Chapter 9, Problem 77A

To determine

The new velocity of the $0.200\text{kg}$ ball.

Answer to Problem 77A

  $0.22\text{m/s}$ in the initial direction of the motion of balls

Explanation of Solution

Given:

The mass of the 1^st ball, $m_1=0.2\text{kg}$

The mass of the 2^nd ball, $m_2=0.10\text{kg}$

Initial velocity of 1^st ball, $v_{1i}=0.30\text{m/s}$

Initial velocity of 2^nd ball, $v_{2i}=0.10\text{m/s}$

After colliding, the new velocity of the 2^nd ball, $v_{2f}=0.26\text{m/s}$

Formula used:

Momentum of an object of mass m and moving with velocity v is given by,

  $p=mv$

And, law of conservation of linear momentum: Momentum is conserved in a closed and isolated system and is given by,

  $p_i=p_f$

Calculation:

Consider both the balls as a system, then it will be a closed and isolated system and law of conservation of momentum can be applied to it.

Let, the final speed of the $0.20\text{kg}$ ball, that is 1^st ball be $v_{1f}$ , then using the law of conservation of linear momentum,

  $\begin{array}{c}{p}_i=p_f\\ m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f}\end{array}$

Using given values in above,

  $\begin{array}{c}\left(0.2\right)\left(0.30\right)+\left(0.1\right)\left(0.10\right)=\left(0.2\right)\left(v_{1f}\right)+\left(0.1\right)\left(0.26\right)\\ v_{1f}=\frac{\left(0.06+0.01-\text{0}\text{.026}\right)}{0.2}\text{m/s}\\ v_{1f}=0.22\text{m/s}\end{array}$

Conclusion:

Thus, the new velocity of the $0.200\text{kg}$ ball is $0.22\text{m/s}$ in the initial direction of the motion of balls.