Answer:
844.9 m/s
Explanation:
We are given that
Mass of bullet,m=4.4 g=[tex]4.4\times 10^{-3} kg[/tex]
1kg=1000 g
Mass of pendulum=m'=2.9 kg
Vertical distance,y=8.3 cm=[tex]0.083 m[/tex]
1 m=100 cm
We have to find the initial speed of bullet.
According to law of conservation of energy
Change in kinetic energy=Change in potential energy
Initial kinetic energy=0,Initial potential energy=0
[tex]K_f=P_f[/tex]
[tex]\frac{1}{2}(m+m')v^2=(m+m')gy[/tex]
Where [tex]g=9.8 m/s^2[/tex]
[tex]v^2=2yg[/tex]
[tex]v=\sqrt{2gy}=\sqrt{2\times 9.8\times 0.083}=1.28 m/s[/tex]
It is final velocity after collision.
According to law of conservation of momentum
[tex]mu+m'u'=(m+m')v[/tex]
Initial speed of pendulum,u'=0
[tex]4.4\times 10^{-3}u=(4.4\times 10^{-3}+2.9)(1.28)[/tex]
[tex]u=\frac{(4.4\times 10^{-3}+2.9)(1.28)}{4.4\times 10^{-3}}[/tex]
u=844.9 m/s
Hence, the initial speed of bullet=844.9 m/s
