What is the Equation for Momentum in a Rifle Firing a Bullet?

[blackout85]

A rifle of mass M is intially at rest but free to recoil. It fires a bullet of mass m and velocity v (relative to the ground). After firing, the velocity of the rifle (relative to the ground) is:

P= m2V2 -m1v1
P= M(v-0) - m(v-0)
M(V-0) = m(V-0)
V-0= mv/M
V= -mv/M
is that right

 

[radou]

Yes, it is. Btw, you can post similar questions into the same thread next time.

 

[kyle8921]

?

The equation for momentum in this scenario is P = m2v2 - m1v1, where m2 is the mass of the bullet, v2 is its velocity, m1 is the mass of the rifle, and v1 is its velocity. This equation is derived from the principle of conservation of momentum, which states that the total momentum of a system remains constant unless acted upon by an external force.

In this case, the initial momentum of the system (rifle and bullet) is zero, as the rifle is at rest. When the bullet is fired, it gains momentum in the forward direction, causing the rifle to recoil in the opposite direction with an equal amount of momentum. This is represented by the negative sign in the equation.

Therefore, the final momentum of the system can be calculated by substituting the values for m2, v2, m1, and v1. The resulting equation is P = m(v - 0), where v is the velocity of the bullet relative to the ground. This equation can also be written as P = M(V - 0), where V is the velocity of the rifle relative to the ground. This shows that the final velocity of the rifle is equal in magnitude to the initial velocity of the bullet, but in the opposite direction.

To summarize, the equation for momentum in a rifle firing a bullet is P = m2v2 - m1v1, and the final velocity of the rifle can be calculated using V = -mv/M. It is important to note that this equation assumes an ideal scenario and does not take into account external factors such as air resistance and friction.