Answer:
423m/s
Explanation:
Suppose after the impact, the bullet-block system swings upward a vertical distance of 0.4 m. That's means their kinetic energy is converted to potential energy:
[tex]E_p = E_k[/tex]
[tex]mgh = mv^2/2[/tex]
where m is the total mass and h is the vertical distance traveled, v is the velocity right after the impact at, which we can solve by divide both sides my m
Let g = 9.81 m/s2
[tex]gh = v^2/2[/tex]
[tex]v^2 = 2gh = 2 * 9.81* 0.4 = 7.848[/tex]
[tex]v = \sqrt{7.848} = 2.8m/s[/tex]
According the law of momentum conservation, momentum before and after the impact must be the same
[tex]m_uv_u + m_ov_o = (m_u + m_o)v[/tex]
where [tex]m_u = 0.01, v_u[/tex] are the mass and velocity of the bullet before the impact, respectively.[tex]m_ov_o[/tex] are the mass and velocity of the block before the impact, respectively, which is 0 because the block was stationary before the impact
[tex]0.01v_u + 0 = (0.01 + 1.5)*2.8[/tex]
[tex]0.01v_u = 4.23[/tex]
[tex]v_u = 4.23 / 0.01 = 423 m/s[/tex]
