Bullet Collision Question

  • #1

Homework Statement


A 6.64-g bullet is moving horizontally with a velocity of +340 m/s, where the sign + indicates that it is moving to the right. 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 1234 g, and its velocity is +0.577 m/s after the bullet passes through it. The mass of the second block is 1595 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.

Homework Equations




The Attempt at a Solution


i've already found the answer for (a) Vf = 0.965 m/s, but i'm stuck in (b)

Steps:
1. KE(before) = 1/2(0.00664kg)(340 m/s)^2 = 3837.92
2. KE(after) = 1/2(1.60164kg)(0.965 m/s)^2 = 0.745743604
 

Answers and Replies

  • #2
23
3
The Kinetic energy after should probably include the KE of the first moving block too.
 
  • #3
So would the KE(after) = 1/2(1.234kg)(0.577m/s)^2 + 1/2(1.60164kg)(0.965 m/s)^2 = 0.951160797
 
  • #4
23
3
Yeah, does that give you the right answer?
 
  • #5
if i got my equation correct, the ratio should be:

KE(after)/KE(before) = 2.48x10^-4 (which i got wrong)
 
  • #6
23
3
2.48 x 10^-3 if we include the key of the first block, 1.943 x 10^-3 if we don't.
 
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  • #7
ah, you're right. i made a mistake on my sig figs. Thanks for the help!

the answer was 2.48x10^-3
 

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