Analyzing Stopping Time & Force of a .22 Rifle Bullet

In summary, a .22 rifle bullet with a mass of 1.80 g, traveling at 350 m/s, penetrates a block of soft wood to a depth of 0.130 m. The block is clamped in place and exerts a constant retarding force on the bullet. To find the time it takes for the bullet to stop, we can use the kinematic equation Vf^2 = Vi^2 + 2ad, since the final speed is 0 and we know the initial speed and distance. The calculated acceleration can then be used to find the force exerted by the wood on the bullet.
  • #1
rsala
40
0

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 :smile:
f=ma
[tex] x = x_{0} + V_{0}T + .5aT^{2} [/tex]

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
 
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • #3
thanks , that worked
 

Related to Analyzing Stopping Time & Force of a .22 Rifle Bullet

1. How is stopping time calculated for a .22 rifle bullet?

The stopping time for a .22 rifle bullet is calculated by dividing the bullet's velocity by the deceleration force. This can be done using the formula t = v/a, where t is the stopping time, v is the initial velocity of the bullet, and a is the deceleration force.

2. What factors affect the stopping time of a .22 rifle bullet?

The stopping time of a .22 rifle bullet can be affected by several factors, including the bullet's velocity, mass, and shape, as well as the density and composition of the material it is hitting. Other factors such as air resistance and the angle of impact can also play a role in the bullet's stopping time.

3. How is the force of a .22 rifle bullet calculated?

The force of a .22 rifle bullet can be calculated using the formula F = m x a, where F is the force, m is the mass of the bullet, and a is the acceleration or deceleration force acting on the bullet. The bullet's initial velocity and stopping time can also be used to calculate the force using the formula F = m x (v/t).

4. What is the typical stopping time and force of a .22 rifle bullet?

The stopping time and force of a .22 rifle bullet can vary depending on the specific bullet and the material it is hitting. However, on average, a .22 rifle bullet has a stopping time of around 0.01 seconds and a force of approximately 200 Newtons.

5. How is the data for analyzing stopping time and force of a .22 rifle bullet collected?

The data for analyzing the stopping time and force of a .22 rifle bullet can be collected through various methods, such as using high-speed cameras to record the bullet's impact and measuring its velocity before and after impact. Other methods may include using ballistic gel or other materials to simulate the impact of a bullet on a human body or other objects.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
6K
  • Introductory Physics Homework Help
Replies
8
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
6K
  • Introductory Physics Homework Help
Replies
1
Views
901
  • Introductory Physics Homework Help
Replies
14
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
3K
Back
Top