Conservation of Momentum Law- Recoil Velocity (Question)

[seeindubble12]

First time posting, I hope I do this correctly!

I am struggling to grasp this concept. This is my attempt at solving. If I am completely off base, will some explain to me again how it works. Our instructor explained it to us with a very detailed formula which, I must confess, has caused me more than a little anxiety because I am not following! Thanks for the help!

1. Rifle (suspended by strings) - 2kg; fires bullet of 0.01 kg at speed of 200 m/s. What is the recoil velocity of the rifle?



2. (mv) + (mv) = (M + m)(v)



3. (0.01)(200) + 0 = (0.01 +2)(v)
2=2.01(v)
divide both sides by 2.01, and the answer is v= .99

 

[Doc Al]

This isn't quite right.

Momentum is conserved. What's the total momentum of the system (rifle + bullet) initially? What's the total momentum of the system after the bullet is fired?

Set that total momentum equal to m1v1 + m2v2.

 

[thomate1]

m/s



Hello! Don't worry, the concept of conservation of momentum can be tricky at first. Let me try to explain it in simpler terms. The law of conservation of momentum states that in a closed system (meaning no external forces acting on it), the total momentum before an event is equal to the total momentum after the event. In the case of a rifle firing a bullet, the system includes both the rifle and the bullet.

To solve for the recoil velocity of the rifle, we can use the equation you provided: (mv) + (mv) = (M + m)(v), where m is the mass of the bullet, v is the velocity of the bullet, M is the mass of the rifle, and v is the recoil velocity of the rifle.

In this problem, we know the mass and velocity of the bullet (0.01 kg and 200 m/s, respectively). We also know the mass of the rifle (2 kg). So, we can plug these values into the equation:

(0.01 kg)(200 m/s) + (0 kg)(0 m/s) = (2 kg + 0.01 kg)(v)

Simplifying, we get:

2 kg*m/s = (2.01 kg)(v)

Now, to solve for v, we divide both sides by 2.01 kg:

v = 2 kg*m/s / 2.01 kg

v = 0.995 m/s

So, the recoil velocity of the rifle would be approximately 0.995 m/s.

I hope this helps! Remember, practice makes perfect in understanding these concepts. Keep at it and don't hesitate to ask for clarification if needed. Good luck!