Respuesta :
Answer:395.6 m/s
Explanation:
Given
mass of bullet [tex]m=5 gm[/tex]
mass of wood block [tex]M=1 kg[/tex]
Length of string [tex]L=2 m[/tex]
Center of mass rises to an height of [tex]0.38 cm[/tex]
initial velocity of bullet [tex]u=450 m/s[/tex]
let [tex]v_1[/tex] and [tex]v_2[/tex] be the velocity of bullet and block after collision
Conserving momentum
[tex]mu=mv_1+Mv_2[/tex] -------------1
Now after the collision block rises to an height of 0.38 cm
Conserving Energy for block
kinetic energy of block at bottom=Gain in Potential Energy
[tex]\frac{Mv_2^2}{2}=Mgh_{cm}[/tex]
[tex]v_2=\sqrt{2gh_{cm}} [/tex]
[tex]v_2=\sqrt{2\times 9.8\times 0.38}[/tex]
[tex]v_2=0.272 m/s[/tex]
substitute the value of [tex]v_2[/tex] in equation 1
[tex]5\times 450=5\times v_1+1000\times 0.272[/tex]
[tex]v_1=395.6 m/s[/tex]
