Bullet passing through block of wood. Find max height of the wood.

[demenius]

Homework Statement

A bullet of mass 0.005kg is shot at a suspended block of wood with mass 1kg. The bullet initially has a speed of 200m/s, and passes through the block of wood. The bullet passes through the wood, and 50J of work is done deforming the shape of the bullet and block. If the block is suspended from a very long string, what is the maximum height to which it will rise?

http://imageshack.us/photo/my-images/266/bulletblock.png/"

Homework Equations

None given.

The Attempt at a Solution

I tried solving it using momentum before and after. But there is no mass given for the block and I do not know what to do with the 50J of work.

 

[PeterO]

Homework Statement

A bullet of mass 0.005kg is shot at a suspended block of wood with mass 1kg. The bullet initially has a speed of 200m/s, and passes through the block of wood. The bullet passes through the wood, and 50J of work is done deforming the shape of the bullet and block. If the block is suspended from a very long string, what is the maximum height to which it will rise?

http://imageshack.us/photo/my-images/266/bulletblock.png/"

Homework Equations

None given.

The Attempt at a Solution

I tried solving it using momentum before and after. But there is no mass given for the block and I do not know what to do with the 50J of work.

In red

 

[demenius]

I cannot believe I did not see that. :S. Thank you.

 

[demenius]

So would the height be equal to (v*mb/(mb+mw))^2 * 1/2g?

So. (200*0.005/(0.005+1))^2 *1/2(9.81) = 0.05m?

 

[asteriatic]

I would approach this problem by first identifying the key variables and equations that can be used to solve it. The variables in this problem are the mass of the bullet (m1 = 0.005kg), the initial velocity of the bullet (v1 = 200m/s), the mass of the block (m2 = 1kg), and the work done on the system (W = 50J). The equation that relates these variables is the work-energy theorem, which states that the work done on a system is equal to the change in the system's kinetic energy.

Using this equation, we can set up the following equation:

W = ΔKE = 0.5*m1*v1^2 - 0.5*m2*v2^2

Where v2 is the final velocity of the block after the bullet passes through it. We know that the bullet's kinetic energy is completely transferred to the block, so we can solve for v2:

v2 = √(2*W/m2)

Now, to find the maximum height that the block will reach, we can use the conservation of energy equation, which states that the initial total energy (kinetic energy of the bullet) is equal to the final total energy (potential energy of the block at maximum height). So, we can set up the following equation:

0.5*m1*v1^2 = m2*g*h

Where g is the acceleration due to gravity (9.8m/s^2) and h is the maximum height of the block. We can now solve for h:

h = (0.5*m1*v1^2)/(m2*g)

Plugging in the given values, we get h = 0.51m or approximately 51cm. This is the maximum height that the block will rise after the bullet passes through it.