Chapter 8, Problem 65A

A 5-kg fish swimming I m/s swallows an absent-minded 1-kg fish at rest. What is the speed of the large fish immediately after lunch? What would its speed be if the small fish were swimming toward it at 4 m/s?

To determine

To Calculate: The speed of the bigger fish in both cases.

Explanation of Solution

Given info:

Mass of the bigger fish, $m_1=5\text{kg}$

Mass of smaller fish, $m_2=1\text{kg}$

(i)

Initial velocity of bigger fish, $u_1=1\text{m/s}$

Initial velocity of smaller fish, $u_2=0$

(ii)

Initial velocity of smaller fish, $u_2\text{'}=4\text{m/s}$

Formula used:

According to law of conservation of momentum,

Initial momentum = final momentum

  $m_1{v}_1+m_2{v}_2=\left(m_1+m_2\right)v$

Calculation:

(i) Substitute the values:

  $\begin{array}{l}{m}_1{v}_1+m_2{v}_2=\left(m_1+m_2\right)v\\ v=\frac{m_1{v}_1+m_2{v}_2}{\left(m_1+m_2\right)}\\ =\frac{5\times 1+0}{6}\\ =\frac{5}{6}\text{m/s}\end{array}$

(ii)

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=\left(m_1+m_2\right)v\\ v=\frac{m_1{u}_1+m_2{u}_2}{\left(m_1+m_2\right)}\\ =\frac{5\times 1-1\times 4}{6}\\ =\frac{1}{6}\text{m/s}\end{array}$

Conclusion

Thus, the speed in the first case is $\frac{5}{6}\text{m/s}$ and the second case is $\frac{1}{6}\text{m/s}$ respectively.