Mastering Physics Assignment #9
13 Pages

Mastering Physics Assignment #9

Course Number: PHYS 040A, Summer 2013

College/University: UC Riverside

Word Count: 466

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Assignment #9 Due: 11:00pm on Tuesday, July 23, 2013 You will receive no credit for items you complete after the assignment is due. Grading Policy A message from your instructor... ***** Note: this assignmnet is due on MONDAY March 10 ***** A Bullet Is Fired into a Wooden Block A bullet of mass mb is fired horizontally with speed vi at a wooden block of mass mw resting on a frictionless table. The bullet hits...

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#9 Due: Assignment 11:00pm on Tuesday, July 23, 2013 You will receive no credit for items you complete after the assignment is due. Grading Policy A message from your instructor... ***** Note: this assignmnet is due on MONDAY March 10 ***** A Bullet Is Fired into a Wooden Block A bullet of mass mb is fired horizontally with speed vi at a wooden block of mass mw resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed vf . Part A Which of the following best describes this collision? Hint 1. Types of collisions An inelastic collision is a collision in which kinetic energy is not conserved. In a partially inelastic collision, kinetic energy is lost, but the objects colliding do not stick together. From this information, you can infer what completely inelastic and elastic collisions are. ANSWER: perfectly elastic partially inelastic perfectly inelastic Correct Part B Which of the following quantities, if any, are conserved during this collision? Hint 1. When is kinetic energy conserved? Kinetic energy is conserved only in perfectly elastic collisions. ANSWER: kinetic energy only momentum only kinetic energy and momentum neither momentum nor kinetic energy Correct Part C What is the speed of the block/bullet system after the collision? Express your answer in terms of vi , mw , and mb . Hint 1. Find the momentum after the collision What is the total momentum pto ta l of the block/bullet system after the collision? Express your answer in terms of vf and other given quantities. ANSWER: pto ta l = (mw + mb )vf Hint 2. Use conservation of momentum The momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for pto ta l , this time expressed as the total momentum of the system before the collision. Express your answer in terms of vi and other given quantities. ANSWER: pto ta l = mb vi ANSWER: mv bi vf = m +m w b Correct A One-Dimensional Inelastic Collision Block 1, of mass m1 = 5.50kg , moves along a frictionless air track with speed v1 = 15.0m/s . It collides with block 2, of mass m2 = 13.0kg , which was initially at rest. The blocks stick together after the collision. Part A Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically. Hint 1. How to approach the problem Find the initial momentum of each block separately; then add the two momenta together. Recall that momentum is equal to mass times velocity: p = mv. ANSWER: pi = 82.5 kg m/s Correct Part B Find vf , the magnitude of the final velocity of the two-block system. Express your answer numerically. Hint 1. How to approach the problem Apply conservation of momentum, keeping in mind that the mass after the collision is the sum of the individual blocks' masses. ANSWER: vf = 4.46 m/s Correct Part C What is the change K = Kfina l Kinitia l in the two-block system's kinetic energy due to the collision? Express your answer numerically in joules. Hint 1. Find the initial kinetic energy Find Kinitia l , the initial kinetic energy of the two-block system. Express your answer numerically in joules. ANSWER: Kinitia l = 619 J ANSWER: K = -435 J Correct A Ball Hits a Wall Elastically i A ball of mass m moving with velocity vi strikes a vertical wall. The angle between the ball's initial velocity vector and the wall is i as shown on the diagram, which depicts the situation as seen from above. The duration of the collision between the ball and the wall is t, and this collision is completely elastic. Friction is negligible, so the ball does not start spinning. In this idealized collision, the force exerted on the ball by the wall is parallel to the x axis. Part A What is the final angle f that the ball's velocity vector makes with the negative y axis? Express your answer in terms of quantities given in the problem introduction. Hint 1. How to approach the problem Relate the vector components of the ball's initial and final velocities. This will allow you to determine f in terms of i . Hint 2. Find the y component of the ball's final velocity What is vfy , the y component of the final velocity of the ball? Express your answer in terms of quantities given in the problem introduction and/or vix and viy , the x and y components of the ball's initial velocity. Hint 1. How to approach this part There is no force on the ball in the y direction. From the impulse-momentum theorem, this means that the change in the y component of the ball's momentum must be zero. ANSWER: vfy = vi cos( i ) Hint 3. Find the x component of the ball's final velocity What is vfx , the x component of the ball' final velocity? Express your answer in terms of quantities given in the problem introduction and/or vix and viy , the and y components of the ball's initial velocity. x Hint 1. How to approach this problem Since energy is conserved in this collision, the final speed of the ball must be equal to its initial speed. ANSWER: vfx = vi sin( i ) Hint 4. Putting it together Once you find the vector components of the final velocity in terms of the initial velocity, use the geometry of similar triangles to determine f in terms of i . Register to View Answer= i Correct Part B What is the magnitude F of the average force exerted on the ball by the wall? Express your answer in terms of variables given in the problem introduction and/or vix . Hint 1. What physical principle to use Use the impulse-momentum theorem, only J case, one force is acting, so = pf pi , along with the definition of impulse, J = F t . In this J = F t. Putting everything together, F = p f p i . t Hint 2. Change in momentum of the ball = i implies that the y component of the ball's momentum does not change during the collision. What is px , the magnitude of the change in the ball's x momentum? The fact that f Express your answer in terms of quantities given in the problem introduction and/or vix . ANSWER: px = 2mvi sin( i ) Register to View Answer= 2 mvisin(f) t Correct Colliding Cars In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses m1 and m2 collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of v1 , and car 2 was traveling northward at a speed of v2 . After the collision, the two cars stick together and travel off in the direction shown. Part A First, find the magnitude of v, that is, the speed v of the two-car unit after the collision. Express v in terms of m1 , m2 , and the cars' initial speeds v1 and v2 . Hint 1. Conservation of momentum Recall that conservation of linear momentum may be expressed as a vector equation, pinitia l = pfina l . Each vector component of linear momentum is conserved separately. Hint 2. x and y components of momentum The momentum of the two-car system immediately after the collision may be written as p = px ^ + py ^, where i j the x and y directions are the eastward and northward directions, respectively. Find px and py Express the two components, separated by a comma, in terms of m1 , m2 , v1 and v2 . ANSWER: px , py = m1 v1 ,m2 v2 Hint 3. A vector and its components Recall that the square of the magnitude of a vector is given by the Pythagorean formula: p2 = p2 + p2 . x y Hint 4. Velocity and momentum Find v, the magnitude of the final velocity. Express v in terms of the magnitude of the final momentum p and the masses m1 and m2 . ANSWER: p v = m +m 1 2 ANSWER: v= (m1v1) +(m2v2) 2 2 m1+m2 Correct Part B Find the tangent of the angle . Express your answer in terms of the momenta of the two cars, p1 and p2 . ANSWER: p tan( ) = p2 1 Correct Part C Suppose that after the collision, tan ANSWER: = 1; in other words, is 45 . This means that before the collision: The magnitudes of the momenta of the cars were equal. The masses of the cars were equal. The velocities of the cars were equal. Correct A message from your instructor... Note: The problem numbers do not correspond to the end-of-chapter problems in the textbook (Knight). These problems are taken from a database provided by the publisher. Problem 8.58 A steel ball with mass 42.0g is dropped from a height of 1.90m onto a horizontal steel slab. The ball rebounds to a height of 1.50m . Part A Calculate the impulse delivered to the ball during impact. Take the free fall acceleration to be g = 9.80m/s2 . ANSWER: 0.484 Ns Correct Part B If the ball is in contact with the slab for a time of 2.40ms , find the average force on the ball during impact. Take the free fall acceleration to be g = 9.80m/s2 . ANSWER: 202 N Correct Problem 8.70 A rifle bullet with mass 8.00 g strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring. The impact compresses the spring 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cm. Part A Find the magnitude of the block's velocity just after impact. ANSWER: 2.60 m/s Correct Part B What was the initial speed of the bullet? ANSWER: 325 m/s Correct Problem 8.72 A movie stuntman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor . Grabbing a rope attached to a chandelier, he swings down to grapple with the movie's villain (mass 70.0 kg), who is standing directly under the chandelier. (Assume that the stuntman's center of mass moves downward 5.0 m. He releases the rope just as he reaches the villain.) Part A With what speed do the entwined foes start to slide across the floor? Take the free fall acceleration to be 9.80 m/s2 . ANSWER: 5.28 m/s Correct Part B If the coefficient of kinetic friction of their bodies with the floor is k = 0.215, how far do they slide? ANSWER: 6.61 m Correct Problem 8.76 A 8.00-kg ball, hanging from the ceiling by a wire 135 cm long, is struck in an elastic collision by a 2.00-kg ball moving with speed v0 just before the collision. Part A If the maximum tension that the wire can have without breaking is 1600 N, what is the largest that v0 can be without the wire breaking? ANSWER: 40.1 m/s Correct Problem 8.78 A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.000 m long. The center of mass of the block rises a distance of 0.45 cm. Part A Find the speed of the bullet as it emerges from the block if its initial speed is 450 m/s. ANSWER: 391 m/s Correct Problem 8.88 A 232 T h (thorium) nucleus at rest decays to a 228 Ra (radon) nucleus with the emission of an alpha particle. The total kinetic energy of the decay fragments is 6.54 10 13 J. An alpha particle has 1.76% of the mass of a 228 Ra nucleus. Part A Calculate the kinetic energy of the recoiling 228 Ra nucleus. ANSWER: 1.13!10"14 J Correct Part B Calculate the kinetic energy of the alpha particle. ANSWER: 6.43!10"13 J Correct Score Summary: Your score on this assignment is 98.4%. You received 98.4 out of a possible total of 100 points.

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