HW5_solutions - PROBLEM 14.5 l~ I ~E-L,tJ T kB A bullet is...

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PROBLEM 14.5 A bullet is fired with a horizontal velocity of 500 mls through a 3-kg T block A and becomes embedded in a 2.5-kg block B. Knowing that blocks A and B start moving with velocities of 3 mls and 5 mis, respectively k B determine (a) the mass of the bullet, (b) its velocity as it travels from ~E;-L,tJ I block A to block B. l~ SOLUTION The masses are m for the bullet and mA and mB for the blocks. (a) The bullet passes through block A and embeds in block B. Momentum is conserved. Initial momentum: Final momentum: Equating, mAvA + mBvB _ (3)(3) + (2.5)(5) = 43.434x 10-3kg m= - Vo- vB 500 - 5 m = 43.4g~ Initial momentum: (b) The bullet passes through block A. Momentum is conserved. Final momentum: Equating, VI= mvo - mAvA = (43.434 x 10-3)(500) - (3)(3) m 43.434 x 10-3 = 292.79 mls VI = 293 mls- ~
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PROBLEM 14.21 I j 130~ In a game of pool, ball A is traveling with a velocity v0 when it strikes balls Band C which are at rest and aligned as shown. Knowing that after the collision the three balls move in the directions indicated and that Vo = 4 mls and ve = 2.1 mis, determine the magnitude of the velocity of (a) ball A, (b) ball B. .-- SOLUTION Velocity vectors: Vo = Vo(cos300i + sin300j) v A = V A (sin 7.4°i + cos 7.4°j ) VB= VB (sin 49.3°i - cos49.3°j) ve = ve(cos45°i + sin45°j) Conservation of momentum: Divide by mA = mB = me and substitute data. 4( cos300i + sin300j) = vA (sin 7.4°i + cos 7.4°j) + VB (sin 49.3°i - cos49.3°j) + 2.1( cos45°i + sin 45°j) Resolve into components and rearrange. i:
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HW5_solutions - PROBLEM 14.5 l~ I ~E-L,tJ T kB A bullet is...

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