Angular momentum: bullet through block with rod and pivot

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Homework Help Overview

The problem involves a wooden block attached to a rigid rod, with a bullet colliding with the block. The context includes angular momentum and kinetic energy considerations in a physics scenario, specifically focusing on the bullet-block system's dynamics during the collision.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of angular momentum for the bullet-block system and the fraction of kinetic energy converted into internal energy during the collision. There are attempts to derive expressions for angular momentum and kinetic energy changes.

Discussion Status

Some participants are providing feedback on the original poster's calculations, with one noting a potential oversight in squaring terms in the kinetic energy equation. There is an acknowledgment of the correctness of certain expressions, but no consensus on the final answers has been reached.

Contextual Notes

Participants mention the importance of correctly identifying the variable for length as "script l," indicating a potential source of confusion in the problem setup.

blackheart
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7. 0/10 points All Submissions Notes Question: SerPSE8 11.P.037.
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A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length script i and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.

(a) What is the angular momentum of the bullet–block system about a vertical axis through the pivot? (Use any variable or symbol stated above as necessary.)
My work: L=mvr sin (theta)

r=l and sin (theta) = sin 90 = 1

so L=mvl However, this answer is wrong.

I also tried L=(M+m)vl but this was incorrect also

(b) What fraction of the original kinetic energy of the bullet is converted into internal energy in the system during the collision? (Use any variable or symbol stated above as necessary.)
ΔK/Ki = ?

My work: delta K/Ki = (Kf-Ki)/Ki
K= (1/2)mv^2
Li=Lf
L=mvrsin(theta)

Li= mv
Lf=(m+M)Vf l
mv=(m+M)Vfl

Vf= mv/l(m+M)

(Kf-Ki)/Ki = [(1/2)(M+m)Vf^2 - mV^2]/mV^2
(Kf-Ki)/Ki = [(1/2)(M+m)(mv/(m+M))^2 - mV^2]/mV^2

This is also wrong.
(In bold are general equtions.)
 
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hi blackheart! welcome to pf! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)
blackheart said:
(a) What is the angular momentum of the bullet–block system about a vertical axis through the pivot? (Use any variable or symbol stated above as necessary.)
My work: L=mvr sin (theta)

r=l and sin (theta) = sin 90 = 1

so L=mvl However, this answer is wrong.

looks right to me. :confused:
(b) What fraction of the original kinetic energy of the bullet is converted into internal energy in the system during the collision? (Use any variable or symbol stated above as necessary.)
ΔK/Ki = ?

your V = mv/(M+m) is correct, but when you squared V, you forgot to square m or (M+m) :redface:
 
So I know it's an old post, but I just had the same problem. While entering it into webassign, I didn't realize that they gave the value of length l as "script l". If you go into the symbols section of the toolbar that pops up when you go to answer, you should find the "script l" which looks like a cursive l and you enter it like you have it.
 
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legnip said:
So I know it's an old post, but I just had the same problem. While entering it into webassign, I didn't realize that they gave the value of length l as "script l". If you go into the symbols section of the toolbar that pops up when you go to answer, you should find the "script l" which looks like a cursive l and you enter it like you have it.

Thanks a bunch man!
 

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