Precalculus II

Time and Date Stamps (logged): 17:12:20 06-10-2020 ｰｶ涓ｱ浤ｯｯｵ涓ｯ浤ｯｱｯ 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 18:35:40

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.

Problem Number 1

The velocity of an object of constant mass 5 Kg is observed to change from 3 m/s to -18.6 m/s in .8 seconds.

Use the Impulse-Momentum Theorem to determine the average force exerted on the object.
Verify your results using your knowledge of uniformly accelerated motion.

Problem Number 2

What is the total momentum of two objects, one moving in the positive x direction at 20 m/s and the other moving at 26 m/s in the negative x direction? The mass of the first object is 3 kg and that of the second is 8 kg.

Solution

The magnitude of the momentum of the first is ( 20 m/s)( 3 kg)= 60 kg m/s, and that of the second is ( 26 m/s)( 8 kg) = 208 kg m/s.

Note: This gives the second object the greater influence in the collision; if the two objects were to become embedded in one another in the collision the final velocity of the system would be in the direction of the second.

The velocity of the second is in the negative direction so its momentum is - 208 kg m/s.

The total momentum is therefore 60 kg m/s + - 208 kg m/s = - 148 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

pTotal = m1 v1 + m2 v2

of their individual momenta m1 v1 and m2 v2.

Problem Number 3

An object of mass 10 kg experiences 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 .2 seconds.

If the average force is known to be Fave = 1490 Newtons:

Use the Impulse-Momentum Theorem to find its change in velocity.
Use Newton's Second Law and your knowledge of uniformly accelerated motion to verify your result.

Problem Number 4

Two objects collide and remain stuck together after collision.

One object has mass 8 kg and is moving in the positive direction at 24 m/s and the other has mass 8 kg and and moves at 14 m/s in the negative direction.

What is their total momentum, and what will be the velocity of this system immediately after collision?

Solution

The momentum of the first is ( 24 m/s)( 8 kg)= 192 kg m/s, and that of the second is ( - 14 m/s)( 7 kg) = -98 kg m/s.

The total momentum is therefore 192 kg m/s + `p2 kg m/s = 94 kg m/s.
The total mass is easily found to be 15 kg. Since the total momentum after collision is the same as that before collision, we see that after collision we have a mass of 15 kg with momentum -98 kg m/s.
Dividing we obtain velocity ( -98 kg m/s)/( 15 kg) = 6.266667 m/s.

Generalized Solution

The total momentum of a mass m1 moving at velocity v1 and a mass m2 moving at velocity v2 is

pTot = m1 v1 + m2 v2.

By Newton's Third Law and the Impulse-Momentum Theorem this momentum will remain unchanged during collision.

After collision we wil have one object of mass m1 + m2 and momentum m1 v1 + m2 v2. The object will therefore have velocity

velocity = momentum / mass = (m1 v1 + m2 v2) / (m1 + m2).

Problem Number 5

By how much does the velocity of an object of mass 9 Kg change under the following conditions:

The object experiences a variable net force F(t) for .09 seconds, if the force has the following characteristics:
The force vs. clock time graph increases linearly from 0 Newtons at the beginning of the .09-second interval to 9990 Newtons, then decreases linearly back to 0 at the end of the time interval.