Analyzing Stopping Time & Force of a .22 Rifle Bullet

[rsala]

Homework Statement

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

How much time is required for the bullet to stop?
What force, in Newtons, does the wood exert on the bullet?

Homework Equations

kg= 1000g
f=ma
$$x = x_{0} + V_{0}T + .5aT^{2}$$

The Attempt at a Solution

i do know that ill have to convert the mass of the bullet to kg,
so mass of bullet = .0018 kg

we haven't covered such of an example in class, i don't understand how to find how much time it stays in the wood moving.

i guess i would have to find the rate of which that it accelerates in the opposite direction for that.
that is as far as my knowledge goes on this, any help?
(ive tried searching but only find on how to find the answer to my first question and in the solution they use the answer to the 2nd question...so I am stuck anyway)

thank you

 

[cemar.]

In each of the equations you posted you have two unknowns which is not cool so neither will help you right off the bat.
Try using the kinematice equation that doesn't include time to kick it off.

Vf^2 = Vi^2 + 2ad

Your final speed is 0 because it has stopped and you know your initial speed and the distance so that will give you accelleration.

 

[rsala]

thanks , that worked