A 28-g rifle bullet travelling\(190\;{\rm{m/s}}\)embeds itself in a 3.1-kg pendulum hanging on a 2.8-m-long string, which makes the pendulum swing upward in an arc. Determine the vertical and horizontal components of the pendulum’s maximum displacement.

The linear momentum of a body is defined as the product of its mass and velocity. It is given by \(p = mv\).

The principle of conservation of linear momentum states that if two bodies collide with each other, the total linear momentum before and after collision remains the same if no external force acts on the system.

\(\begin{array}{c}{p_{{\rm{before}}}} = {p_{{\rm{after}}}}\\{\left( {mv} \right)_{{\rm{before}}}} = {\left( {mv} \right)_{{\rm{after}}}}\end{array}\)

The mass of the bullet is\(m = 28\;{\rm{g}} = 0.028\;{\rm{kg}}\).

The speed of the bullet is\(v = 190\;{\rm{m/s}}\).

The mass of the pendulum is\(M = 3.1\;{\rm{kg}}\).

The length of the string is \(l = 2.8\;{\rm{m}}\).

Suppose the pendulum does not move at the moment when the bullet collides and embeds in it. The momentum of the system before the collision is\({p_{{\rm{before}}}} = mv\).

The speed of the system just after the collision is\({p_{{\rm{after}}}} = \left( {m + M} \right)v'\).

Apply the conservation of linear momentum as the collision is inelastic at the bottom of the arc.

\(mv = \left( {m + M} \right)v'\) … (i)

After the collision, the swinging motion of the bullet and pendulum combined will conserve the mechanical energy. Suppose the zero level of the gravitational potential energy is at the bottom of the arc.

For the pendulum to swing exactly above the arc, the potential energy at the top of the arc must be equal to the kinetic energy at the bottom.

\(\begin{array}{c}K{E_{{\rm{bottom}}}} = P{E_{{\rm{top}}}}\\\frac{1}{2}\left( {m + M} \right){{v'}^2} = \left( {m + M} \right)gh\end{array}\) … (ii)

Rearranging equation (2), determine the value of height h reached by the pendulum as it swings up.

\(v' = \sqrt {2gh} \) … (iii)

Substitute equation (iii) in equation (i).

\(\begin{array}{c}mv = \left( {m + M} \right)\left( {\sqrt {2gh} } \right)\\h = {\left( {\frac{m}{{m + M}}} \right)^2}\frac{{{v^2}}}{{2g}}\\ = {\left( {\frac{{0.028\;{\rm{kg}}}}{{0.028\;{\rm{kg}} + 3.1\;{\rm{kg}}}}} \right)^2}\frac{{{{\left( {190\;{\rm{m/s}}} \right)}^2}}}{{2\left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)}}\\ = 0.15\;{\rm{m}}\end{array}\)

This height is the vertical component of the displacement and has a value of 0.15 m.

The horizontal displacement of the pendulum can be determined using the Pythagorean theorem.

\(\begin{array}{c}{x^2} + {\left( {L - h} \right)^2} = {L^2}\\x = \sqrt {2Lh - {h^2}} \\ = \sqrt {2\left( {2.8\;{\rm{m}}} \right)\left( {0.15\;{\rm{m}}} \right) - {{\left( {0.15\;{\rm{m}}} \right)}^2}} \\ = 0.90\;{\rm{m}}\end{array}\)

Thus, the horizontal component of displacement is 0.90 m.