Hi, Need help, please, step bystep. A 3.54-g bullet is fired horizontally at two

Question

Hi,

Need help, please, step bystep.

A 3.54-g bullet is fired horizontally at two blocks resting on a frictionless tabletop, as shown in Fig. 38a. The bullet passes through the first block, with mass 1.22 kg, and embeds itself in the second, with mass 1.78 kg. Speeds of 0.630 m/s and 1.48 m/s, respectively, are thereby imparted to the blocks, as shown in Fig. 386. Neglecting the mass removed from the first block by the bullet, find the speed of the bullet immediately after emerging from the first block and the original speed of the bullet.

Explanation / Answer

    Mass of the bullet, m = 3.54 g                                     = 0.00354 kg     Mass of the first block, M1 = 1.22 kg     Mass of the second block, M2 = 1.78 kg     Original speed of the bullet, U = ?     Speed of the bullet after emerging from thefirst block , V = ?     Case I : ( Collision with first block )     Velocity of the first block after collision, V1 = 0.63 m/s     Total initial momentum = Total finalmomentum     m U = M1 V1 + m V     m ( U - V ) = M1 V1     U - V = 1.22 * 0.63 / 0.00354              = 217.1 m/s     Case II : ( Collision with the second ball )     Velocity of the second block + bullet aftercollision, V2 = 1.48 m/s     Total initial momentum = Total finalmomentum     m V = ( M2 + m ) V2         V = ( 1.78 + 0.00354 ) *1.48 / 0.00354            = 745.7 m/s     Speed of the bullet after emerging from thefirst block , V = 745.7 m/s     Original speed of the bullet, U = 217.1 +V                                                   = 962.8 m/s

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