Question

5. You fire a 15-g bullet straight into a 2-kg wood block hanging from a 2-meter long string causing the block with the bullet totally embedded in it to swing out to a 30-degree angle. Find the magnitude of the velocity of the bullet as it was fired into the wood block.

Answer 1

mass of bullet, m_b = 15 g = 0.015 kg

mass of wood block, m_w = 2 kg

Initially block is at rest, (u_w=0)

Let the initial velocity of the bullet be 'u_b', and final velocity of the (block + bullet) system be 'v'

The collision between bullet and the wooden block is completely inelastic, as they 'stick' together after collision

applying conservation of momentum

m_b*u_b + m_w*u_w = (m_b + m_w)*v {Let, m_b+m_w = M}

putting the values

0.015*u_b + 2*0 = (0.015+2)*v-------(1)

After the collision the system's kinetic energy is converted into potential energy,

Let the system rise to a height of 'h' after the collision

0.5Mv² = Mgh

v = sqrt(2gh) ---------(2)

and also from geometry, L = L*cos($\theta$) + h

h = L*(1- cos($\theta$)) = 2*(1- cos(30))

h = 0.268 m

putting this in equation 2

v = sqrt(2*9.8*0.268) = 2.292 m/s

putting this in equation 1

0.015*u_b + 2*0 = (0.015+2)*2.292

u_b = 307.892 m/s