105 - exam 4 - Fall 2009 - solutions

105 - exam 4 - Fall 2009 - solutions - Instructions Record...

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

View Full Document Right Arrow Icon
Instructions: Record your answers on the bubble sheet. The Testing Center no longer allows students to see which problems they got right & wrong, so I strongly encourage you to mark your answers in this test booklet . You will get this test booklet back (but only if you write your CID at the top of the first page ). You may write on this exam booklet, and are strongly encouraged to do so. In all problems, ignore friction, air resistance, and the mass of all springs, pulleys, ropes, cables, strings etc., unless specifically stated otherwise. Use g = 9.8 m/s 2 only if there are “9.8” numbers in the answer choices; otherwise use g = 10 m/s 2 . Problem 1. A 10 kg pendulum bob passes through the lowest part of its path at a speed of 6 m/s. What is the tension in the pendulum cable at this point if the pendulum is 2 m long? a. Less than 235 N b. 235 – 245 c. 245 – 255 d. 255 – 265 e. 265 – 275 f. 275 – 285 g. More than 285 N 1. FBD at lowest point: Σ F = ma c T – mg = mv 2 /r T = mg + mv 2 /r = 10*10 + 10*36/2 = 100 + 180 = 280 N. Choice F Problem 2. A 3 kg mass moving east at 6 m/s on a frictionless horizontal surface collides with a 2 kg mass that is initially at rest. After the collision, the first mass moves due south at 4 m/s. What is the magnitude of the velocity of the second mass after the collision? a. Less than 7 m/s b. 7 – 8 c. 8 – 9 d. 9 – 10 e. 10 – 11 f. 11 – 12 g. More than 12 m/s 2. Momentum is conserved in both x- and y-directions. x-direction y-direction ( Σ p x ) bef = ( Σ p x ) aft ( Σ p y ) bef = ( Σ p y ) aft 3*6 = 2*v x 0 = -3*4 + 2*v y v x = 9 v y = 6 combine components: v = sqrt(v x 2 + v y 2 ) = sqrt(81+36) = sqrt(117). Since sqrt(100) = 10 and sqrt(121) = 11, this answer is between 10 and 11. Choice E. Problem 3. A string attached to a bucket (mass 6 kg) is wound over a large pulley having a mass of 20 kg (not zero mass!). The pulley can be considered to be a solid cylinder of radius 0.6 m. The pulley turns as the block is allowed to fall from rest. No energy is lost to friction. If the bucket falls 2 m, how fast will it be going? a. 9 m/s b. 12 c. 15 d. 18 e. 21 1
Background image of page 1

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

View Full Document Right Arrow Icon
f. 24 g. 27 m/s 3. PE = KE + rotKE mgh = ½ mv 2 + ½ I ϖ 2 = ½ mv 2 + ½ (½ MR 2 ) (v 2 /R 2 ) = (1/2 m + ¼ M) v 2 Solve for v. v = sqrt(mgh/(1/2 m + ¼ M)) = sqrt(6*10*2/(3+5)) = sqrt(120/8) = sqrt(15). Choice C Problem 4. A pool is filled half with water and half with a light oil (density 400 kg/m 3 ). The oil floats on the water. When a diver comes up from the bottom of the pool, from the water into the oil, she will experience a buoyant force in the oil that is ____________ the buoyant force she felt in the water. a.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 9

105 - exam 4 - Fall 2009 - solutions - Instructions Record...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online