Answer:
290.4 m/s
Explanation:
Let g = 10m/s2
15g = 0.015 kg
After getting momentum from the bullet, the block-bullet system has a kinetic energy that raises itselft upward, this kinetic energy is converted to potential energy as the mass stops at 15cm = 0.15m high:
[tex]E_p = E_k[/tex]
[tex]mgh = mv^2/2[/tex]
where m is the system mass and h is the vertical distance traveled, v is the system initial velocity, which is what we are looking for. We can divide both sides of the equation by m:
[tex]v^2 = 2gh = 2*10*0.15 = 3[/tex]
[tex]v = \sqrt{3} = 1.732 m/s[/tex]
At the impact, according to the conservation of momentum:
[tex]m_uv_u + m_bv_b = (m_u + m_b)v[/tex]
where [tex]m_u[/tex] = 0.015 kg is the mass of the bullet, [tex]v_u[/tex] is the velocity of the bullet before the impact, [tex]m_b[/tex] = 2.5 kg is the mass of the block, [tex]v_b[/tex] = 0 is the velocity of the block before the impact. v = 1.732 m/s is the velocity of the system after the impact.
[tex]0.015v_u + 0 = (0.015 + 2.5)1.732[/tex]
[tex]v_u = \frac{(0.015 + 2.5)1.732}{0.015} = 290.4 m/s[/tex]
