High school physics momentum

Question

If a 22 g bullet, shot straight up into the air, reaches an altitude of 4.59 km, what is the recoil velocity of the 3 kg rifle.

answer: -2.2 m/s

Answer 1

Let

m = 22g = 0.022 kg be the mass of the bullet,

M = 3kg be the mass of the rifle,

v be the muzzle velocity of the bullet (unknown),

V be the recoil velocity of the rifle (to be calculated),

h = 4.59 km = 4590 m be the highest altitude,

g = 9.81 m/s² be gravitational acceleration.

By conservation of momentum

MV = mv

So

V = mv/M (1)

From conservation of energy the initial kinetic energy of the bullet gets converted

to its potential energy at the highest altitude:

mv²/2 = mgh

From that

v = √(2gh)

Plugging that into (1)

V = m√(2gh)/M

Plugging in actual value

V = 0.022×√(2×9.81×4590)/3 = 2.2 m/s

In your answer you have what looks like a minus sign.

You would write the answer that way if you considered upward motion positive

and downward motion negative.

It would be the natural result of expressing the conservation of momentum by

MV + mv = 0