Bullet Through Pendulum Bob-Inelastic Collision

[DubbTom]

Homework Statement

A bullet of mass m and speed v passes completely through a pendulum bob of mass
M. The bullet emerges on the other side of the pendulum bob with half its original
speed. Assume that the pendulum bob is suspended by a stiff rod of length L and
negligible mass. What is the minimum value of v such that the pendulum bob will
barely swing through a complete vertical cycle

Homework Equations

m₁v_1i² + m₂v_2i² = m₁v_1f² + m₂v_2f²

w=Δk+Δp

k = 1/2mv²

p = mgh

The Attempt at a Solution

initial kinetic energy of bullet = potential energy of bob @ max height + final kinetic energy of bullet

1/2mv² = Mg2L + 1/2 m(1/2v)²
1/2mv² - 1/8 mv² = 2MgL
3/8mv²=2MgL
v² = (16MgL)/(3m)
v = 4[(MgL)/(3m)]^(1/2)

Answer is [4M(gL)^(1/2)]/m

The key sets
1/2MV_b² = Mg2L

if V_b = velocity of the bob, then at max height the bob is not moving hence velocity is 0 and all of the energy that the bob carried at the bottom is now potential energy.

and then solves via the equation:
mv = MV_b + mv/2

initial momentum = momentum of bob + momentum of bullet after going through bob

If everything that I have said about the professors method is true, then I feel like I am beginning to understand why the professors way works, but why doesn't my method come to the same conclusion.

 

[TSny]

Hello, and welcome to PF!

The Attempt at a Solution

initial kinetic energy of bullet = potential energy of bob @ max height + final kinetic energy of bullet

Is the collision of the bullet with the bob elastic or inelastic?

 

[DubbTom]

Hello, and welcome to PF!
Is the collision of the bullet with the bob elastic or inelastic?

Looking back on the pdf, it is inelastic because kinetic energy is being transformed into potential energy.
But why can't we say that
initial kinetic energy = sum of final energies?

 

[TSny]

Looking back on the pdf, it is inelastic because kinetic energy is being transformed into potential energy.
But why can't we say that
initial kinetic energy = sum of final energies?

What type of energy does an inelastic collision produce? Did you include that type of energy in your sum of final energies?

 

[DubbTom]

What type of energy does an inelastic collision produce? Did you include that type of energy in your sum of final energies?

An inelastic collision will take kinetic energy and transform it into some other types of energy like potential or thermal. (from my textbook). Am I not allowed to say that the initial kinetic energy of the bullet is equal to the potential energy imparted on the bob + what is left of the kinetic energy of the bullet?

K_A + K_B = K'_A + K'_B + thermal & other forms of energy

initially, the bullet has all of the energy of the system, so K_B would be 0, after the bullet will have 1/2 the velocity so it will have 1/4 the kinetic energy and the kinetic energy of the bob should be 0 because it has no velocity at the top of the loop. So potential energy of the bob must be 3/4 K_A.
I feel like something in my logic isn't correct though, as it gives me an entirely different answer.

 

[consciousness]

You simply can't conserve energy before and after the collision here. Not all of the kinetic energy lost by the bullet will be converted into kinetic energy in the bob. There will be some losses.

 

[TSny]

K_A + K_B = K'_A + K'_B + thermal & other forms of energy

This equation is correct. Note that you need to include the thermal energy on the right side. But you don't know the amount of thermal energy created in the inelastic collision, so this equation has too many unknowns.