What Is the Exit Velocity of a Projectile After Passing Through a Wooden Block?

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Homework Help Overview

The discussion revolves around a physics problem involving a projectile and a wooden block on a frictionless table. The original poster seeks to determine the exit velocity of the projectile after it passes through the block, given specific masses and initial velocities.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the conservation of momentum as a principle to solve the problem. Some suggest setting up equations based on initial and final momentum. Others question the assumptions made about the scenario, such as the existence of a frictionless table and the realism of the projectile's behavior.

Discussion Status

There are multiple interpretations of the problem, with some participants providing guidance on using conservation of momentum. However, there is no explicit consensus on the correct approach or the realism of the scenario presented.

Contextual Notes

Participants note constraints such as the unrealistic nature of a frictionless table and the implications of using a heavy projectile in the context of the problem. There are discussions about the physical plausibility of the situation described.

shawonna23
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A 0.165 kg projectile is fired with a velocity of +705 m/s at a 2.00 kg wooden block that rests on a frictionless table. The velocity of the block, immediately after the projectile passes through it, is +55.0 m/s. Find the velocity with which the projectile exits from the block.

I used the equation vf1=(m1-m2)/(m1+m2) but I got the wrong answer. Can you please help me?
 
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Conservation of momentum is the answer.

Before the impact, the total momentum is:

0.165 * 705 + 0 * 0 kg m/s (bullet + wood)

After the impact, the total momentum is (vp is the bullet residual velocity):

0.165 * vp + 2 * 55 kg m/s (bullet + wood)

Since momentum is conserved, these two momentum should be equal:

0.165 * 705 + 0 * 0 = 0.165 * vp + 2 * 55

You can solve this equation for vp and get the result.

You can check that the bullet final velocity is 38.33 m/s (right ?) since:

0.165 * 705 + 0 * 0 = 0.165 * 38.33 + 2 * 55

(hint: never apply an equation, apply the principles)
 
shawonna23 said:
A 0.165 kg projectile is fired with a velocity of +705 m/s at a 2.00 kg wooden block that rests on a frictionless table. The velocity of the block, immediately after the projectile passes through it, is +55.0 m/s. Find the velocity with which the projectile exits from the block.

I used the equation vf1=(m1-m2)/(m1+m2) but I got the wrong answer. Can you please help me?

Try using law of conservation of momentum i.e.

mv1+MV1=mv2+MV2
By using it we get velocity of projectile as 38.33
 
First of all:
Forum Rules: Do not post your homework/school-type questions here!

So here you may got a common sense physics answer rather than a solution to most unrealistic school exercise:
1. there is no such thing as 'frictionless table';
2. "0.165 kg projectile is fired with a velocity of 705 m/s" means you deal with a bullet of heaviest (20mm) machine gun;
3. have you ever seen a wooden plank hit by a rifle bullet? So try to imagine what may remain after being hit by 20mm bullet, then try to count the splinters and estimate momentum carried by them; If you haven't seen - watch Discovery Channel military programmes ;)
4. If you listen to answers like "velocity of projectile as 38.33" try to picture it - the bullet 'passes through it...exits from the block' at the speed lower than block has itself.

EDITED>>>
Sorry - I've underestimated caliber. Heaviest 20mm bullets have only 0.11 kg.
0.165kg may be an armour-piercing bullet for 25mm cannons, used in assault helicopters. Their typical muzzle velocity is 720m/s.
So now picture yourself a wooden block hit by such bullet...
 
Last edited:

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