Question

Your answer is partially correct. Try again. A 6.98-9 bullet is moving horizontally with a velocity of +364 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 1219 g, and its velocity is +0.717 m/s after the bullet passes through it. The mass of the second block is 1620 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. +364 m/s = Block 1 Block 2 (a) Before collision +0.717m/s block 2 m block 1 = 12199 mblock 2 = 16208 mbullet = 6.988 (b) After collision (a) Vblock2 = Number 7--238.785 238.785 units. m/s Units m/s units im/s (b) KEfter/KEbefore = Number 307.77 Click if you would like to Show Work for this question: Open Show Work

Answer 1

Initial velocity velocity of bullet- before collision Voi = 364015 velocity of book I before collision, o, ;= omis Velocity of Hocil a before colligion Obs,i= omis mars of bullet, mo= 6.989= 0.00698 kg Maes of blocki ,mo,= 12199= 1.219 kg Mass of Hock 2 , inby = 1620g = 1.620 kg velocity of Hock 1 after collision, belf = 0.717 mi past a. - According to Principle of confervation of Momentum Total linees Momentum Lefüle collision = Total lineer - Mohon tum after Collision maloi = (ma+mbe} Uplockat my of 0.00698X964 = 10.006987 1.620 16slovcet 1.219X0.717 Urlocky = 1.02 mis . part. 6. Past. 6. to Eafter K.E defel ex mat mode ) Uslocka + £mo, 1967 e malo? H0.006987 1.62.1X1.02?+ 4x1.2.1980.7032 ax 0.00698x3692 K.Eafter 2.51 -- 2.0 K.E before