Question

Figure p9.55 shows a collision between three balls of clay.The three hit simultaneously and stick together. What are the speedand direction of the resulting blob of clay?

Answer 1

Concepts and reason

The required concepts to solve the problem are conservation of momentum and vector algebra.

First, find the speed of the resulting blob using the conservation of momentum and then find the direction of the resulting blob using vector algebra.

Fundamentals

According to the law of conservation of momentum, the momentum before collision is equal to the momentum after collision.

When two objects collide, according to law of conservation of momentum

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

Here, ${m_1}$is the mass of the first particle, ${m_2}$is the mass of the second particle, ${u_1}$is the velocity of the first mass before collision, ${u_2}$is the velocity of the second mass before collision, ${v_1}$is the velocity of the first mass after collision ${v_2}$is the velocity of the second mass after collision.

If a vector is represented by,

$\vec X = x\hat i + y\hat j$

Then the direction of the vector is given by,

$\theta = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)$

Here, $x$is the $x$component of the velocity and$y$is the $y$component of the velocity.

Applying conservation of momentum in the horizontal direction,

${m_1}{u_1}\cos {\theta _1} + {m_2}{u_2}\cos {\theta _2} + {m_3}{u_3}\cos {\theta _3} = M{v_x}$

Here, ${\theta _1}$is the angle made by the velocity of the first particle with the horizontal, ${\theta _2}$is the angle made by the velocity of the second particle with the horizontal,${\theta _3}$is the angle made by the velocity of the third particle with the horizontal, $M$is the mass of the resulting blob and${v_x}$is the horizontal component of the velocity of the resulting blob.

Substitute$40\,{\rm{g}}$for${m_1}$, $4\,{\rm{m/s}}$for${u_1}$and$45^\circ$for${\theta _1}$in ${m_1}{u_1}\cos {\theta _1}$.

$\begin{array}{c}\\{m_1}{u_1}\cos {\theta _1} = \left( {\left( {40\,{\rm{g}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right)\left( {4\,{\rm{m/s}}} \right)\cos 45^\circ \\\\ = 0.113\,{\rm{kg}} \cdot {\rm{m/s}}\\\end{array}$

Substitute$20\,{\rm{g}}$for${m_2}$, $2\,{\rm{m/s}}$for${u_2}$and$90^\circ$for${\theta _2}$in${m_2}{u_2}\cos {\theta _2}$.

$\begin{array}{c}\\{m_2}{u_2}\cos {\theta _2} = \left( {\left( {20\,{\rm{g}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right)\left( {2\,{\rm{m/s}}} \right)\cos 90^\circ \\\\ = 0\,{\rm{kg}} \cdot {\rm{m/s}}\\\end{array}$

Substitute $30\,{\rm{g}}$for${m_3}$, $3\,{\rm{m/s}}$for${u_3}$, $180^\circ$for${\theta _3}$in ${m_3}{u_3}\cos {\theta _3}$.

$\begin{array}{c}\\{m_3}{u_3}\cos {\theta _3} = \left( {\left( {30\,{\rm{g}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right)\left( {3\,{\rm{m/s}}} \right)\cos 180^\circ \\\\ = - 0.09\,{\rm{kg}} \cdot {\rm{m/s}}\\\end{array}$

Substitute$90\,{\rm{g}}$for$M$.

$M{v_x} = \left( {\left( {90\,{\rm{g}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right){v_x}$

Substitute $0.113\,{\rm{kg}} \cdot {\rm{m/s}}$ for ${m_1}{u_1}\cos {\theta _1}$, $0\,{\rm{kg}} \cdot {\rm{m/s}}$ for ${m_2}{u_2}\cos {\theta _2}$, $- 0.09\,{\rm{kg}} \cdot {\rm{m/s}}$ for${m_3}{u_3}\cos {\theta _3}$ and $\left( {\left( {90\,{\rm{g}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right){v_x}$ for $M{v_x}$ in the equation of conservation of momentum, ${m_1}{u_1}\cos {\theta _1} + {m_2}{u_2}\cos {\theta _2} + {m_3}{u_3}\cos {\theta _3} = M{v_x}$ becomes,

$\begin{array}{c}\\\left( {0.113\,{\rm{kg}} \cdot {\rm{m/s}}} \right) + \left( {0\,{\rm{kg}} \cdot {\rm{m/s}}} \right) + \left( { - 0.09\,{\rm{kg}} \cdot {\rm{m/s}}} \right) = \left( {\left( {90\,{\rm{g}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right){v_x}\\\\{v_x} = 0.26\,{\rm{m/s}}\\\end{array}$

Applying conservation of momentum in the vertical direction,

${m_1}{u_1}\sin {\theta _1} + {m_2}{u_2}\sin {\theta _2} + {m_3}{u_3}\sin {\theta _3} = M{v_y}$

Here, ${v_y}$is the vertical component of the velocity of the resulting blob.

Substitute$40\,{\rm{g}}$for${m_1}$, $- 4\,{\rm{m/s}}$for${u_1}$and$45^\circ$for${\theta _1}$in ${m_1}{u_1}\sin {\theta _1}$.

$\begin{array}{c}\\{m_1}{u_1}\sin {\theta _1} = \left( {\left( {40\,{\rm{g}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right)\left( { - 4\,{\rm{m/s}}} \right)\sin 45^\circ \\\\ = - 0.113\,{\rm{kg}} \cdot {\rm{m/s}}\\\end{array}$

Substitute$20\,{\rm{g}}$for${m_2}$, $2\,{\rm{m/s}}$for${u_2}$and$90^\circ$for${\theta _2}$in${m_2}{u_2}\sin {\theta _2}$.

$\begin{array}{c}\\{m_2}{u_2}\sin {\theta _2} = \left( {\left( {20\,{\rm{g}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right)\left( {2\,{\rm{m/s}}} \right)\sin 90^\circ \\\\ = 0.04\,{\rm{kg}} \cdot {\rm{m/s}}\\\end{array}$

Substitute $30\,{\rm{g}}$for${m_3}$, $3\,{\rm{m/s}}$for${u_3}$, $180^\circ$for${\theta _3}$in ${m_3}{u_3}\sin {\theta _3}$.

$\begin{array}{c}\\{m_3}{u_3}\sin {\theta _3} = \left( {\left( {30\,{\rm{g}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right)\left( {3\,{\rm{m/s}}} \right)\sin 180^\circ \\\\ = 0\,{\rm{kg}} \cdot {\rm{m/s}}\\\end{array}$

Substitute$90\,{\rm{g}}$for$M$.

$M{v_y} = \left( {\left( {90\,{\rm{g}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right){v_y}$

Substitute $- 0.113\,{\rm{kg}} \cdot {\rm{m/s}}$ for ${m_1}{u_1}\sin {\theta _1}$, $0.04\,{\rm{kg}} \cdot {\rm{m/s}}$ for ${m_2}{u_2}\sin {\theta _2}$, $0\,{\rm{kg}} \cdot {\rm{m/s}}$ for${m_3}{u_3}\sin {\theta _3}$ and $\left( {\left( {90\,{\rm{g}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right){v_y}$ for $M{v_y}$ in the equation of conservation of momentum, ${m_1}{u_1}\sin {\theta _1} + {m_2}{u_2}\sin {\theta _2} + {m_3}{u_3}\sin {\theta _3} = M{v_y}$ becomes,

$\begin{array}{c}\\\left( { - 0.113\,{\rm{kg}} \cdot {\rm{m/s}}} \right) + \left( {0.04\,{\rm{kg}} \cdot {\rm{m/s}}} \right) + \left( {0\,{\rm{kg}} \cdot {\rm{m/s}}} \right) = \left( {\left( {90\,{\rm{g}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right){v_y}\\\\{v_y} = - 0.81\,{\rm{m/s}}\\\end{array}$

The magnitude of the vertical component of the velocity is,

${v_y} = 0.81\,{\rm{m/s}}$

The speed of the resulting blob is,

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

Substitute $0.26\,{\rm{m/s}}$for${v_x}$and $0.81\,{\rm{m/s}}$for${v_y}$.

$\begin{array}{c}\\v = \sqrt {{{\left( {0.26\,{\rm{m/s}}} \right)}^2} + {{\left( {0.81\,{\rm{m/s}}} \right)}^2}} \\\\ = 0.85\,{\rm{m/s}}\\\end{array}$

The direction of the velocity vector is given by the equation

$\theta = {\tan ^{ - 1}}\left( {\frac{{{v_y}}}{{{v_x}}}} \right)$

Substitute $0.26\,{\rm{m/s}}$for${v_x}$and $0.81\,{\rm{m/s}}$for${v_y}$.

$\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{0.81}}{{0.26}}} \right)\\\\ = 72.2^\circ \\\end{array}$

Ans:

The speed of the resulting blob is$0.85\,{\rm{m/s}}$