A pellet gun is discharged at a cardboard box of mass m2 = 0.75 kg on a frictionless surface. The pellet has a mass of m1 = 0.016 kg and flys at a velocity of v1 = 86 m/s. It is observed that the box is moving at a velocity of v2 = 0.19 m/s after the pellet goes through it.
1. Write an expression for the magnitude of the pellet's velocity as it exits the box vf.
2. What is the pellet's final velocity vf, in meters per second?

1. According to the law of conservation of momentum, the momentum of the system before and after the collision remains the same because no external forces are acting on the system:

initial momentum = final momentum

The momentum of the system is calculated as the sum of the momentum vectors of each object. The equation of momentum is the following:

p = m · v

Where:

p = momentum.

m = mass of the object.

v = velocity of the object.

Let´s calculate the initial momemtum of the system:

pI = initial momentum of the system

pp = momentum of the pellet

pc = momentum of the cardboard box

pI = pp + pc

pI = m1 · v1 + m2 · v2i

Where v2i is the initial velocity of the box.

Since the velocity of the box is zero, the initial momentum of the system will be:

pI = m1 · v1

The final momentum (pf) of the system can be expressed as follows:

pf = m1 · vf + m2 · v2

Where vf is the final velocity of the pellet.

Let´s write the law of conservation of momentum:

intial momentum = final momentum

m1 · v1 = m1 · vf + m2 · v2

Solving the equation for vf

(m1 · v1 - m2 · v2)/m1 = vf

2. Now, replacing with the data we have, we can calculate the final velocity of the pellet (vf):

v1f = (0.016 kg · 86 m/s - 0.75 kg · 0.19 m/s)/0.016 kg

v1f = 77 m/s

The final velocity of the pellet is 77 m/s.