Answer:
The velocity is [tex]v = 796 \ m/s [/tex]
Explanation:
From the question we are told that
The mass of the bullet is [tex]m_b = 5.00\ g = 0.005 \ g[/tex]
The mass of the hand gun is [tex]m_g = 1.00 \ kg[/tex]
The speed of the gun is [tex]v_b = 8.00 * 10^{2} \ m/s[/tex]
Generally from the law of momentum conservation
[tex]m_b * u_b + m_g * u_g = m_b* v_b + m_g * - v_g[/tex]
here [tex]u_g\ and \ u_b[/tex] are zero given that before the bullet was shot the gun was at rest
[tex]v_g[/tex] is negative given because from Newtons third law action and reaction is equal and opposite so if the bullet is moving forward the gun will be moving backward
So
[tex] 0.005 * 0 + 1* 0= 0.005 * 8.00 * 10^{2} - 1.00* v_g[/tex]
=> [tex] v_g = 4 \ m/s [/tex]
Hence the velocity of the bullet with respect to the loose gun is
[tex]v = v_b - v_g[/tex]
=> [tex]v = 8.00 * 10^{2} - 4[/tex]
=> [tex]v = 796 \ m/s [/tex]
