How Long Does It Take for a Bullet to Stop in Wood?

[envoystud]

momentum question...REALLY NEED HELP!

Homework Statement

A .22 rifle bullet, traveling at 350 m/s, strikes a block of soft wood, which it penetrates to a depth of 0.130m . The block of wood is clamped in place and doesn't move. The mass of the bullet is 1.80g. Assume a constant retarding force.

??-How much time is required for the bullet to stop? (in seconds)

??-What force, in Newtons, does the wood exert on the bullet?

Homework Equations

The Attempt at a Solution

well honestly, i have no idea where to begin... can some one please help me

 

[Dick]

Kinematics, think x(t)=x0+v0*t+(1/2)*a*t^2. What acceleration is required to stop the bullet? Once you have a, what force is needed to stop the bullet? Now think of Newton's third law.

 

[m2003]

out?!

I would approach this problem by using the principles of momentum and force. First, I would calculate the initial momentum of the bullet using the formula p=mv, where p is momentum, m is mass, and v is velocity. In this case, the initial momentum would be (0.0018 kg)(350 m/s) = 0.63 kg m/s.

Next, I would use the equation F=ma to calculate the force acting on the bullet. Since the bullet is experiencing a constant retarding force, the acceleration (a) would be constant. We can rearrange the equation to solve for force, which would be F=m*a. Plugging in the mass of the bullet (0.0018 kg) and the acceleration (which we will solve for), we can calculate the force acting on the bullet.

To find the acceleration, we can use the equation v^2 = v0^2 + 2ax, where v is the final velocity (0 m/s), v0 is the initial velocity (350 m/s), and x is the distance the bullet travels (0.130 m). Solving for a, we get a=-12692 m/s^2.

Now, we can plug in this value for acceleration into the equation F=m*a. This gives us a force of -0.023 N (note the negative sign indicates that the force is acting in the opposite direction of motion, which makes sense since the bullet is being slowed down).

To find the time required for the bullet to stop, we can use the equation v=v0+at. Since we know the initial velocity (350 m/s), final velocity (0 m/s), and acceleration (-12692 m/s^2), we can solve for t. This gives us a time of approximately 0.028 seconds.

In summary, the bullet experiences a force of -0.023 Newtons and it takes approximately 0.028 seconds to stop.