Time and Date Stamps (logged): 03:42:06 08-29-2008 ¯²Ÿ³±Ÿ¯µ¯·Ÿ±¸Ÿ±¯¯· Precalculus II

Principles of Physics (Phy 121) Test_Set_4

Completely document your work and your reasoning.

You will be graded on your documentation, your reasoning, and the correctness of your conclusions.

Date and Time are 02-15-2001 17:14:04

Signed by Learning Lab Attendant: ______________________

Date and Time: ______________________

Attendant:

Test is to be taken without reference to text or outside notes.

Calculator is allowed.

No time limit but test is to be taken in one sitting.

Please place test in Dave Smith's folder when completed.

Student:

Completely document your work.

Undocumented and unjustified answers may be counted wrong.

Besides undocument and unjustified answers, if wrong, never get partial credit.

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Problem Number 1

A system consists of an object of mass 25 kg moving in the positive x direction at 6 m/s,  and another of mass 29 kg moving at 27 m/s in the negative x direction.  What is the total momentum of this system?

Solution

The magnitude of the momentum of the first is ( 6 m/s)( 25 kg)= 150 kg m/s, and that of the second is ( 27 m/s)( 29 kg) = 783 kg m/s.

The velocity of the second is in the negative direction so its momentum is - 783 kg m/s.

We see that the total momentum is in the direction (negative) of the more influential object.

Generalized Solution

The total momenum of two masses is the sum of their individual momenta.

Thus the total momentum of masses m1 and m2, moving with respective velocities v1 and v2, is the total

of their individual momenta m1 v1 and m2 v2.

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Problem Number 2

What is the velocity of an object of mass 3 kg if its momentum is known to be 12 kg m/s?

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Problem Number 3

A variable force F(t) (here F(t) indicates function notation, not multiplication of F by t; F(t) is the force at clock time t) is applied to an object of mass 8 kg for 7 seconds.

If the average force is Fave = 1024 Newtons,

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Problem Number 4

An object of mass 13 kg is subjected to a variable force F(t) (here F(t) indicates function notation, not multiplication of F by t; F(t) is the force at clock time t) for .06 seconds.

During the .06-second time interval, the force increases linearly from 0 to 118.3 Newtons, then decreases linearly back to 0.

Find the change in the object's velocity.

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Problem Number 5

An object of mass 3 kg is moving to the right at 4 m/s. It collides with a second mass of 7 kg which is moving at -4 m/s.  The first object ends up with a velocity of -5 m/s.