1. A 3.50 g bullet is firedhorizontally at two blocks resting on a smooth surfac

Question

1.      A 3.50 g bullet is firedhorizontally at two blocks resting on a smooth surface. The

     bullet passes through the first block,with mass 1.20 kg and embeds itself in the

     second block, with mass 1.80 kg. The first block has a velocity of 0.630 m/s after the

     bullet passes through it. Thesecond block has a velocity of 1.40 m/s with the bullet

     embedded in it. Neglect the massremoved from the first block as the bullet passes

     through. What is the initialspeed of the bullet?

Please show ALL work and Formulas used to get the answer. Thankyou!

Explanation / Answer

from the conservation of the momentum we have               initial momentum = final momentum                             0.0035 kg * V = ( 1.20kg *0.630m/s ) + (1.80kg +0.003kg) *1.4 m/s                                         = 0.756 kg m/s +2.5242 kg m/s                                      V = ( 3.2802kg m/s ) / 0.0035 kg                                         = 932.7 m/s V is the velocity of th e bullet

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