Chapter 11, Problem 95A

To determine

The speed of the ball A and B after collision.

Answer to Problem 95A

The speed of the ball A after collision is $v_{\text{A2}}=\left(\frac{m_{\text{A}}-m_{\text{B}}}{m_{\text{A}}+m_{\text{B}}}\right)v_{\text{A1}}+\left(\frac{2m_{\text{B}}}{m_{\text{A}}+m_{\text{B}}}\right)v_{\text{B1}}$ .

The speed of the ball B after collision is $v_{\text{B2}}=\left(\frac{2m_{\text{A}}}{m_{\text{A}}+m_{\text{B}}}\right)v_{\text{A1}}+\left(\frac{m_{\text{B}}-m_{\text{A}}}{m_{\text{A}}+m_{\text{B}}}\right)v_{\text{B1}}$ .

Explanation of Solution

Given:

The mass of ball A is $m_{\text{A}}$ .

The mass of ball B is $m_{\text{B}}$ .

The speed of ball A is $v_{\text{A}}$ .

The speed of ball B is $v_{\text{B}}$ .

Formula used:

As per the law of conservation of momentum, the initial momentum is equal to final momentum.

Calculation:

If the collision is elastic, then the momentum and kinetic energy is conserved.

Apply the conservation of momentum, according to which the total momentum of the two balls before and after collision are equal.

  $m_{\text{A}}{v}_{\text{A1}}+m_{\text{B}}{v}_{\text{B1}}=m_{\text{A}}{v}_{\text{A2}}+m_{\text{B}}{v}_{\text{B2}}$ …… (1)

  $\begin{array}{l}{m}_{\text{A}}{v}_{\text{A1}}-m_{\text{A}}{v}_{\text{A2}}=-m_{\text{B}}{v}_{\text{B1}}+m_{\text{B}}{v}_{\text{B2}}\\ m_{\text{A}}\left(v_{\text{A1}}-v_{\text{A2}}\right)=-m_{\text{B}}\left(v_{\text{B1}}-v_{\text{B2}}\right)\end{array}$ …… (2)

Solve equation (1) for $v_{\text{A2}}$ and $v_{\text{B2}}$ .

  $\begin{array}{l}\left(1\right)\Rightarrow m_{\text{A}}{v}_{\text{A1}}+m_{\text{B}}{v}_{\text{B1}}=m_{\text{A}}{v}_{\text{A2}}+m_{\text{B}}{v}_{\text{B2}}\\ m_{\text{A}}{v}_{\text{A2}}=m_{\text{A}}{v}_{\text{A1}}+m_{\text{B}}{v}_{\text{B1}}-m_{\text{B}}{v}_{\text{B2}}\\ v_{\text{A2}}=v_{\text{A1}}+\frac{m_{\text{B}}}{m_{\text{A}}}\left(v_{\text{B1}}-v_{\text{B2}}\right)\end{array}$ …… (3)

  $\begin{array}{l}\left(1\right)\Rightarrow m_{\text{A}}{v}_{\text{A1}}+m_{\text{B}}{v}_{\text{B1}}=m_{\text{A}}{v}_{\text{A2}}+m_{\text{B}}{v}_{\text{B2}}\\ m_{\text{B}}{v}_{\text{B2}}=m_{\text{A}}{v}_{\text{A1}}+m_{\text{B}}{v}_{\text{B1}}-m_{\text{A}}{v}_{\text{A2}}\\ v_{\text{B2}}=v_{\text{B1}}+\frac{m_{\text{A}}}{m_{\text{B}}}\left(v_{\text{A1}}-v_{\text{A2}}\right)\end{array}$ …… (4)

According to law of conservation of energy,

  $\begin{array}{l}\frac{1}{2}{m}_{\text{A}}{v}_{\text{A1}}^2+\frac{1}{2}{m}_{\text{B}}{v}_{\text{B1}}^2=\frac{1}{2}{m}_{\text{A}}{v}_{\text{A2}}^2+\frac{1}{2}{m}_{\text{B}}{v}_{\text{B2}}^2\\ m_{\text{A}}{v}_{\text{A1}}^2-m_{\text{A}}{v}_{\text{A2}}^2=-m_{\text{B}}{v}_{\text{B1}}^2+m_{\text{B}}{v}_{\text{B2}}^2\\ m_{\text{A}}\left(v_{\text{A1}}^2-v_{\text{A2}}^2\right)=-m_{\text{B}}\left(v_{\text{B1}}^2-v_{\text{B2}}^2\right)\end{array}$ …… (5)

Divide equation (2) by equation (5).

  $\begin{array}{l}\frac{m_{\text{A}}\left(v_{\text{A1}}-v_{\text{A2}}\right)}{m_{\text{A}}\left(v_{\text{A1}}^2-v_{\text{A2}}^2\right)}=\frac{-m_{\text{B}}\left(v_{\text{B1}}-v_{\text{B2}}\right)}{-m_{\text{B}}\left(v_{\text{B1}}^2-v_{\text{B2}}^2\right)}\\ \frac{\left(v_{\text{A1}}-v_{\text{A2}}\right)}{\left(v_{\text{A1}}+v_{\text{A2}}\right)\left(v_{\text{A1}}-v_{\text{A2}}\right)}=\frac{\left(v_{\text{B1}}-v_{\text{B2}}\right)}{\left(v_{\text{B1}}+v_{\text{B2}}\right)\left(v_{\text{B1}}-v_{\text{B2}}\right)}\\ v_{\text{A1}}+v_{\text{A2}}=v_{\text{B1}}+v_{\text{B2}}\end{array}$ …… (6)

The velocity of ball A using equation (4) and equation (6) is,

  $\begin{array}{l}{v}_{\text{A1}}+v_{\text{A2}}=v_{\text{B1}}+v_{\text{B1}}+\frac{m_{\text{A}}}{m_{\text{B}}}\left(v_{\text{A1}}-v_{\text{A2}}\right)\\ v_{\text{A2}}=2v_{\text{B1}}+\frac{m_{\text{A}}}{m_{\text{B}}}{v}_{\text{A1}}-\frac{m_{\text{A}}}{m_{\text{B}}}{v}_{\text{A2}}-v_{\text{A1}}\\ v_{\text{A2}}+\frac{m_{\text{A}}}{m_{\text{B}}}{v}_{\text{A2}}=2v_{\text{B1}}+\left(\frac{m_{\text{A}}}{m_{\text{B}}}-1\right)v_{\text{A1}}\\ \left(\frac{m_{\text{B}}+m_{\text{A}}}{m_{\text{B}}}\right)v_{\text{A2}}=2v_{\text{B1}}+\left(\frac{m_{\text{A}}-m_{\text{B}}}{m_{\text{B}}}\right)v_{\text{A1}}\end{array}$

  $\begin{array}{l}{v}_{\text{A2}}=2\left(\frac{m_{\text{B}}}{m_{\text{B}}+m_{\text{A}}}\right)v_{\text{B1}}+\left(\frac{m_{\text{A}}-m_{\text{B}}}{m_{\text{B}}}\right)\times \left(\frac{m_{\text{B}}}{m_{\text{B}}+m_{\text{A}}}\right)v_{\text{A1}}\\ v_{\text{A2}}=\left(\frac{2m_{\text{B}}}{m_{\text{B}}+m_{\text{A}}}\right)v_{\text{B1}}+\left(\frac{m_{\text{A}}-m_{\text{B}}}{m_{\text{B}}+m_{\text{A}}}\right)v_{\text{A1}}\end{array}$

The velocity of ball B using equation (3) and equation (6) is,

  $\begin{array}{l}{v}_{\text{A1}}+v_{\text{A1}}+\frac{m_{\text{B}}}{m_{\text{A}}}\left(v_{\text{B1}}-v_{\text{B2}}\right)=v_{\text{B1}}+v_{\text{B2}}\\ 2v_{\text{A1}}+\frac{m_{\text{B}}}{m_{\text{A}}}{v}_{\text{B1}}-v_{\text{B1}}=v_{\text{B2}}+\frac{m_{\text{B}}}{m_{\text{A}}}{v}_{\text{B2}}\\ 2v_{\text{A1}}+\left(\frac{m_{\text{B}}}{m_{\text{A}}}-1\right)v_{\text{B1}}=v_{\text{B2}}\left(1+\frac{m_{\text{B}}}{m_{\text{A}}}\right)\\ 2v_{\text{A1}}+\left(\frac{m_{\text{B}}-m_{\text{A}}}{m_{\text{A}}}\right)v_{\text{B1}}=v_{\text{B2}}\left(\frac{m_{\text{A}}+m_{\text{B}}}{m_{\text{A}}}\right)\end{array}$

  $\begin{array}{l}{v}_{\text{B2}}=2\left(\frac{m_{\text{A}}}{m_{\text{A}}+m_{\text{B}}}\right)v_{\text{A1}}+\left(\frac{m_{\text{B}}-m_{\text{A}}}{m_{\text{A}}}\right)\left(\frac{m_{\text{A}}}{m_{\text{A}}+m_{\text{B}}}\right)v_{\text{B1}}\\ v_{\text{B2}}=\left(\frac{2m_{\text{A}}}{m_{\text{A}}+m_{\text{B}}}\right)v_{\text{A1}}+\left(\frac{m_{\text{B}}-m_{\text{A}}}{m_{\text{A}}+m_{\text{B}}}\right)v_{\text{B1}}\end{array}$

Conclusion:

Thus, the speed of the ball A after collision is $v_{\text{A2}}=\left(\frac{m_{\text{A}}-m_{\text{B}}}{m_{\text{A}}+m_{\text{B}}}\right)v_{\text{A1}}+\left(\frac{2m_{\text{B}}}{m_{\text{A}}+m_{\text{B}}}\right)v_{\text{B1}}$ and the speed of the ball B after collision is $v_{\text{B2}}=\left(\frac{2m_{\text{A}}}{m_{\text{A}}+m_{\text{B}}}\right)v_{\text{A1}}+\left(\frac{m_{\text{B}}-m_{\text{A}}}{m_{\text{A}}+m_{\text{B}}}\right)v_{\text{B1}}$ .