A 19 gram bullet is travelling at 750 m/s towards a 2 kgblock. After striking th

Question

A 19 gram bullet is travelling at 750 m/s towards a 2 kgblock. After striking the block, the bullet becomes embeddedin the block. How much does the block move? Here's how I did it: p1 =mbulletvbullet = (.019kg)(750 m/s) = 14.25 kg-m/s p2 = 14.25 kg-m/s = (.019 kg + 2kg)vboth vboth = 14.25/2.02 = 7.05m/s vf2 =vi2 +2a(s) 7.052 = 0 - 19.6(s) 49.7/-19.6 = s -2.53 = s Where did I go wrong? Why is the change in positionnegative when the block would have to move in the positivey-direction (which in the case is up)? Thanks in advance foryour help. vboth = 14.25/2.02 = 7.05m/s vf2 =vi2 +2a(s) 7.052 = 0 - 19.6(s) 49.7/-19.6 = s -2.53 = s Where did I go wrong? Why is the change in positionnegative when the block would have to move in the positivey-direction (which in the case is up)? Thanks in advance foryour help.

Explanation / Answer

                Given that the mass of the buller is m = 19 g = 19*10-3kg                  Initialvelocity of the bullet is U = 750 m/s                 The mass of the block is M = 2 kg ------------------------------------------------------------------------------------------           Fromthe conservation of momentum                  mU + M(0) = (m +M ) V                                 V = mU / (m+M )                                      =--------- m/s     From the conservation of energy just afterthe collision and at the            (1/2)(m+M) V2 = (m+M)gh                                V2 =  2gh                                  h = V2/2g                                                                                                             = ------------ m      

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