Question

A 20 g ball of clay traveling east at 3.6 m/s collides with a 30 g ball of clay traveling north at 2.0 m/s. What are the speed and the direction of the resulting 50 g ball of clay?

Answer 1

Concepts and reason

The concept required to solve this problem is Conservation of momentum.

Initially, calculate the momentum of each clay ball before the collision and add them to get the total initial momentum in horizontal and vertical direction.

Later, use the conservation of momentum in both the directions to solve for final velocity of the clay. Then use the magnitude equation to solve for the speed of the clay.

Finally, use the tangent function to solve for the direction of final velocity of the clay.

Fundamentals

The momentum of an object is the product of its mass and velocity. The magnitude of momentum is expressed as,

$p = mv$

Here, $p$ is the momentum, $m$ is the mass and $v$ is the speed.

The conservation of momentum principle says that total momentum of an isolated system is always conserved. This can be expressed as,

${p_{{\rm{initial}}}} = {p_{{\rm{final}}}}$

Here, ${p_{{\rm{initial}}}}$ is the total initial momentum of the system and ${p_{{\rm{final}}}}$ is the total final momentum.

The speed is the magnitude of the velocity that is,

$v = \sqrt {v_x^2 + v_y^2}$

Here, ${v_x}$ is the velocity in horizontal direction and ${v_y}$ is the velocity in the vertical direction.

The direction of the velocity is given by inverse of tangent function as,

$\theta = \arctan \left( {\frac{{{v_y}}}{{{v_x}}}} \right)$

Here, ${v_x}$ is the velocity in horizontal direction and ${v_y}$ is the velocity in the vertical direction.

The expression for momentum is,

$P = mv$

Substitute ${p_{1{\rm{ix}}}}$ for $p$, $20{\rm{ g}}$ for $m$, and $3.6{\rm{ m/s}}$ for $v$ in the above equation to calculate the initial momentum of the clay 1, ${p_{1{\rm{ix}}}}$ in horizontal direction.

$\begin{array}{c}\\{p_{1{\rm{ix}}}} = \left( {{\rm{20 g}}} \right)\left( {3.6{\rm{ m/s}}} \right)\\\\ = \left( {{\rm{20 g}}\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right)} \right)\left( {3.6{\rm{ m/s}}} \right)\\\\ = 0.072{\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}$

Substitute ${p_{1{\rm{iy}}}}$ for $p$, $20{\rm{ g}}$ for $m$, and $0{\rm{ m/s}}$ for $v$ in the equation $p = mv$ to calculate the initial momentum of the clay 1, ${p_{1{\rm{iy}}}}$ in vertical direction.

$\begin{array}{c}\\{p_{1{\rm{iy}}}} = \left( {{\rm{20 g}}} \right)\left( {0{\rm{ m/s}}} \right)\\\\ = 0{\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}$

Substitute ${p_{{\rm{2ix}}}}$ for $p$, ${\rm{30 g}}$ for $m$, and $0{\rm{ m/s}}$ for $v$ in the equation $p = mv$ to calculate the initial momentum of the clay 2, ${p_{{\rm{2ix}}}}$in the horizontal direction.

$\begin{array}{c}\\{p_{{\rm{2ix}}}} = \left( {{\rm{30 g}}} \right)\left( {0{\rm{ m/s}}} \right)\\\\ = 0{\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}$

Substitute ${p_{{\rm{2iy}}}}$ for $p$, ${\rm{30 g}}$ for $m$, and $0{\rm{ m/s}}$ for $v$ in the equation $p = mv$ to calculate the initial momentum of the clay 2, ${p_{{\rm{2iy}}}}$in the vertical direction.

$\begin{array}{c}\\{p_{{\rm{2iy}}}} = \left( {{\rm{30 g}}} \right)\left( {{\rm{2}}{\rm{.0 m/s}}} \right)\\\\ = \left( {{\rm{30 g}}\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right)} \right)\left( {{\rm{2}}{\rm{.0 m/s}}} \right)\\\\ = 0.06{\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}$

The total initial momentum is the sum of initial momentum of clay 1 and clay 2.

${p_{\rm{i}}} = {p_{{\rm{1i}}}} + {p_{{\rm{2i}}}}$

Substitute $0.072{\rm{ kg}} \cdot {\rm{m/s}}$ for ${p_{{\rm{1i}}}}$ and $0$ for ${p_{{\rm{2i}}}}$ in the equation ${p_{\rm{i}}} = {p_{{\rm{1i}}}} + {p_{{\rm{2i}}}}$ to calculate the total initial momentum in horizontal direction ${p_{{\rm{ix}}}}$.

$\begin{array}{c}\\{p_{{\rm{ix}}}} = 0.072{\rm{ kg}} \cdot {\rm{m/s}} + 0\\\\ = 0.072{\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}$

Substitute ${\rm{0 kg}} \cdot {\rm{m/s}}$ for ${p_{{\rm{1i}}}}$ and $0.06{\rm{ kg}} \cdot {\rm{m/s}}$ for ${p_{{\rm{2i}}}}$ in the equation ${p_{\rm{i}}} = {p_{{\rm{1i}}}} + {p_{{\rm{2i}}}}$ to calculate the total initial momentum in vertical direction, ${p_{{\rm{iy}}}}$.

$\begin{array}{c}\\{p_{{\rm{iy}}}} = 0 + 0.06{\rm{ kg}} \cdot {\rm{m/s}}\\\\ = 0.06{\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}$

The total final momentum is the sum of final momentum of clay 1 and 2.

${p_{{\rm{final}}}} = {p_{1{\rm{f}}}} + {p_{2{\rm{f}}}}$

Here, ${p_{1{\rm{f}}}}$ is the final momentum of the clay 1 and ${p_{{\rm{2f}}}}$ is the final momentum of clay 2.

Substitute ${m_1}{v_{\rm{f}}}$ for ${p_{{\rm{1f}}}}$ and ${m_2}{v_{\rm{f}}}$ for ${p_{2{\rm{f}}}}$ in the above equation ${p_{{\rm{final}}}} = {p_{1{\rm{f}}}} + {p_{2{\rm{f}}}}$.

${p_{{\rm{final}}}} = {m_1}{v_{\rm{f}}} + {m_2}{v_{\rm{f}}}$

Substitute ${m_1}{v_{\rm{f}}} + {m_2}{v_{\rm{f}}}$ for ${p_{{\rm{final}}}}$ in the conservation of momentum equation ${p_{{\rm{initial}}}} = {p_{{\rm{final}}}}$ and solve for ${v_{\rm{f}}}$.

$\begin{array}{c}\\{p_{{\rm{initial}}}} = {m_1}{v_{\rm{f}}} + {m_2}{v_{\rm{f}}}\\\\{v_{\rm{f}}} = \frac{{{p_{{\rm{initial}}}}}}{{{m_1} + {m_2}}}\\\end{array}$

Substitute ${v_x}$ for ${v_{\rm{f}}}$, $0.072{\rm{ kg}} \cdot {\rm{m/s}}$ for ${p_{{\rm{initial}}}}$, ${\rm{20 g}}$ for ${m_1}$, and ${\rm{30 g}}$ for ${m_2}$ in the equation ${v_{\rm{f}}} = \frac{{{p_{{\rm{initial}}}}}}{{{m_1} + {m_2}}}$ and calculate ${v_x}$ in the horizontal direction.

$\begin{array}{c}\\{v_x} = \frac{{0.072{\rm{ kg}} \cdot {\rm{m/s}}}}{{20{\rm{ g}} + 30{\rm{ g}}}}\\\\ = \frac{{0.072{\rm{ kg}} \cdot {\rm{m/s}}}}{{50{\rm{ g}}\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right)}}\\\\ = 1.44{\rm{ m/s}}\\\end{array}$

Substitute ${v_y}$ for ${v_{\rm{f}}}$, $0.06{\rm{ kg}} \cdot {\rm{m/s}}$ for ${p_{{\rm{initial}}}}$, ${\rm{20 g}}$ for ${m_1}$, and ${\rm{30 g}}$ for ${m_2}$ in the equation ${v_{\rm{f}}} = \frac{{{p_{{\rm{initial}}}}}}{{{m_1} + {m_2}}}$ and calculate ${v_y}$ in the vertical direction.

$\begin{array}{c}\\{v_y} = \frac{{0.06{\rm{ kg}} \cdot {\rm{m/s}}}}{{20{\rm{ g}} + 30{\rm{ g}}}}\\\\ = \frac{{0.06{\rm{ kg}} \cdot {\rm{m/s}}}}{{50{\rm{ g}}\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right)}}\\\\ = 1.2{\rm{ m/s}}\\\end{array}$

Use the magnitude equation to calculate speed.

Substitute $1.44{\rm{ m/s}}$ for ${v_x}$ and $1.2{\rm{ m/s}}$ for ${v_y}$ in the equation $v = \sqrt {v_x^2 + v_y^2}$.

$\begin{array}{c}\\v = \sqrt {{{\left( {1.44{\rm{ m/s}}} \right)}^2} + {{\left( {1.2{\rm{ m/s}}} \right)}^2}} \\\\ = 1.87{\rm{ m/s}}\\\end{array}$

Substitute $1.44{\rm{ m/s}}$ for ${v_x}$ and $1.2{\rm{ m/s}}$ for ${v_y}$ in the equation $\theta = \arctan \left( {\frac{{{v_y}}}{{{v_x}}}} \right)$.

$\begin{array}{c}\\\theta = \arctan \left( {\frac{{1.2{\rm{ m/s}}}}{{1.44{\rm{ m/s}}}}} \right)\\\\ = 39.8^\circ \\\end{array}$

Ans:

The magnitude of the final speed of the resulting 50 g clay is 1.87 m/s.

The direction of the final speed of the resulting 50 g clay is $39.8^\circ$ north of east.