A 31 kg gun is standing on a frictionless surface. the gun fires a 52.8 g bullet with a muzzle velocity of 317 m/s. the positive direction is that of the bullet. calculate the momentum of the bullet immediately after the gun was fired. answer in units of kg · m/s. calculate the momentum of the gun immediately after the gun was fired. answer in units of kg · m/s.

The momentum of the bullet is P=mv = 0.0528*317=16.7376 kg·m/s in the positive direction.
Since the surface is frictionless and momentum is always conserved, we can say this is the same momentum of the gun afterward, but negative. This is because the gun and bullet were initially at rest so the initial momentum was zero. The sum of momenta must then be zero after as well.
Pgun = -16.7376 kg·m/s

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moya1316 moya1316 · Answer 2 · 2021-10-14T12:31:06+02:00

The definition of momentum allows to find the result for the bodies's moment are:

The momentum bullet isp =
The momentum gun

The momentum is defined by the product of the mass and the velocity of the body

p = m v

The bold letters indicate vectors, p is the momentum, m the mass and v the velocity of the body.

They indicate that the velocity of the bullet is va = 317 m / s and the mass of the bullet is

m = 52.8 m ( [tex]\frac{1kg}{1000g}[/tex] ) = 52.8 10⁻³ kg

Let's calculate

p = 52.8 10⁻³ 317

[tex]p_{bullet}[/tex] = p = 16.7 kg m / s

The positive sign indicates that the momentum is directed to the right

As the gun is on a surface without friction, for the isolated bullet-gun system the momentum is conserved.

[tex]p_{gun} + p_{bullet}= 0[/tex]

[tex]p_{gun} = - p_{bullet}[/tex]

[tex]p_{gun}[/tex] = - 16.7 m / s

The negative sign indicates that the gun goes back in the opposite direction to the bullet, in the attachment we have a scheme of the situation.

In conclusion they use the definition of momentum we could find the result for the momentum of the bodies

Momentum bullet [tex]p_{bullet}[/tex] = 16.7 kg m / s =
Momentum gun [tex]p_{gun}[/tex] = - 16.7 m / s

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