# Momentum/Conservation of Energy problem.

• Sefrez
In summary, the problem is asking for the minimum value of initial velocity (v) for a bullet of mass m and speed v to pass through a pendulum bob of mass M suspended by a stiff rod of length script i. The bullet emerges with a speed of v/2 and the bob should barely swing through a complete vertical circle. The minimum value of v is derived as v = (2M/m)*sqrt(5gL), which is different from the commonly found solution of 4M/m*sqrt(gL). This discrepancy can be explained by equating the initial kinetic energy (K_i) to the potential energy (P.E.) at the top of the swing. However, this would imply a zero velocity at the top of
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

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

Sefrez said:
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.

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

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

## 1. What is the definition of momentum?

Momentum is a measure of an object's motion and is equal to its mass multiplied by its velocity.

## 2. How is momentum conserved in a closed system?

In a closed system, the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction. This is known as the law of conservation of momentum.

## 3. Can momentum be transferred between objects?

Yes, momentum can be transferred between objects through collisions or interactions. The total momentum of the system remains constant, but the individual objects may experience changes in their momentum.

## 4. How does an object's mass and velocity affect its momentum?

The greater an object's mass and velocity, the greater its momentum will be. This is because momentum is directly proportional to both mass and velocity.

## 5. What is the relationship between momentum and energy?

Momentum and energy are both conserved quantities in a closed system. This means that the total momentum and total energy before a collision or interaction will be equal to the total momentum and energy after the collision or interaction.

• Introductory Physics Homework Help
Replies
55
Views
2K
• Introductory Physics Homework Help
Replies
20
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
832
• Introductory Physics Homework Help
Replies
21
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
993
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
417
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
11
Views
3K
• Introductory Physics Homework Help
Replies
7
Views
1K