = __________(number and unit).
(b)
the speed of the satellite is vs = ast = (______/______)
t = (______/______) (0.8s) = __________(number and unit).
5.1- Using
Diagram:
Fill in the blank: Due to gravity, the
acceleration down a frictionless inclined surface of angle θ is a= ________________(formula)= ___________(number);
on the other hand the kinematic equation for constant acceleration is, x = ½ a t2 , so t =
_________________(formula)=________________(numbers and unit)
5.2- Multiple
Objects. A 10kg dog is hanging on
the bottom of the tablecloth with all his weigh, a 2.5 kg cat is standing in
the middle of the table. Neglecting the tablecloth mass, what is the
acceleration of the cat?
Diagram:
Fill in the blank: Assuming no friction
between the cloth and the table, the system of cat-dog-tablecloth is falling
under the weigh of the dog, which is mdog
g = ___________(number and units), with the tablecloth mass negligible, the cat
acceleration is acat= _______/(mcat + mdog) =
______________________(numbers and unit), and of course the dog acceleration
is adog= ___________(number
and unit).
5.3- Circular
Motion. A bucket of water is
whirled in a vertical circle of 1.6m, what is the minimum speed that will keep
the water from falling out?
Diagram:
Fill in the blank: If the acceleration at the
top of the circle is greater than ___________, the water will remain in the
bucket, i.e. if the linear speed is greater than __________________(formula) =
_________________(numbers and unit)