Minimum Initial Speed for Complete Circular Loop in Inelastic Collision

[anubis01]

Homework Statement

A 18.00 Kg lead sphere is hanging from a hook by a thin wire 3.80 m long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 4.50kg steel dart that embeds itself in the lead sphere.

a)What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision

Homework Equations

v₁=(m_s+m_d/m_d)v₂

The Attempt at a Solution

d=dart s=sphere

mdvd=(ms+md)v2
vd=(ms+md/md)v2

v2 makes a complete circular loop so

1/2(md+ms)v2²=(md+ms)2gr
v2=sqrt(4gr)
vd=(ms+md/md)sqrt(4gr)
vd=(18+4.5/4.5)sqrt(4*9.8*3.8)
=5*12.204=61.02m/s

vd=61.02m/s

Now my problem with this is the TA said I took all the right steps to solve this equation but I'm still getting an error using masteringphysics and I have gone over my work for errors 10 times by now. Is the work I did correct or am I missing something in my calculations.

Thanks in advanced for any assistance.

 

[alphysicist]

Hi anubis01,

Homework Statement

A 18.00 Kg lead sphere is hanging from a hook by a thin wire 3.80 m long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 4.50kg steel dart that embeds itself in the lead sphere.

a)What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision

Homework Equations

v₁=(m_s+m_d/m_d)v₂

The Attempt at a Solution

d=dart s=sphere

mdvd=(ms+md)v2
vd=(ms+md/md)v2

v2 makes a complete circular loop so

1/2(md+ms)v2²=(md+ms)2gr

This equation is incorrect. Do you see what's wrong with it, since the object is going in a circular path (and attached to a wire instead of something like a rod)?

 

[anubis01]

oh I see now, I didn't take into account that the velocity at the top of the loop needed to make one complete loop, so then the V2 comes out to sqrt(5gr). Thanks for the help.

 

[Perillux]

mdvd=(ms+md)v2
vd=(ms+md/md)v2

Why did you put the md inside the parenthesis "md/md" it should be on the outside:
vd = [(ms+md)v2]/md
OR
vd = ms*v2/md + v2

 

[alphysicist]

Hi Perillux,

Why did you put the md inside the parenthesis "md/md" it should be on the outside:
vd = [(ms+md)v2]/md
OR
vd = ms*v2/md + v2

It would definitely be less ambiguous to put the md outside the parenthesis, but I believe anubis01 had it interpreted correctly since he had:

(18+4.5/4.5) = 5

I've seen some textbooks use a+b/c to mean $$\frac{a+b}{c}$$. It can be confusing (since it doesn't follow the order of operations we all learn in elementary school), and so is not a good idea, but I think in this case anubis01 was just being a bit sloppy in typing his post.

 

[freakinclown]

oh I see now, I didn't take into account that the velocity at the top of the loop needed to make one complete loop, so then the V2 comes out to sqrt(5gr). Thanks for the help.

why does V2 become sqrt(5gr) after taking into account the velocity at the top of the loop? how do you take into account the velocity at the top?