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0.0250-kg bullet is accelerated from rest to a speed of

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli ISBN: 9780130606204 3

Solution for problem 38PE Chapter 8

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 38PE

0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse if you hold the gun loosely a few centimeters from your shoulder rather than holding it tightly against your shoulder. (a) Calculate the recoil velocity of the rifle if it is held loosely away from the shoulder. (b) How much kinetic energy does the rifle gain? (c) What is the recoil velocity if the rifle is held tightly against the shoulder, making the effective mass 28.0 kg? (d) How much kinetic energy is transferred to the rifle shoulder combination? The pain is related to the amount of kinetic energy, which is significantly less in this latter situation. (e) Calculate the momentum of a 110-kg football player running at 8.00 m/s. Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s. Discuss its relationship to this problem.

Step-by-Step Solution:

Step-by-step solution In this problem a bullet is fired so,we have to calculate the recoil velocity of the rifle if it is held loosely away from the shoulder, the gain in the kinetic energy , the recoil velocity if the rifle is held tightly against the shoulder, the kinetic energy transfer to the rifle shoulder combination Step 1 o f 12 (a) The expression of conservation of momentum is: Here, is mass of rifle, is mass of bullet, is initial velocity of rifle, is initial velocity of bullet, is recoil velocity of rifle and is final velocity of bullet. Substitute, 3.00 kg for , 0.0250 kg for , 0 m/s for , 550 m/s for and 0 m/s for in the expression of momentum. Hence the recoil velocity of the rifle is Step 2 o f 12 (b) The expression of initial kinetic energy is: Here, m is the mass and is the initial velocity. Substitute, 3.00 kg for and 0 for in the above expression, The expression of final kinetic energy is: Here, m is the mass and is the final velocity of rifle. Substitute 3.00 kg for and 4.58 m/s for , Step 3 o f 12 Gain of kinetic energy by rifle is: Substitute, 0 J for and 31.5 J for , Hence, the gain of kinetic energy by the rifle is Step 4 o f 12 (c) The expression of conservation of momentum is: Here, is effective mass of rifle, is mass of bullet, is initial velocity of rifle, is initial velocity of bullet, is recoil velocity of rifle when effective mass is considered and is final velocity of bullet. Substitute, 28.00 kg for , 0.0250 kg for , 0 m/s for , 550 m/s for and 0 m/s for in the expression of momentum. Hence the recoil velocity of the rifle is .negative sign shows that the velocity is away from bullet. Step 5 o f 12 (d) The expression of initial kinetic energy is: Here, is the mass and is the initial velocity Substitute, 28.0 kg for and 0 for , The expression of final kinetic energy is when effective mass is considered is: Here, is the mass and is the final velocity Substitute, 28.0 kg for and 0.491 m/s for in the above expression, Step 6 o f 12 Gain of kinetic energy by rifle when effective mass is considered: Substitute, 0 J for and 3.38 J for in the above expression, Hence the gain of kinetic energy by the rifle is .

Step 7 of 12

Chapter 8, Problem 38PE is Solved
Step 8 of 12

Textbook: Physics: Principles with Applications
Edition: 6
Author: Douglas C. Giancoli
ISBN: 9780130606204

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