Answer:
The unknown mass of the bullet is [tex]m1=3.751x10^{-3} kg[/tex]
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:
[tex]F=\frac{P}{dt}[/tex]
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:
[tex]m_{2}=5.79x10^{-3}kg \\m_{3}=5.79x10^{-3}kg\\v_{2}=v_{3}=392 \frac{m}{s}\\[/tex]
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:
[tex]m_{1}*v_{1}*cos(180)+m_{2}*v_{2}*cos(60)+m_{3}*v_{3}*cos(-60)=0\\[/tex][tex]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\\[/tex]
[tex]-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[/tex]
