solution_5

# solution_5 - ouseph(jao888 HW05 tsoi(92940 1 This print-out...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ouseph (jao888) HW05 tsoi (92940) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points An object having an initial momentum that may be represented by the vector below strikes an object that is initially at rest. Which of the following sets of vectors may represent the momenta of the two objects af- ter the collision? Note carefully: The original vector above and the following vectors are all drawn to the same length scale. 1. 2. correct 3. 4. 5. 6. 7. Explanation: There is no external force for the two-object system, so the total momentum is a constant. From the choices, applying the vector sum- mation for the momenta of the two objects, we can easily identify the correct choice. The figure below shows the sum of the x- and y-components of the vectors which repre- sent the correct answer. initial momentum The horizontal vectors add to be the same length as the vector presented in the question. The vertical vectors cancel, as expected, since there is no vertical momentum. 002 10.0 points The graph below shows the force on an object of mass M as a function of time. Time (s) Force(N) 1 2 3 4 10- 10 For the time interval 0 to 4 s, the total change in the momentum of the object is ouseph (jao888) HW05 tsoi (92940) 2 1. p = 20 kg m / s . 2. Indeterminable unless the mass M of the object is known 3. p = 0 kg m / s . correct 4. p = 40 kg m / s . 5. p =- 20 kg m / s . Explanation: The Newtons second law of motion, in one dimension, is F = M a = M dv dt . From this, we obtain F dt = M dv , = ( M v ) = integraldisplay F dt, where M v is the momen- tum of the object. So from the graph above, the change in the momentum is zero . 003 10.0 points A constant force is exerted for a short time interval on a cart that is initially at rest on an air track. This force gives the cart a certain final speed. Suppose we repeat the experiment but, instead of starting from rest, the cart is already moving with constant speed in the direction of the force at the moment we begin to apply the force. After we exert the same constant force for the same short time interval, the increase in the carts speed 1. is equal to two times its initial speed. 2. is equal to the square of its initial speed. 3. is the same as when it started from rest. correct 4. is equal to four times its initial speed. 5. cannot be determined from the informa- tion provided. Explanation: The increase in speed is proportional to both the force on the cart and the time over which it acts. 004 10.0 points A child bounces a 55 g superball on the side- walk. The velocity change of the superball is from 27 m / s downward to 12 m / s upward....
View Full Document

## This note was uploaded on 07/09/2011 for the course PHY 302 taught by Professor Staff during the Summer '08 term at University of Texas.

### Page1 / 11

solution_5 - ouseph(jao888 HW05 tsoi(92940 1 This print-out...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online