A bullet of mass $0.012kg$ and horizontal speed $70m{s}^{-1}$ strikes a block of wood of mass $0.4kg$ and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Stwp 1: Since on striking the wooden block, the bullet comes to rest w.r.t. the block of wood, the collision is inelastic in nature.

Let $v$ be the velocity acquired by the combination of the bullet and block.

Applying principle of conservation of linear momentum for before and after the collision we can write,

$(m_1+m_2)v=m_1{u}_1+m_2{u}_2$

i.e. $v=\frac{(m_1{u}_1+m_2{u}_2)}{(m_1+m_2)}=\frac{0.012\times 70}{0.012+0.4}=2.04m/s$

Step 2: Because of the bullet impact, the block rises to height $h.$

So using the conservation of energy theorem,

$\frac{1}{2}(m_1+m_2)v^2=(m_1+m_2)gh$

i.e. $h=\frac{v^2}{2g}=\frac{(2.04{)}^2}{2\times 9.8}=21.2m$


Step 3: Here, the heat is produced due to loss of kinetic energy of the bullet after the collision.

Heat produced = Initial K.E.of the bullet − Final K.E.of the combination
Therefore,

Heat produced $=\frac{1}{2}{m}_1{u}_1^2-\frac{1}{2}(m_1+m_2)v^2$

$=\frac{1}{2}\times 0.012\times {70}^2-\frac{1}{2}(0.012+0.4)(2.04{)}^2$

$=29.4-0.86=28.54J$