# Bullet and a mass on a string. conservation problem.

1. Oct 6, 2009

### Dissonance in E

A bullet with mass m and velocity v, penetrates a mass M suspended on a piece of string lenght R.
The bullet loses half of its velocity in the impact and carries on in a straight line.
What must the minimum magnitude of v be so that the mass would spin a full circle.

p = mv, 1/2mv^2, mgh,

3. The attempt at a solution
Ok so im not sure whether this is a moment of kinetic energy problem but im leaning towards the Ke option. Im assuming that the mass needs only to get to the top of the circle as a result of the collision where all of its kinetic energy has turned into potential energy which then allows it to fall back down thus making a full revolution.

Bullet's initial Ke = 1/2mv^2
Bullet's final Ke = 1/2m(v/2)^2 = 1/2m(v^2/4) = (mv^2)/8
Ke gained by mass M = (mv^2)/8
Energy needed for mass M to get to the top of the circular path = Mgh = Mg2R
Based on the assumption that work done in raising a mass to a certain height is independent of the chosen path.
Thus (mv^2)/8 = Mg2R and solving for v we get
v = SQRT(16MgR/m)

A swing and a miss?

2. Oct 6, 2009

### rl.bhat

To spin a full circle, the mass M must have some velocity at the top of the circle. What should be this minimum velocity? What will be the total energy of M at the top of the circle?

3. Oct 6, 2009

### Dissonance in E

Well now consideing that the string isnt a rigid body I guess the mass would need to have some velocity at the top so that it would travel along the circular path rather than drop straightdown. Not sure about how to go about speculating its magnitude though, I would guess its got something to do with the velocity´s x component (it´s only component at the top) being large enough to keep the mass on the circular path while gravity pulls it down.

That said, the total energy at the top is not Mg2R but, Mg2R + 1/2Mv^2 where v = is the unknown velocity at the top.

4. Oct 6, 2009

### rl.bhat

Mv^2/R is the centripetal force which keeps the mass in the circular motion. It is provided by the weight of the mass.
So Mv^2/R = Mg.
This gives you the minimum velocity at the top of the circle.

5. Oct 6, 2009

### Dissonance in E

Ok so Mv^2=Mg, v^2 = RG v = SQRT(Rg)
Total energy at the top:
= Mg2R + 1/2M(SQRT(Rg))^2
= 5MgR/2

We know that the kinetic energy lost by the bullet = total energy of M at the top.
Thus:
(mv^2)/8 = 5MgR/2
Rearranging for v:
v = Sqrt(20MgR/m)

6. Oct 6, 2009

### rl.bhat

To find the energy of the bullet and the mass M at the bottom of the circle, you have to find the velocity V of m and M after impact.

7. Oct 6, 2009

### Dissonance in E

1/2mv^2 - 1/2m(v/2)^2 = 3mv^2/8 = Ke lost by the bullet during impact.
3mv^2/8 needs to satisfy 5MgR/2
Rearranging for v gives v = Sqrt(20MgR/3m)

Last edited: Oct 6, 2009
8. Oct 6, 2009

### DocZaius

Isn't the centripetal force (the force that aims towards the center of the circle) provided by the tension of the string? The only time the weight of the mass is pointing towards the center of the circle is at the very top of the circle. All other times, it seems to me, it is the string's tension that keeps the mass in circular motion? What am I missing?

9. Oct 6, 2009

### Dissonance in E

The tension is equal to Mg though. If the mass were just hanging on the string it would have a downward force equal to Mg, since it is suspended on the string the string tension has to provide an equal and opposite force for the mass to stay where it is. No?

10. Oct 6, 2009

### DocZaius

I should probably start a new thread for this and not hijack your homework thread so I'll stop here, but my understanding is: I do agree that the tension of the string is equal in magnitude and opposite in direction when the weight is hanging down, but in terms of magnitude, I am not sure that is necessarily the case at other points on the circle. Considering the point on the very top of the circle for example, it seems that the weight of the mass contributes to the (centripetal) force necessary to keep the mass going in a circle, and that the contribution of the tension could be less then, and is not necessarily equal to Mg.

11. Oct 6, 2009

### rl.bhat

At the top of the circle total mass of the system is (m+M) and its velocity V at the bottom is such that,
V^2 = 5gR
So after impact kE of m and M = 1/2*(m+M)*5gR = 1/8*mv^2
Now find v.

12. Oct 7, 2009

### Dissonance in E

The bullet goes throught the mass M though and carries on in a straight line with velocity v/2. Why would the mass of the system at the top be m + M?

13. Oct 7, 2009

### rl.bhat

Sorry. I didn't notice that.