Angular momentum: bullet through block with rod and pivot

[blackheart]

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.)

 

[tiny-tim]

hi blackheart! welcome to pf!

(try using the X² and X₂ icons just above the Reply box )

(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.

(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)

 

[legnip]

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.

 

[Ticklez_Panda]

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!

 

[Nickie]

I would like to commend you on your attempt to solve this problem. However, there are a few mistakes in your calculations that may have led to the incorrect answers.

Firstly, in part (a), the correct equation for angular momentum in this scenario is L = (M+m)vl, where v is the velocity of the bullet before the collision. This is because the bullet and block both contribute to the angular momentum of the system, not just the bullet alone.

Secondly, in part (b), the correct equation for the fraction of kinetic energy converted into internal energy is ΔK/Ki = (Kf-Ki)/Ki, where Kf is the final kinetic energy of the system after the collision and Ki is the initial kinetic energy of the bullet. In your calculation, you have used the initial kinetic energy of the entire system (bullet + block) instead of just the bullet.

Additionally, when calculating Kf, you should use the final velocity of the bullet after the collision, which can be found by applying conservation of momentum in the horizontal direction. This will give you a more accurate answer for the fraction of kinetic energy converted into internal energy.

Overall, it is important to carefully consider the variables and equations that are relevant to the specific scenario in order to accurately solve a physics problem. I would suggest reviewing the concepts of angular momentum and conservation of energy to better understand this situation. Keep up the good work!