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### p0515

Course: AME 301, Fall 2010
School: USC
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Word Count: 452

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CHAPTER 5.15. 5, PROBLEM 15 249 5.15 Chapter 5, Problem 15 Problem: A bullet of mass mB is fired into a wooden block of mass mA and becomes embedded in it. The block and bullet then move up the frictionless incline for a time before they come to a stop. (a) Determine the magnitude of the initial bullet velocity, vo , as a function of mA , mB , , and gravitational acceleration, g . (b) Using the Principle of...

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CHAPTER 5.15. 5, PROBLEM 15 249 5.15 Chapter 5, Problem 15 Problem: A bullet of mass mB is fired into a wooden block of mass mA and becomes embedded in it. The block and bullet then move up the frictionless incline for a time before they come to a stop. (a) Determine the magnitude of the initial bullet velocity, vo , as a function of mA , mB , , and gravitational acceleration, g . (b) Using the Principle of Impulse and Momentum, compute the magnitude of the impulse of the force exerted by the bullet on the block, |I|, as a function of mB , g , , , and vo . (c) Compute vo and |I| for mB g = 1 ounce, mA g = 8 lb, = 1.2 sec and = 15o . Solution: Since the bullet and block move together after the impact, the motion of both is constrained. Letting the x axis be parallel to the incline, momentum is conserved in the x direction. . . . . . . . . . . ........................ ........................ .. .. .. . . . . ....... . ....... . . . ...................... . ....... ..... . ... . ....................... . .... . ... . . ... .... . .. . . . .. ...... . . .. .... ....... . ....... . . . . . . . ....... ....... o . .. . .. .... . ....... .. ....... . . . ....... ... . ....... . ... . ....... .. . ... . . .. ........ . ... . . . .. .. . . ......... . .............. ...... ..... . . . ........... ......... .... ...................... . . . ..... . .............................. . . ...... . . ...... ... . ...... . . . ............. . ......................... . .. ........................................... ...................................... . .. . . . .... . ......... . .............................. . . .......................................... ..... ....... . . . .......... . . . . . . ..... . ....... . ............. ... . .. . . . ................... ..... ..... . . .......................... . ...................................... . . . ............... . .................................... ..... . . . ................................................... .......... .... . . . .... . . . ...................... . . ....................................... ............... ........................................... . ... . . .............. .. . .... ..... .......................... . . . .... .. ...... ...................................... . ................... ..................................... ..... . . . ....... . .. ............... ..... . . . ..................................... . . ....... ........ . . ......................... ..... . .. . ........................................... . . . ................................................... . ... . ..... ....... . ....... ... ........................................................................................................................................... ................. ........ . ..... ... ....................................................................................................................... . ............ . . .. . . ...... . . g = g sin i g cos j x y v 2 Thus, we can express the velocity vo as follows. vo = (a) Denoting the velocity of the block and bullet after the impact by u, then x-momentum conservation tells us that mB vo cos 2 + mA 0 = (mA + mB ) u mA + mB u mB cos 2 After the impact, the only force acting is gravity. So, in the x direction, Newtons Second Law tells us that (mA + mB ) dvx = (mA + mB ) g sin dt Dividing through by (mA + mB ) and integrating once over time yields vx = u gt sin where we have used the fact that the initial velocity is u. The block and bullet stop when vx = 0, so that u = g sin 250 Therefore, the initial velocity of the bullet is vo = CHAPTER 5. IMPULSE AND MOMENTUM mA + mB g sin mB cos 2 (b) Focusing on the bullet, the Principle of Impulse and Momentum is mB vo + I = mB u where vo and u are Thus, solving for I yields vo = vo cos 2 i vo sin 2 j and u = ui I = mB (u vo ) = mB (u vo cos 2) i + mB vo sin 2 j Therefore, the magnitude of the impulse is |I| = mB 2 (u vo cos 2)2 + vo sin2 2 = mB 2 u2 + vo 2uvo cos 2 Finally, since u = g sin , we conclude that |I| = mB 2 g 2 2 sin2 + vo 2g vo sin cos 2 (c) The given values are mB g = 1 ounce, mA g = 8 lb, = 1.2 sec and = 15o . Consequently, g sin = (32.2 ft/sec2 )(1.2 sec) sin 15o = 10.0 ft/sec, and the values of vo and |I| are as follows. vo mA g + mB g g sin mB g cos 2 8 lb + 1/16 lb (32.2 ft/sec2 )(1.2 sec) sin 15o = 1/16 lb cos 30o ft = 1490 sec = |I| = = mB g 2 g 2 2 sin2 + vo 2g vo sin cos 2 g 1/16 lb (10 ft/sec)2 + (1490 ft/sec)2 2(10 ft/sec)(1490 ft/sec) cos 30o 32.2 ft/sec2 = 2.87 lb sec
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