Question

A 15g bullet traveling at 500m/s strikes an 0.8kg block of wood that is balanced on the table edge 0.8m above the ground. If the bullet buries itself in the block . Find the distance D at wich the block hits the floor.

Answer 1

The given parameters:

mass of the bullet, m₁ = 0.015 kg

speed of the bullet, u₁ = 500 m/s

mass of block wood, m₂ = 0.8 kg

height of the table, h = 0.8 m

The final velocity of the bullet-block system after the collision is calculated by applying the principle of conservation of linear momentum:

$m_1u_1=v(m_1+m_2)$

$v=\frac{m_1u_1}{m_1+m_2}=\frac{0.015\cdot500}{0.015+0.8}=9.2$ m/s

The time taken for the bullet-block system to fall to the floor after collision is calculated as follows:

$h=v_{0y}t+gt^2/2=gt^2/2$

$t=\sqrt{2h/g}=\sqrt{2\cdot0.8/g}=0.4$ s

The horizontal distance where the block hits the floor is calculated as follows:

$x=vt=9.2\cdot0.4=3.68$ m