Honors Physics Assessment 5^th Week

 

Standard:  1C Velocity and Acceleration

 

1.     How long does it take to travel 375 meters at 24 m/s?

 

d = v/t so t = d/v (check the algebra!).  time = 375 m÷24 m/s = 15.6 s

 

Most common mistake was units.  The question is about time so the units had better be in seconds.  Note that meters cancel.  Don’t be so hung up on equations that you don’t pay attention to what’s going on.

 

 

2.     How far does a rock fall from rest in 4.0 seconds?

 

Initial velocity 0.  It gains 10 m/s each second for 4 seconds.  4 • 10 = 40 m/s down, or -40 m/s at the end.  Its average velocity is (0+-40)÷2 = -20 m/s for 4 seconds we get -80 meters.

 

or use d = 0.5 at² and get the same answer.

 

 

3.     A car is traveling 20.0 m/s when it begins to accelerate at uniform rate.  5.0 seconds after it begins accelerating it is traveling 30.0 m/s.

a.     What is its rate of acceleration?

 

a = ∆V/t = (30 – 20)÷5 = 2 m/s/s

 

 

b.     How far does it travel during those 5.0 seconds?

 

Average speed is (20 + 30)÷2 = 25 m/s for 5 seconds means 125 meters.

 

Once again, pay attention to the question.  If you used the formula and got 25 m, consider that it never moved slower than 20 m/s for 5 seconds means it has to be bigger than 100 m.  Don’t check your common sense at the door.

 

 

Standard:  4b Measurement and Analysis

 

There are four stations in the room.  Each is identified by index card labeled with a capital letter A or B or C or D. 

Note the letter:  __________________

 

Place the front edge (the end nearest the two upright posts) of the cart at the 25 cm mark.  Release the cart so that it rolls through the photogate.

 

 

 

Use this space to identify, label and record all relevant data:

 

Varies by station, your data will look like this:

 

time

0.0000

0.012

0.092              ∆t = .103 – 0 = .103 seconds

0.103

 

 

Use this space to show your work in determining the average speed of the cart as it passes through the gate.

 

distance is the distance between outside of the posts:  one of them was 7.5 cm or .075 meters.

 

so speed is d÷∆t = .075/.103 = .73 m/s

 

 

 

Standard:  5a  Simple Momentum and Impulse

 

 

1.     Find the momentum of a 5 kg mass traveling at 12 m/s to the east.

 

 

 

5 • 12 = 60 kg•m/s

 

 

 

2.     A force of 4.0 N acts against the object in number one (it exerts a force towards the west).  How long (seconds) will it need to stop the object?

 

 

∆p = -60 kg•m/s

 

∆p = F•t so F = ∆p/F = -60 kg•m/s ÷ -4.0 N = 15 seconds

 

(the negative sign on the ∆p is because it is being reduced, the negative sign on the force is because its “retarding” the motion of the object)  or (the original momentum is plus for east so reducing it makes the change negative and the force is negative because it’s west).  What is important is that the force and the resulting ∆p will always have the same sign.

 

 

 

 

 

 

3.     In a head on auto collision, the airbag inflates in a tiny fraction of a second using the gases produced by an explosive chemical reaction.  The resulting bag of air has a fist-sized hole in it.  In a couple of sentences and using the vocabulary of physics, explain how an airbag saves you from injury and why the hole is found in the bag.

 

First, what the hole does.  The bag, when inflated would be quite rigid and would not compress enough to extend the time for change in momentum enough.  Hitting it would not be much better than hitting the steering wheel.  By filling a bag with a large hole the air will be forced out when the driver’s head plows into it.  The hole size is engineered to give the maximum deflation TIME for an average sized driver.  By extending the time for the change in the driver’s momentum the force is reduced and injury or death avoided.

 

In this class we’ll study a lot of basic physics.  Example the relationship between force, time and changing momentum.  It’s very important that you be prepared to recognize those principles in the world around you.  It’s not necessary (or shouldn’t be) for me to specifically address every single imaginable application of those principles.  It’s important that you begin to broaden your thinking so that these principles become apparent to you.

 

4.     A 0.5 kg cart traveling at +2.0 m/s collides with a 1.0 kg car initially at rest.  The two carts couple and continue as a unit.  How fast?

 

p_before = p_after

 

m₁v₁ + m₂v₂ = (m₁+m₂)v   Perfect score on this question if you get to this line.

 

.5•2 + 1•0 = (.5 +1)v

 

1 kg•m/s = 1.5 v

 

v = 1/1.5 = .67 kg•m/s