Calculate Force Exerted by Wood on Bullet

[PhysicslyDSBL]

Homework Statement

A bullet of mass 0.0022 kg initially moving at 504 m/s embeds itself in a large fixed piece of wood and travels 0.72 m before coming to rest. Assume that the acceleration of the bullet is constant.

What force is exerted by the wood on the bullet?

F=N

Homework Equations

Ok, so I need to find acceleration to plug into the equation for force, which is F=m*a. The equation I am using is x=x0 + v0*t + .5*a*t^2.

The Attempt at a Solution

x=x0 + v0*t + .5*a*t^2 so this equation need to be set to solve for a. This is what I am plugging in:

x0 = 0
v0 = 504
t = 0.0014 (from d/s = t)

My answer is a = 0.706

The equation vf^2 = v0^2 + 2a (delta x), and set to solve for a but I am a bit confused about delta x? In this equation, vf = 0.72 and v0 = 504

I am either having trouble setting up the first equation for a, or I am using the wrong one in which case, I am not sure by what is meant by delta x in the second. Some direction is appreciated!

 

[Rake-MC]

It was looking good, but you said this:

"x0 = 0
v0 = 504
t = 0.0014 (from d/s = t)"

This assumes constant velocity, which it's not.

You have to use this formula instead : v^2 = u^2 + 2as where v = final velocity
u = initial velocity
a = acceleration
s = displacement.

Solve for acceleration, then multiply by mass.

 

[Doc Al]

x=x0 + v0*t + .5*a*t^2 so this equation need to be set to solve for a. This is what I am plugging in:

x0 = 0
v0 = 504
t = 0.0014 (from d/s = t)

My answer is a = 0.706

You could use this method if you found the right time. To find the time you need to use the average speed, not the initial speed.

The equation vf^2 = v0^2 + 2a (delta x), and set to solve for a but I am a bit confused about delta x? In this equation, vf = 0.72 and v0 = 504

This equation is a better choice, but 0.72m is the distance (Δx) not the final speed. (The final speed is zero, of course.)

 

[PhysicslyDSBL]

Thanks for the help!

I got the correct answer of 388.08 N by taking the equation v^2=u^2 + 2as and solving for a and then multiplying by mass. Both of your suggestions were very helpful in helping me to understand these concepts better!

 

[Andrew Mason]

Thanks for the help!

I got the correct answer of 388.08 N by taking the equation v^2=u^2 + 2as and solving for a and then multiplying by mass. Both of your suggestions were very helpful in helping me to understand these concepts better!

It often is helpful in this kind of problem to use a graph. A graph of speed vs. time makes it very easy. The slope is the acceleration (constant -) and the distance is the area under the graph.

AM