Find the ratio of the total kinetic energy after the collision

[emerica1254]

Homework Statement

A 5.39-g bullet is moving horizontally with a velocity of +354 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 1249 g, and its velocity is +0.555 m/s after the bullet passes through it. The mass of the second block is 1504 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

m1Vf1+m2vf2=m1v01+m2v02
KEafter/KEbefore

The Attempt at a Solution

i got 1.26 for the velocity of the second block but it was wrong and my answer was wrong for the ratio of the collision too. This question has me really stumped

 

[collinsmark]

i got 1.26 for the velocity of the second block but it was wrong and my answer was wrong for the ratio of the collision too. This question has me really stumped

Use conservation of momentum for part (a).

What is the total momentum of the entire system (one bullet and two blocks) before any collisions happen?

What is the final momentum of the system, after the bullet is embedded in the second block?

 

[kuruman]

Kinetic energy is not conserved because there is friction between the bullet and the blocks. You need to conserve linear momentum only. There are two instances when you conserve momentum

Before 1 = bullet has not collided with first block
After 1 = bullet has gone through first block but has collided with second block

Before 2 = bullet has gone through first block but has collided with second block
After 2 = bullet has fully embedded itself in block 2

For each instance, you need to say

P_before = P_after

 

[emerica1254]

So do i need to figure out the velocity of the bullet after it goes through the first block, then use that to find the velocity of the second block? And do i use MbVb+MBVB=MbfVbf+MBfVBf
b= bullet, B=block, f= final

 

[kuruman]

So do i need to figure out the velocity of the bullet after it goes through the first block,

Yes.

then use that to find the velocity of the second block? And do i use MbVb+MBVB=MbfVbf+MBfVBf
b= bullet, B=block, f= final

You need to write two separate equations for the two instances I pointed out.

 

[collinsmark]

So do i need to figure out the velocity of the bullet after it goes through the first block,

Yes.

Actually, since momentum is conserved throughout, calculating the velocity of the bullet between the first and second block is unnecessary. It doesn't do any harm to do so, but it's not really needed. The total momentum of the system before either collision take place is the same as the total linear momentum of the system after both collisions take place.

 

[emerica1254]

For some reason i get -611 for the velocity of the bullet, am i doing my algebra wrong?
Vb=(Vb_before/VB_after)-MB
When i solve for just the second block it get Mb*Vb_before/MB and i get 1.26 which is not the right answer

 

[collinsmark]

For some reason i get -611 for the velocity of the bullet, am i doing my algebra wrong?
Vb=(Vb_before/VB_after)-MB
When i solve for just the second block it get Mb*Vb_before/MB and i get 1.26 which is not the right answer

You've got something switched around and not right, if you're trying to find the velocity of the bulled after the first collision, but before the second.
m_bv_{b_before} = m_bv_{b_after} + m_Bv_B

But like I said in my previous post, finding the speed of the bullet between the first and second collisions is unnecessary if you're only trying to find the velocity of the second block after the second collision (and if you're looking for the kinetic energies before and after both collisions [and if you don't care about the kinetic energy in-between the two collisions]).