Question

Chapter 07, Problem 42 GO Chalkboard Video A 6.80-g bullet is moving horizontally with a velocity of +346 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 1185 g, and its velocity is +0.688 m/s after the bullet passes through it. The mass of the second block is 1607 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 +346m/s Block 1 Block 2 (a) Before collision +0.688m/s block 2 mblock 2 1607 Mbullet 6.80g mblock I 1185 (b) After collision

Answer 1

al Let the velouty at block & be v. Momentum before the collision Pi= 6.80 Xo 3 x 346 Kymis. P; = 2353 kgmis - U Momentum after collision PE 118570-3x0.688 +(1607 +6.8 D x V X10-3 °@ = 08153+ 1.614V 22) from (1) 4 (2) 0-8153 +1614 V = 2-353 71.810 v= 1.5377 = 1.5377 co.953 mis lbiy N=0.953 mis e K = *x1185X10-3x6.668)2+{(1607+00) (0-95333x10-3 2x X10-3 *348) 560.91 3+ 1465.668 814,06878 21026.5 f = 0.0025 - OOOLS 14,068 v6 This = 0,0025)