# Linear momentum bullet problem

• brendan3eb
In summary, the conversation discusses solving a physics problem involving a bullet and a block. There is confusion about whether kinetic energy should be conserved and how to calculate the initial velocity of the bullet. It is concluded that the initial velocity cannot be determined accurately due to factors such as friction and heat generation during the collision. The final two parts are considered to be straightforward.
brendan3eb
yeah, part c was for the block :)

ahh..I thought I had the right answer originally but it didn't say if variables were supposed to be left in the problem, so I kind of doubted myself. Thanks for the help you guys.

I am learng physixguru, maybe slowly, but learning. lol

Last edited:
come on buddy...these kinds of ques have been discussed a million times...

it seems u have not paid attention to them...
anywayz...

calculate the loss in kinetic energy of bullet...this equals to the work done by the block on the bullet...
then u can easily calculate the velocity of the block...using simple kinematic equations...

The third part is not understandable.:p

Your vf for the bullet is correct. As physixguru says, you do not need conservation of kinetic energy to solve the other two parts. Part (c) probably means the kinetic energy gained by the block, or else (b) and (c) contradict each other.

If you want to assume that kinetic energy is conserved (i.e. the collision between the bullet and the block is elastic), you will find that v0 does not cancel:

$$\text{From part (a), }v_{\text{block}} = \frac{v_0}{6}$$

$$\text{From the assumption that kinetic energy is conserved, }$$

$$mv_0 &=& m\cdot\left(\frac{v_0}{3}\right)^2 + 4m\cdot\left(\frac{v_0}{6}\right)^2 = \frac{mv_0^2}{9} + \frac{4mv_0^2}{36} = \frac{2mv_0^2}{9} \implies v_0^2 - \frac{9v_0}{2} = v_0\left(v_0 - \frac{9}{2}\right) = 0 \implies v_0 = 0 \text{ or } \frac{9}{2}$$

So if the given conditions are met, kinetic energy is conserved when the initial velocity of the bullet is either 0 (it isn't fired) or 9/2.

However, it is unrealistic that kinetic energy is conserved (i.e. the collision is elastic), because the impact of the bullet against the wood and the friction on the bullet when it travels through the wood will generate heat, so some of the initial kinetic energy will be lost (i.e. the collision is inelastic).

Therefore, you really cannot find out what the initial velocity v0 is. For the given block-bullet system, there may be a particular v0 which will cause the exit speed of the bullet to be v0/3, but that speed is impossible to calculate and depends, among other factors, on the length of the block and the materials used.

so when I get v0/6 for part A, that is precisely what they want..if so, awesome :)

then the last two parts are fairly simple

## 1. What is linear momentum?

Linear momentum is a measure of the motion of an object, taking into account both its mass and its velocity. It is defined as the product of an object's mass and its velocity, and is a vector quantity with direction in the same direction as the object's velocity.

## 2. How does a bullet's linear momentum change after being fired?

When a bullet is fired, it experiences a change in linear momentum due to the force exerted by the gunpowder explosion. This force accelerates the bullet forward, increasing its velocity and therefore its linear momentum.

## 3. How does the mass of a bullet affect its linear momentum?

The linear momentum of a bullet is directly proportional to its mass. This means that a bullet with a larger mass will have a greater linear momentum, assuming all other factors such as velocity remain constant.

## 4. Does the shape of a bullet affect its linear momentum?

The shape of a bullet can affect its linear momentum in certain situations. For example, a pointed bullet will have a higher linear momentum compared to a flat-tipped bullet, as it experiences less air resistance and can maintain its velocity better. However, the overall mass and velocity of the bullet will still have a greater impact on its linear momentum.

## 5. How is the conservation of linear momentum applied in a bullet problem?

The conservation of linear momentum states that the total linear momentum of a system remains constant, as long as there are no external forces acting on the system. In a bullet problem, this means that the linear momentum of the bullet before being fired must be equal to the linear momentum of the bullet after being fired, taking into account any changes in mass or velocity.

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