Question

Three guns are aimed at the center of a circle, and each fires a bullet simultaneously. The directions in which they fire are 120° apart. Two of the bullets have the same mass of 5.79 x 10-3 kg and the same speed of 392 m/s. The other bullet has an unknown mass and a speed of 605 m/s. The bullets collide at the center and mash into a stationary lump. What is the unknown mass?

Answer 1

Answer:

The unknown mass of the bullet is  $m1=3.751x10^{-3} kg$

Explanation:

According to Newton's laws of motion, when a net external force acts on a body of mass  m , it results in change in momentum of the body and is given by:

$F=\frac{P}{dt}$

Where:

P

is the linear momentum of the body

As a consequence, when there are no external forces acting on the body the total momentum remains conserved i.e.

Given:

$m_{2}=5.79x10^{-3}kg \\m_{3}=5.79x10^{-3}kg\\v_{2}=v_{3}=392 \frac{m}{s}$

For momentum along the y-direction to be zero, it is achieved when the equal masses are moving at angles of  

θ1=180°, θ2=60°, θ3=-60°

Therefore, from conservation of momentum along x - direction:

$m_{1}*v_{1}*cos(180)+m_{2}*v_{2}*cos(60)+m_{3}*v_{3}*cos(-60)=0$$m_{1}*605\frac{m}{s}*cos(180)+5.79x10^{-3}kg *392\frac{m}{s}*cos(60)+5.79x10^{-3}kg*392\frac{m}{s}*cos(-60)=0$

$-m_{1}*605+5.79x10^{-3}kg*196\frac{m}{s}+5.79x10^{-3}kg*196\frac{m}{s}=0\\m_{1}*605kg= 2.26968\frac{kg*m}{s}\\m_{1}=3.75 x10^{-3} kg$