Question

A 7.55-g bullet is moving horizontally with a velocity of +360 m/s, where the sign + indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1247 g, and its velocity is +0.619 m/s after the bullet passes through it. The mass of the second block is 1640 g. (a) What is the velocity of the second block after the bullet imbeds itself? (b) Find the ratio of the total kinetic energy after the collision to that before the collision. +360m/s Block 1 Block 2 (a) Before collision +0.6191/s block 2 m block 1 = 12478 mblock 2 = 16408 mbullet = 7.55g (b) After collision (a) Vblocky = Number Units (b) KEaſter/KEbefore = Number Units

Answer 1

and after collision Answer: Conserving the momentum of the system before of bullet with the blocks mvo=minit (m+m2) Vo= mvo-mini m+m2 0.00755 X 360 -1.247 x 0.619 0.00755+ 1.640 2.718 -0.771893 1.64755 V=1.18 mis 1 (6) kinetice kinetic energy of the system bebole collision. KE;= £ £ more = x(0.00755)(360) KE;= 489.245 Kinetic energy of the system abter collision KE f = 1 ex murt 1 - 3 (Mths) va *(1.242)(0.619) 756-00755+ 1•640) 118 ) = 0.29684835 +1.14702431 KEL=10445 KEL - 1044 KE; 489.24 = 2.948163