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PHY 303K (3)
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Midterm

6 Pages
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School
University of Texas at Austin
Department
Physics
Course
PHY 303K
Professor
Rodenborn
Semester
Spring

Description
Version 001 – Test 3 Additional Review – ﬂorin – (57850) 1 This print-out should have 13 questions. From the setup, the ﬁnal velocity of the col- Multiple-choice questions may continue on lision is the initial velocity of the subsequent the next column or page – ﬁnd all choices swing. During the swinging process the total before answering. energy is conserved Ballistic Pendulum 02 Ei= E f 1 001 (part 1 of 2) 10.0 points (m 1 m )v2=( m + m 1gh 2 2 f Auntrsainm experiment using an apparatus similar to that ▯ shown in the ﬁgure. vf= 2gh Initially the bullet is ﬁred at the block while ▯ = 2(9 .8m /s )(0 .047 m) the block is at rest (at its lowest swing point). After the bullet hits the block, the block rises to its highest position, see dashed block in the = 0.959792 m/s . ﬁgure, and continues swinging back and forth. The following data is obtained: the maximum height the pendulum rises is 002 (part 2 of 2) 10.0 points Find v1,theinitialspeedof m 1 4.7cm, the mass of the bullet is 79 g, and the mass of the pendulum bob is 945 kg. Correct answer: 11482 m/s. Explanation: The linear momentum is conserved in a collision 47 ▯ pi= p f m v =( m + m )v 1 1 1 2 f m 1 m 2 v1= m vf 1 (0.079 kg) + (945 kg) = (0.079 kg) 4.7cm ▯ (0.959792 m/s) v 945 kg v = 11482 m/s . 79 g i f Find the ﬁnal velocity of the system (1 + m 2immediatelyafterthecollisionandbefore CanoeCenterofMass the pendulum starts to swing upward. The 003 (part 1 of 2) 10.0 points 2 Am nwhsmsi5kgndawmn acceleration of gravity is 9.8m /s . whose mass is 60 kg sit at opposite ends of a Correct answer: 0.959792 m/s. canoe 4 m long, whose mass is 25 kg. Assume the man is seated at x =0a d Explanation: the boat extends along the positive x axis with the woman at the other end. Where is the center of mass of the system consisting of ▯ Let : ▯ =46 .6332 , man, woman, and canoe? m 179g=0 .079 kg , m 2945kg , and Correct answer: 1.70588 m. h =4 .7cm=0 .047 m. Explanation: Version 001 – Test 3 Additional Review – ﬂorin – (57850) 2 With the man (m )1ttheorigiehe SkaterRealAndPP m 1 1 m x 2 2 x 3 3 005 10.0 points x CM = Aserhstayrmaw ll m 1 m + 2 3 She pushes on the wall with a force whose 0+(60kg)(4m)+(25kg)(2m) = magnitude is F,ostelsnhr 85 kg + 60 kg + 25 kg with a force F (in the direction of her motion). =1 .70588 m. As she moves away from the wall, her center of mass moves a dista htedis.C 004 (part 2 of 2) 10.0 points following statements regarding energy. Suppose that the man moves quickly to the center of the canoe and sits down there, while I▯ K trans+▯ E internal Fd the woman moves quickly 1/3 the length of II. ▯K trans▯ E internal ▯Fd the canoe towards the man. How far does III. ▯K trans▯ E internal0 the canoe move in the water? Assume force IV. ▯K trans Fd of friction between water and the canoe is V. ▯K trans= ▯Fd negligible. What is the correct form of the energy prin- Correct answer: 0.529412 m. ciple for the skater as a real system and as a point particle (PP) system? Explanation: Since friction is negligible, the net horizon- tal force on the system (the man, woman and canoe) is zero. Therefore, its center of mass 1. Real: IV, PP: III momentum (and velocity) remain constant (which is zero in this case). So the center of 2. Real: III,PP: IV correct mass of the system remains stationary. 3. Real: II, PP: V For this problem, let us take the origin to be at the left end of the canoe (note that it is not a stationary point). The new center of 4. Real: IV, PP: IV mass is at 5. Real: V, PP: I xCM = m 1 1 m x2 2m x 3 3 6. Real: I, PP: V m 1 m +2m 3 (85 kg)(2 m) + (60 kg)(2.66667 m) = 7. Real: I, PP: IV 85 kg + 60 kg + 25 kg (25 kg)(2 m) 8. Real: II, PP: IV + 85 kg + 60 kg + 25 kg =2 .23529 m. 9. Real: III,PP: V 10. Real: V, PP: IV At ﬁrst, the center of mass was .70588 m from the left end of the canoe. Now the center of mass is 2.23529 m from the left end of the Explanation: canoe. This means the boat moved As a real system, the wall does no work on the skater since the contact point of the 2.23529 m ▯ 1.70588 m = 0.529412 m skater’s hand and the wall does not move. to the left as the man walked 0.529412 m to Also, the wall does not transfer energy to the the right and the woman walked 1.33333 m to skater, i.e. by cooling down. The right hand the left. side of the energy principle must therefore Version 001 – Test 3 Additional Review – ﬂorin – (57850) 3 ▯ ▯ be zero for the real systemFurthermore, 1+ k 2 the left hand side must contain the skater’s 4. K = mv . ▯ 2 ▯ change in internal energy for a real system. k 2 III is correct. 5. K = 1+ mv . As a point particle system, the skater’s cen- 2 6. K =[1+2 k]mv . ter of mass moves a distance d while a force F is applied. Thus, the work done is Fd. Translational kinetic energy is the only type7. K =[1+ k]mv . of energy a point particle system can have. IV ▯ ▯ k is correct. 8. K = 1+ mv. 2 Wheel and a Weight 01 9. K =[1+2 k]mv. 006 (part 1 of 2) 10.0 points ▯ ▯ Consider a circular wheel with a mass 2 m, 10. K = 1+2 k mv . correct and a radius R (as seen in the ﬁgure below), 2 mounted on a horizontal frictionless axle paExplanation: ing through its center (not shown in ﬁgure). Basic Concepts: Rotational kinetic en- The moment of inertia about the center of thergy is wheel is I =2 kmR 2, where 0.5
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