.Calculating Time for .22 Rifle Bullet to Stop in Soft Wood

[theonerealazn]

Ok, so here's my question:

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.

I know this has already been posted in another thread, but i have an additional question that i can't get.
1. How much time is required for the bullet to stop?
i tried 0.130m/350m/s and got .00037143 seconds as an answer. This was not correct. Any suggestions on what to do?

I already know that the wood exerts 848 N on the bullet

 

[steve23063]

F=ma where F=net force on the bullet=848N and m=mass of the bullet=.0018kg. Find a

Vf=Vi+a*t
Vf, Vi and a are all known at this point. Solve for time.

Your answer is close. You are off by a factor of 2.

 

[kyle8921]

for 4.3ms, but how much time does it really take for the bullet to come to a stop?

I would first like to commend you for your efforts in attempting to calculate the time required for the bullet to stop in soft wood. It is important to note that this type of calculation can be quite complex, as there are several factors that can affect the outcome. However, I will provide some guidance on how to approach this problem.

To calculate the time required for the bullet to stop, we need to consider the forces acting on the bullet. In this case, we have the initial velocity of the bullet (350 m/s) and the retarding force from the wood (848 N). Using the equation F=ma, we can calculate the acceleration of the bullet as a= F/m = 848 N/0.0018 kg = 471111.11 m/s^2.

Next, we can use the equation v=u+at to calculate the time (t) required for the bullet to stop, where u is the initial velocity (350 m/s) and v is the final velocity (0 m/s). Plugging in the values, we get:

0=350 m/s + 471111.11 m/s^2 * t

Solving for t, we get t= -0.000743 s. However, this answer may seem strange since time cannot be negative. This is because we have assumed a constant retarding force, but in reality, the force will decrease as the bullet penetrates deeper into the wood. Therefore, the bullet will not come to a complete stop in the calculated time, but it will continue to slow down until it reaches a final velocity of 0 m/s.

In conclusion, the time required for the bullet to stop in soft wood is dependent on several factors and cannot be accurately calculated without taking into account the changing retarding force. However, using the initial velocity and the retarding force, we can estimate the time required for the bullet to significantly slow down. I hope this helps in your understanding of this problem.