# Momentum of a Bullet Contacting a Block

• LastXdeth
In summary, a 4.0 g bullet with a speed of 650 m/s collided with a 0.095 kg block of wood at rest on a frictionless surface. After the collision, the bullet exited the block and the block had a speed of 23 m/s. To calculate the speed of the bullet when it exited the block, the formula (m1)(v1o)+(m2)(v2o)=(m1)(v1)+(m2)(v2) was used, resulting in a speed of approximately 103.75 m/s. Energy was conserved in this collision.
LastXdeth

## Homework Statement

A bullet with a mass of 4.0 g and a speed of 650 m/s is fired at a block of wood with a mass of 0.095 kg. The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 23 m/s.

(a) What is the speed of the bullet when it exits the block?

(b) Is energy conserved in this collision? (yes or no)

## Homework Equations

(m1)(v1o)+(m2)(v2o)=(m1)(v1)+(m2)(v2)

## The Attempt at a Solution

(4000)(650)+(0.095)(0)=(4000)(v1)+(0.095)(23)

v1≈650.00 (Apparently this is wrong.)

Last edited:

lewando said:
I don't see any mass unit that needs attention. The only thing I did with mass was convert m1 (4.0 g) to the SI unit kg.

4g is not 4000kg.

lewando said:
4g is not 4000kg.

Aww crap! Dang, I feel so ridiculous for reversing it up!

We all have our moments. That would be some big bullet. No wonder it didn't slow down.

With the correct unit conversions and your same formula, I got 296.94

PotentialE said:
With the correct unit conversions and your same formula, I got 296.94

I got 103.75.

MV+MV = MV+MV
.004kg*650m/s + .095kg*0 = .004kg*VF + .095kg*23
2.6 = .004*VF + 2.185
.0415 = .004VF
Vf= 103.75

PotentialE said:
MV+MV = MV+MV
.004kg*650m/s + .095kg*0 = .004kg*VF + .095kg*23
2.6 = .004*VF + 2.185
.0415 = .004VF
Vf= 103.75

No problem! Thanks for confirming!

## 1. What is momentum?

Momentum is a physical quantity that describes the tendency of an object to keep moving in the same direction with the same speed. It is calculated by multiplying an object's mass by its velocity.

## 2. How does the momentum of a bullet contacting a block affect the block?

When a bullet contacts a block, its momentum is transferred to the block. This results in the block gaining momentum and moving in the same direction as the bullet, but with a lower velocity due to its larger mass. The block may also experience a change in shape or be pushed in a different direction depending on the force and angle of impact.

## 3. What factors affect the momentum of a bullet contacting a block?

The momentum of a bullet contacting a block is affected by the mass and velocity of the bullet, as well as the mass and velocity of the block. The angle and force of the impact also play a role in determining the resulting momentum of the block.

## 4. How is momentum conserved in a bullet-block collision?

In accordance with the law of conservation of momentum, the total momentum of the system (bullet and block) remains constant before and after the collision. This means that the initial momentum of the bullet is equal to the final momentum of the block and bullet combined.

## 5. Can the momentum of a bullet contacting a block be calculated?

Yes, the momentum of a bullet contacting a block can be calculated using the formula p = mv, where p is momentum, m is mass, and v is velocity. This can be applied to both the bullet and the block to determine their individual momentums or the total momentum of the system.

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