# Bullet, block, kinetic energy

1. Dec 2, 2007

### Dorney

[SOLVED] Bullet, block, kinetic energy

1. A small block of mass 2m initially rests on a track at the bottom of a circular, vertical loop-the-loop, which has a radius r. The surface contact between the block and the loop is frictionless. A bullet of mass m strikes the block horizontally from the left with initial speed Vo and remains embedded in the block as the block and bullet circle the loop. Determine the following in terms of m, Vo, r, and g. Diagram is the same as in the link, except that radius is just r.

http://dev.physicslab.org/img/9e6268fd-0b0a-41c5-83f2-591f0f56aa70.gif

a) the speed of the block and bullet immediately after impact.
b) the kinetic energy of the block and bullet when they reach a point on the loop that is a height of r units from the bottom (far right side of the loop).
c) the minimum initial speed of the bullet if the block and bullet are to successfully execute a complete circuit of the loop.

2. m1v1 + m2v2 = (m1+m2)Vf
KE = (1/2)mv^2
PEg = -mgh

3. a) I found the speed equals Vo/3 from the first equation.

m1=m, v1=Vo, m2=2m, v2=0.
mVo=3mVf
Vf= Vo/3.

b) I found that the KE (final) = (Vo^2m / 6) + 3mgr by using the second and third equations.

change in KE= -change in PEg (because Ff=0)
I substituted in Vo/3 for Vi and r for the height.
I'm not sure if I have the signs correct or even the method.

c) Vo min = square root of (gr)??

Can someone please tell me if this is correct? Thanks in advance!! :)

Last edited: Dec 3, 2007
2. Dec 4, 2007

### Dorney

Last edited: Dec 4, 2007
3. Dec 11, 2007

### vector3

$$\underbrace{m^{b}_{1}\cdot v^{b}_{1} + m^{b}_{2}\cdot v^{b}_{2}}_{Before\;Colision} = \underbrace{m^{a}_{1}\cdot v^{a}_{1} + m^{a}_{2}\cdot v^{a}_{2}}_{After\;Colision}$$

Where superscripts b = efore and a = [a]fter the colision respectively.

$$m_2\cdot v^b{_2} = 0$$ and the total mass after the collision, $M_t = m_1 + m_2$

The velocity of $m_1$ before the colision is therefore:

$V^b_{1} = \frac{m_1}{M_t} \cdot V_a$ in agreement with your solution.

The initial (and total) kinetic energy $E_o$ of the result system is $\frac{1}{2}M{_t} V^2_a =\frac{1}{6} M_t V^2_1$

The total energy at circle right mid height is equal to the sum of the potential and kinetic energies at that point:

$M_t g r + \frac{1}{2} M_t V^2_r$ however; the total energy is input to the system is $\frac{1}{6} M_t V^2_1$ thus

$M_t g r + \frac{1}{2} M_t V^2_r = \frac{1}{6} M_t V^2_1$or $\frac{1}{2} M_t V^2_r = \frac{1}{6} M_t V^2_1- M_t g r$

At the top of the circle apply Newton's law and assume the radial force = the gravitational force
$mg \Longrightarrow \sum{F} = ma[itex] \Longrightarrow M_t g = M_t\frac{ V^2}{r}$

again, in agreement with your solution.

Looks like we end up in the same place...

Last edited: Dec 11, 2007