Question

Chapter D3, Problem D3/142 A 55-g projectile traveling at 670 m/s strikes and becomes embedded in the 35-kg block, which is initially stationary. Compute the energy lost during the impact. Express your answer as an absolute value |AE| and as a percentage n of the original system energy E. 55 g 670 m/s 35 kg Answers: ΙΔΕΙ = n = %

Answer 1

M U-670 s [35kg IMAM yv m=man of bullet a 55g=0.055kg ME Mass of block - - 35 kg ta ini tial velocity of blueet : 670m/s. V = Final combined 'velocity of bullet and block Applying conservation of momentum principle :- mu + MCO) - (m +M)v =) 0.05546 70 +0 - (35 +0.055)V 36-89 = 35.055V - v -1.051 m/s AF = Energy lost during impact 1 mu? - (m +M) va { [0.055X(670) - 3 5.055 (1.051)2] $(24689.75-38-721) - 12325.5J. :. LAE! - 12.325 KT n = percentage change energy a AE *100 initial Energy 12325.5 4x0.055X670 no 99.84% in -X100 2