Answer:
The magnitude of the BB's velocity is 74.29 m/s.
Explanation:
Given that,
Mass of BB [tex]m_{1}= 0.0185\ kg[/tex]
Mass of box [tex]m_{2}=0.85\ kg[/tex]
Velocity of gun [tex]v_{1}=89\ m/s[/tex]
Velocity of box [tex]v_{2}=0.32\ m/s[/tex]
Suppose write an expression for the magnitude of the BB's velocity as it exits the box [tex]v_{f}[/tex].
Using conservation of momentum
[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}[/tex]
Put the value into the formula
[tex]0.0185\times89+0=0.0185\times v_{1}+0.85\times0.32[/tex]
[tex]\dfrac{1.6465-0.272}{0.0185}=v_{1}[/tex]
[tex]v_{1}=74.29\ m/s[/tex]
Hence, The magnitude of the BB's velocity is 74.29 m/s.
