# A bullet embedded in a sliding block

• Identity
In summary, the conversation discusses the problem of finding the depths of penetration when a bullet of mass m is fired with a horizontal velocity u at a block of mass M, which embeds itself inside the block. The force of resistance of the block is constant. The solution involves using simple kinematics for the case where the block is held fixed, but a conservation of energy/conservation of momentum approach is needed for the case where the block is free to move on a smooth horizontal surface. The final solution is found by comparing the initial and final kinetic energy of the system.
Identity

## Homework Statement

A bullet of mass m is fired with horizontal velocity u at a block of mass M. The bullet embeds itself inside the block. The force of resistance of the block is constant.
Find the depths of penetration in the cases:
a) The block is held fixed.
b) The block is free to move on a smooth horizontal surface

## Homework Equations

Conservation of momentum? Kinematic equations? Conservation of energy? I don't know.

## The Attempt at a Solution

Using simple kinematics for a), with the resistive force $$F = -k$$ of the block on the bullet, I get the depth, x, as

$$x = \frac{mu^2}{2k}$$

However, I'm having more trouble with part b)

I've tried a conservation of energy/conservation of momentum approach:

$$mu = (M+m)v \Rightarrow v = \frac{m}{M+m}u$$

Now, the work done by the bullet on the block is $$Fx$$, which translates into $$\frac{1}{2}Mv^2$$.

$$Fx = \frac{1}{2}Mv^2$$

$$kx = \frac{1}{2}Mv^2$$

$$x = \frac{Mv^2}{2k}=\frac{M}{2k}\left(\frac{M}{M+m}u\right)^2$$

However, this isn't correct, what's the proper way to do this?

Thanks

Identity said:
However, I'm having more trouble with part b)

I've tried a conservation of energy/conservation of momentum approach:

$$mu = (M+m)v \Rightarrow v = \frac{m}{M+m}u$$
This is good; v is the final speed of the 'bullet + block'.

Hint: Compare the initial and final kinetic energy of the system. (How much is 'lost'?)

Ah right!
$$Fx = \frac{1}{2}mu^2 - \frac{1}{2}(M+m)v^2$$

Thanks Doc Al :)

## 1. What happens to the bullet when it is embedded in the sliding block?

When a bullet is embedded in a sliding block, it will transfer its kinetic energy to the block. This will cause the block to move in the direction of the bullet's initial velocity.

## 2. Can the sliding block still move after the bullet is embedded?

Yes, the sliding block can still move after the bullet is embedded. However, the velocity of the block will be reduced due to the transfer of kinetic energy from the bullet.

## 3. How does the mass of the bullet affect the motion of the sliding block?

The mass of the bullet will have a direct impact on the motion of the sliding block. A heavier bullet will transfer more kinetic energy to the block, resulting in a greater velocity and movement of the block.

## 4. Is there any friction between the bullet and the sliding block?

There will be some friction between the bullet and the sliding block, depending on the materials and surfaces involved. This friction will affect the velocity and movement of the block.

## 5. What factors determine the distance the sliding block will move after the bullet is embedded?

The distance the sliding block will move after the bullet is embedded is determined by a variety of factors, including the initial velocity of the bullet, the mass of the bullet, the mass of the block, and the amount of friction between the bullet and the block.

• Introductory Physics Homework Help
Replies
21
Views
1K
• Introductory Physics Homework Help
Replies
28
Views
1K
• Introductory Physics Homework Help
Replies
31
Views
3K
• Introductory Physics Homework Help
Replies
9
Views
6K
• Introductory Physics Homework Help
Replies
17
Views
1K
• Introductory Physics Homework Help
Replies
30
Views
970
• Introductory Physics Homework Help
Replies
12
Views
2K
• Introductory Physics Homework Help
Replies
33
Views
408
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
24
Views
1K