# Why are we applying conservation of momentum in the following question?

A bullet of mass 0.01kg is fired horizontally into a 4kg wooden block at rest on a horizontal surface. The coefficient of friction between the block and the surface is 0.25. The bullet remains embedded in the block and the combination moves 20m before coming to rest. With what speed did the bullet strike the block?

To get the answer, we are supposed to apply the conservation of momentum.

Why? Isn't the frictional force acting as an external force in the given situation?

• Your problem has two stages: a) bullet vs block; b) move. Use conservation to analyze first one. Dec 18, 2020 at 13:15
• The process of the bullet hitting the block is very fast, so you neglect momentum loss Dec 18, 2020 at 13:22
• More of a hint: conservation of momentum allows you to associate the initial speed of the block/bullet combination with the initial speed of the bullet. Dec 18, 2020 at 16:30

As mentioned in the comment section, there are two processes in action here.

1. Bullet hitting the block
2. Motion after the impact

And the most important point is that friction plays its role only after the motion starts since it is its job(😁) to oppose relative motion between surfaces in contact. And in your question, the relative motion starts only after the impact.

During the first instant of the impact, the block was considered to be at rest.

So there is no friction in action in the first part and hence you can blindly apply conservation of linear momentum for the impact.

As mentioned in the comment below by Luke Pritchett, it is important to note that it is true only if we assume the momentum transfer process is completed instantly after the collision.

Hope it helps 🙂.

• So the criteria to apply the conservation of momentum is when no external forces are acting on the system before the collision and not during the collision. Am I right? Dec 18, 2020 at 13:39
• @Jeeshaan yes we could conserve the momentum since there was no friction before the impact and just just after the impact... Dec 18, 2020 at 13:41
• I think you should clarify that this is an approximation. The approximation is assuming that the collision happens so quickly that it is over before the friction has any time to act. In reality the block will be sliding at the same time as the collision, and hence the total momentum of the block and bullet will be less at the end of the collision than at the start. But when the collision happens fast this change in momentum will be tiny compared to the change that happens after the collision. Dec 18, 2020 at 14:18
• @Luke Pritchett thanks that's important here. I will edit . Dec 18, 2020 at 14:23

The force of friction is an external force on the block. During the short time of the impulsive impact $$\Delta t$$ the change in momentum due to the force of friction is $$F_{fric} \enspace \Delta t = m_{total} \enspace \Delta v$$ where $$m_{total}$$ is the mass of the block and the imbedded bullet, and $$\Delta v$$ is the change in velocity of the block\bullet due to the force of friction $$F_{fric}$$. Since $$\Delta t$$ is very short, $$\Delta v$$ is negligible. After the impact, the force of friction will then retard the motion of the block and bullet.

If it assumed the block/bullet is a rigid body, the force of friction reduces the kinetic energy of the block/bullet with no "heating" effects, since for a rigid body there can be no dissipation of energy within a rigid body. In reality the block/bullet is not perfectly rigid and "heating" occurs.

When you lose energy (because of friction) you cannot use conservation of energy, but you can still use conservation of momentum. Common examples are when the two bodies stick to each other (like the one you mentioned)

You can't use conservation of momentum WHEN THERE IS AN EXTERNAL FORCE acting on the system. Friction is a five between the object and is not external so it does not mess up the conservation of momentum. An example of this type of motion is pendulum.