## Muzzle Velocity

Two cardboard target discs are mounted in a straight line with a 3 m gap between the discs’ flat faces. The discs both spin clockwise about their center point at a speed of 1,500 rpm. If a rifle is mounted directly in line with the targets and fires a bullet towards the discs such that the resulting bullet hole in the second disc is displaced 40° with respect to the hole made in the first disc, what is the approximate muzzle velocity in m/s? Ignore friction and gravity.

Hint
Muzzle velocity is the projectile/bullet speed right when it leaves the gun barrel’s end.
Hint 2
First, convert the 1,500 rotations per minute to rotations per second.
$$1,500\:\frac{rotations}{min}\cdot \frac{min}{60sec}=25\:rotations\:per\:sec$$$Muzzle velocity is the projectile/bullet speed right when it leaves the gun barrel’s end. First, convert the 1,500 rotations per minute to rotations per second. $$1,500\:\frac{rotations}{min}\cdot \frac{min}{60sec}=25\:rotations\:per\:sec$$$
To complete a full rotation, the disc needs to rotate 360°. Since the second disc displaced 40°, then:
$$\frac{40^{\circ}}{360^{\circ}}=0.11\:of\:a\:complete\:rotation$$$Therefore, the time it takes the second disc to rotate 40° is: $$\frac{0.11rotation}{25\:rotations\:per\:sec}=0.00444444\:sec$$$
Since the bullet travels 3m in this time, the muzzle velocity is:
$$\frac{3m}{0.004444sec}= 675\:m/s$$\$
675 m/s