Question

A 3.8 kg block of wood sits on a table. A 3.0 g bullet, fired horizontally at a speed of 460 m/s, goes completely through the block, emerging at a speed of 250 m/s. What is the speed of the block immediately after the bullet exits?

Answer 1

Concepts and reason

The concept required to solve this problem is the conservation of momentum. First, derive the expression for the final speed by using the conservation of momentum. Finally, calculate the speed of the block by using the expression for the velocity.

Fundamentals

The conservation of momentum states that the momentum of the system remains conserved if there are no external forces acting on the system.

The equation of the conservation of momentum is given as follows:

${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$

Here, ${m_1}$is the mass of block moving with initial velocities ${u_1}$and ${m_2}$is the mass of bullet moving with initial velocity ${u_2}$. ${v_1}$and ${v_2}$are the final velocities of block and bullet after the collision.

The equation of the conservation of momentum is given as follows:

${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$

Here, ${m_1}$is the mass of block and ${m_2}$is the mass of bullet.

Re-arrange the above expression for ${v_1}$.

${v_1} = \frac{{\left( {{m_1}{u_1} + {m_2}{u_2}} \right) - {m_2}{v_2}}}{{{m_1}}}$

Substitute 3.8 kg for ${m_1}$, 3.0 g for ${m_2}$, 0 m/s for ${u_1}$, 460 m/s for ${u_2}$and 250 m/s for ${v_2}$in the equation ${v_1} = \frac{{\left( {{m_1}{u_1} + {m_2}{u_2}} \right) - {m_2}{v_2}}}{{{m_1}}}$.

$\begin{array}{c}\\{v_1} = \frac{{\left( {\left( {3.8{\rm{ kg}}} \right)\left( {0{\rm{ m/s}}} \right) + \left( {3.0{\rm{ g}}\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1.0{\rm{ g}}}}} \right)} \right)\left( {460{\rm{ m/s}}} \right)} \right) - \left( {3.0{\rm{ g}}\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1.0{\rm{ g}}}}} \right)} \right)\left( {250{\rm{ m/s}}} \right)}}{{3.8{\rm{ kg}}}}\\\\ = 0.166{\rm{ m/s}}\\\end{array}$

Ans:

The speed of the block immediately after the bullet exits is 0.166 m/s.