Billiard ball A of mass \({m_{\rm{A}}} = 0.120\;{\rm{kg}}\) moving with speed \({v_{\rm{A}}} = 2.80\;{\rm{m/s}}\) strikes ball B, initially at rest, of mass \({m_{\rm{B}}} = 0.140\;{\rm{kg}}\) As a result of the collision, ball A is deflected off at an angle of 30.0° with a speed \({v'_{\rm{A}}} = 2.10\;{\rm{m/s}}\)

(a) Taking the x axis to be the original direction of motion of ball A, write down the equations expressing the conservation of momentum for the components in the x and y directions separately.

(b) Solve these equations for the speed, \({v'_{\rm{B}}}\), and angle, \({\theta '_{\rm{B}}}\), of ball B after the collision. Do not assume the collision is elastic.

For a two-dimensional collision, the momentum along both x and y axes is conserved.

The mass of ball A is \({m_{\rm{A}}} = 0.120\;{\rm{kg}}\) and the mass of ball B is \({m_{\rm{B}}} = 0.140\;{\rm{kg}}\).

The initial speed of ball A is \({v_{\rm{A}}} = 2.80\;{\rm{m/s}}\) and the speed of ball B is \({v_{\rm{B}}} = 0\) as ball B is at rest.

After the collision, the speed of ball A is \({v'_{\rm{A}}} = 2.10\;{\rm{m/s}}\) and its direction is \({\theta '_{\rm{A}}} = {30.0^ \circ }\)with the initial direction of the ball A.

Let \({v'_{\rm{B}}}\) be the speed of ball B after the collision and \({\theta '_{\rm{B}}}\) its angle with the initial direction of ball A.

Part (a)

Before the collision, the total momentum along the x-axis is \({m_{\rm{A}}}{v_{\rm{A}}}\) and the total momentum along the y-axis is zero.

After the collision, the total momentum along the x-axis is \({m_{\rm{A}}}{v'_{\rm{A}}}\cos {\theta '_{\rm{A}}} + {m_{\rm{B}}}{v'_{\rm{B}}}\cos {\theta '_{\rm{B}}}\) and the total momentum along the y-axis is \(\left( {{m_{\rm{A}}}{{v'}_{\rm{A}}}\sin {{\theta '}_{\rm{A}}} - {m_{\rm{B}}}{{v'}_{\rm{B}}}\sin {{\theta '}_{\rm{B}}}} \right)\).

Now, using momentum conservation along the x-axis, you get:

\({m_{\rm{A}}}{v_{\rm{A}}} = {m_{\rm{A}}}{v'_{\rm{A}}}\cos {\theta '_{\rm{A}}} + {m_{\rm{B}}}{v'_{\rm{B}}}\cos {\theta '_{\rm{B}}}\) … (i)

Using momentum conservation along the y-axis, you get:

\(0 = {m_{\rm{A}}}{v'_{\rm{A}}}\sin {\theta '_{\rm{A}}} - {m_{\rm{B}}}{v'_{\rm{B}}}\sin {\theta '_{\rm{B}}}\) … (ii)

Hence, equations (i) and (ii) express the momentum conservation along the x and y axes.

Part (b)

From the first equation, you get:

\({m_{\rm{B}}}{v'_{\rm{B}}}\cos {\theta '_{\rm{B}}} = {m_{\rm{A}}}{v_{\rm{A}}} - {m_{\rm{A}}}{v'_{\rm{A}}}\cos {\theta '_{\rm{A}}}\) … (iii)

From the second equation, you get:

\({m_{\rm{B}}}{v'_{\rm{B}}}\sin {\theta '_{\rm{B}}} = {m_{\rm{A}}}{v'_{\rm{A}}}\sin {\theta '_{\rm{A}}}\) … (iv)

Now, dividing equation (iv) by equation (iii), you get:

\(\begin{array}{c}\frac{{{m_{\rm{B}}}{{v'}_{\rm{B}}}\sin {{\theta '}_{\rm{B}}}}}{{{m_{\rm{B}}}{{v'}_{\rm{B}}}\cos {{\theta '}_{\rm{B}}}}} = \frac{{{m_{\rm{A}}}{{v'}_{\rm{A}}}\sin {{\theta '}_{\rm{A}}}}}{{{m_{\rm{A}}}{v_{\rm{A}}} - {m_{\rm{A}}}{{v'}_{\rm{A}}}\cos {{\theta '}_{\rm{A}}}}}\\tan{{\theta '}_{\rm{B}}} = \frac{{{{v'}_{\rm{A}}}\sin {{\theta '}_{\rm{A}}}}}{{{v_{\rm{A}}} - {{v'}_{\rm{A}}}\cos {{\theta '}_{\rm{A}}}}}\\{{\theta '}_{\rm{B}}} = {\tan ^{ - 1}}\left( {\frac{{{{v'}_{\rm{A}}}\sin {{\theta '}_{\rm{A}}}}}{{{v_{\rm{A}}} - {{v'}_{\rm{A}}}\cos {{\theta '}_{\rm{A}}}}}} \right)\end{array}\)

Substituting all the values in the above equation, you get:

\(\begin{array}{c}{{\theta '}_{\rm{B}}} = {\tan ^{ - 1}}\left( {\frac{{\left( {2.10\;{\rm{m/s}}} \right)\sin {{30.0}^ \circ }_{\rm{A}}}}{{2.80\;{\rm{m/s}} - \left( {2.10\;{\rm{m/s}}} \right)\cos {{30.0}^ \circ }_{\rm{A}}}}} \right)\\{{\theta '}_{\rm{B}}} = {46.94^ \circ }\end{array}\)

Now, substituting the value \({\theta '_{\rm{B}}}\) and other values in equation (iv), you get:

\(\begin{array}{c}\left( {0.140\;{\rm{kg}}} \right) \times {{v'}_{\rm{B}}} \times \sin {46.94^ \circ } = \left( {0.120\;{\rm{kg}}} \right) \times \left( {2.10\;{\rm{m/s}}} \right) \times \sin {30.0^ \circ }\\{{v'}_{\rm{B}}} = 1.23\;{\rm{m/s}}\end{array}\)

Hence, the speed of ball B is \({v'_{\rm{B}}} = 1.23\;{\rm{m/s}}\) and the angle of ball B is \({\theta '_{\rm{B}}} = {46.94^ \circ }\) after the collision.