Question

A bullet of mass 0.01 kg moving horizontally strikes a block of wood of mass 1.5 kg which is suspended as a pendulum. The bullet lodges in the wood, and together they swing upwards a distance of 0.40 m. What was the velocity of the bullet just before it struck the wooden block

Answer 1

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:

$E_p = E_k$

$mgh = mv^2/2$

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

$gh = v^2/2$

$v^2 = 2gh = 2 * 9.81* 0.4 = 7.848$

$v = \sqrt{7.848} = 2.8m/s$

According the law of momentum conservation, momentum before and after the impact must be the same

$m_uv_u + m_ov_o = (m_u + m_o)v$

where $m_u = 0.01, v_u$ are the mass and velocity of the bullet before the impact, respectively.$m_ov_o$ are the mass and velocity of the block before the impact, respectively, which is 0 because the block was stationary before the impact

$0.01v_u + 0 = (0.01 + 1.5)*2.8$

$0.01v_u = 4.23$

$v_u = 4.23 / 0.01 = 423 m/s$