Question

Question #57796

Answer 1

The recoil velocity of the rifle: `-"0.9 m/s"`
The total kinetic energy after the bullet is fired is `"677 J"`

Explanation:

So, you have a 5-kg rifle that fires a 15-g bullet with a velocity of 30,000 cm/s east.

Before the shot is fired, the rifle-bullet system has a total net momentum equal to zero. After the shot is fired, the bullet will have a momentum given by its mass and initial velocity.

`p_"bullet" = m_"bullet" * v_0`

Assuming that no external forces act upon the rifle-bullet system, the total net momentum must be preserved. This means that the momentum the system had before the bullet was fired must equal to momentum it has after the bullet is fired.

`p_"before" = p_"after" = 0`

The rifle will also have a momentum, but it will be oriented in the opposite direction, i.e. west, which implies that it will carry a negative sign.

`p_"after" = 0 = p_"bullet" + p_"rifle"`

`p_"rifle" = 0 - p_"bullet" = -p_"bullet"`

`m_"rifle" * v_"rifle" = -m_"bullet" * v_0`

This means that the recoil velocity of the rifle will be

`v_"rifle" = -m_"bullet"/m_"rifle" * v_0 = (15 * 10^(-3)cancel("kg"))/(5 * cancel("kg")) * "30,000 cm/s"`

`v_"rifle" = -"90 cm/s"`

You'll need to convert this to meter per second for the following subpoints

`v_"rifle" = -90cancel("cm")/"s" * "1 m"/(100cancel("cm")) = color(green)(-"0.90 m/s")`

The total kinetic energy before the bullet was fired is equal to zero.

`E_(k0) = "0 J"`

After the bullet is fired, the total kinetic energy will be

`E_k = 1/2 * m_"rifle" * v_"rifle"^2 + 1/2 * m_"bullet" * v_0^2`

`E_k = 1/2 * "5 kg" * (-0.90)^2 "m"^2/"s"^2 + 1/2 * 15 * 10^(-3)"kg" * (3 * 10^(2))^2"m"^2/"s"^2`

`E_k = 2.025 + 675 = color(green)("677 J")`

Notice that you have a difference between the initial kinetic energy, which was zero, and the final kinetic energy. This difference can be attributed to the gun powder explosion and the expansion of the hot gases present in the barrel, which is what actually propels the bullet out of the barrel.