Momentum/Conservation of Energy problem.

[Sefrez]

I am given the problem:

As shown below, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length script i and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? (Use L for script i, g for gravity, and M and m as appropriate.)

And I derive minimum velocity initial as v = (2M/m)*sqrt(5gL) where as every solution I find is 4M/m*sqrt(gL). I am only able to get this when saying that K_i = 2m*g*L. That is, the initial kinetic energy of the suspended mass is the potential at the top of the swing.

Though, wouldn't this imply that the kinetic energy at the top of the swing is zero, and thus zero velocity? And if that was the case, wouldn't the mass free fall (no radial acceleration)? What I did was say the velocity at the top should be such that the downward radial acceleration is g, or v_top = sqrt(g*L).

EDIT:
Wait, I just caught something. It says stiff rod. They should bold that... So is my other solution correct given a "negligible" massed string then? :D

 

[grzz]

Can you show how you got v = (2M/m)*sqrt(5gL)?

 

[NascentOxygen]

As shown below, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length script i and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? (Use L for script i, g for gravity, and M and m as appropriate.)

What is meant by "script i"?

And I derive minimum velocity initial as v = (2M/m)*sqrt(5gL) where as every solution I find is 4M/m*sqrt(gL). I am only able to get this when saying that K_i = 2m*g*L. That is, the initial kinetic energy of the suspended mass is the potential at the top of the swing.

K.E. = P.E. sounds good. 4M/m*sqrt(gL) seems right.

Though, wouldn't this imply that the kinetic energy at the top of the swing is zero, and thus zero velocity?

Yes.

 

[grzz]

I got the same as NascentOxygen got. That is the reason for suggesting to the OP to show the method.

 

[asteriatic]

Hello,

Thank you for sharing your thoughts on this problem. I would like to provide some clarification and further explanation for the minimum velocity initial that you derived and the solutions that you have come across.

Firstly, let's consider the conservation of energy in this problem. The bullet initially has a kinetic energy of 1/2mv^2 and when it passes through the pendulum bob, it transfers some of its energy to the bob. This results in the bob gaining a velocity of v/2. However, since the bullet continues to move with a velocity of v/2, it still has some kinetic energy which is not transferred to the bob. This means that the total kinetic energy of the system (bullet + bob) after the collision is not equal to the total kinetic energy before the collision.

Now, let's look at the pendulum bob. At the top of its swing, it has maximum potential energy (MgL) and zero kinetic energy. As it swings down, it loses potential energy and gains kinetic energy. At the bottom of its swing, it has maximum kinetic energy (1/2Mv^2) and minimum potential energy (zero). From conservation of energy, we can equate the initial kinetic energy of the bullet to the total kinetic energy of the system at the bottom of the swing (bullet + bob). This gives us the equation 1/2mv^2 = 1/2Mv^2 + 1/2M(v/2)^2.

Simplifying this, we get v = (2M/m)*sqrt(5gL). This is the minimum velocity initial that you have derived and is also the correct solution. The other solution that you have come across, v = 4M/m*sqrt(gL), is incorrect. This is because it assumes that all the kinetic energy of the bullet is transferred to the bob, which is not the case as explained earlier.

Now, coming to your question about the velocity at the top of the swing being zero. This is not the case. The velocity at the top of the swing is not zero because the pendulum bob is still in motion. It has a velocity of v/2 which is the result of the bullet transferring some of its energy to the bob. The downward radial acceleration of the bob is not equal to g at the top of the swing because the bob is still moving horizontally and not just vertically. This results in a slightly