I got part a and part e, but need help on b,c,d. Thanks so much! Problem 8-92a:

Question

I got part a and part e, but need help on b,c,d. Thanks so much!

Problem 8-92a:
A 3.14-kg block is traveling in the -x direction at 5.48 m/s, and a 1.15-kg block is traveling in the +x direction at 3.47 m/s. Calculate the velocity vcm of the center of mass.
-3.08 m/s
You are correct.

Problem 8-92b:
Subtract vcm from the velocity of each block to find the velocity of each block in the center-of-mass reference frame. Give velocity of the 3.14-kg block first, then the 1.15-kg block.

Incorrect.

Problem 8-92c:
After they make a head-on elastic collision, the velocity of each block is reversed (in the center-of-mass frame). Calculate the velocity of each block in the center-of-mass frame after the collision. Calculate the velocity for the 3.14-kg block first, then the 1.15-kg block.

Incorrect.

Problem 8-92d:
Transform back into the original frame by adding vcm to the velocity of the 3.14-kg block, then do the same calculation for the 1.15-kg block.


Problem 8-92e:
Check your result by calculating the initial and final kinetic energies of the blocks in the original frame. Enter the initial kinetic energy first.
5.41×101 J 5.41×101 J
You are correct.

Explanation / Answer

b. the first block move along -x axis so its velocity is abregalic is -5.48m/s 2nd block velocity is 3.47 m/s. so in the center of mass reference frame. v1=-5.48-(-3.08)=-2.4(m/s) v2=3.47-(-3.08)=6.55(m/s) c. Because the collision is head on and elastic so v1'=2.4(m/s) and v2'=-6.55(m/s) (reverse the direction of movement of both particles) d. So in normal frame. v1'_nor=2.4+(-3.08)=-0.68(m/s) v2'_nor=-6.55+(-3.08)=-9.63(m/s)

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