Question

Three bullets are fired simultaneously by three guns aimed toward the center of a circle where they mash into a stationary lump. The angle between the guns is 120°. Two of the bullets have a mass of 5.30 10-3 kg and are fired with a speed of 301 m/s. The third bullet is fired with a speed of 554 m/s and we wish to determine the mass of this bullet.

Answer 1

Answer:

2,87 * $10^{-3}$

Explanation:

When the bullets meet at the center and collide, since momentum is a vectoral quantity, their momentum vectors even up and are sumof zero. Formula of momentum is P = m.v , where m is mass and v is velocity. Let’s name the first two bullets as x,y and the one which mass is unknown as z. Then calculate momentum of x and y:

Px= 5,30 * $10^{-3}$ * 301 = 1,5953 kg*m/s

Py= 5,30 * $10^{-3}$ * 301 = 1,5953 kg*m/s

The angle between x and y bullets is 120°, and we know that if the angle between two equal magnitude vectors is 120°, the magnitude of the resultant vector will be equal to first two and placed in exact middle of two vectors. So we can say total momentum of x and y (Px+Py) equals to 1,5953 kg*m/s as well (Shown in the figure).  

For z bullet to equalize the total momentum of x and y bullets, it needs to have the same amount of momentum in the opposite way.

Pz = 1,5953 = m * 554

m = 2,87 * $10^{-3}$ kg