Chapter 9.2, Problem 22PP

To determine

The speed of $1.00\text{kg}$ ball and its direction after the collision.

Answer to Problem 22PP

  $2\text{m/s}$ in the opposite direction of 1^st ball’s initial motion.

Explanation of Solution

Given:

The mass of the 1^st ball, $m_1=0.5\text{kg}$

The mass of the 2^nd ball, $m_2=1.00\text{kg}$

Initial velocity of 1^st ball, $v_{1i}=6.0\text{m/s}$

Initial velocity of 2^nd ball, $v_{2i}=12.0\text{m/s}$ in the opposite direction of 1^st ball’s motion.

After colliding, the final velocity of the 1^st ball, $v_{1f}=14\text{m/s}$ backwards.

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.

Assume the initial direction of 1^st ball’s motion as positive and backward direction as negative, then the given velocities can be written as,

  $\begin{array}{l}{v}_{1i}=6.0\text{m/s}\\ v_{2i}=-12.0\text{m/s}\\ v_{1f}=-14.0\text{m/s}\end{array}$

Let, the final speed of the $1.00\text{kg}$ ball, that is 2^nd ball be $v_{2f}$ , 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,

  $\begin{array}{c}\left(0.5\right)\left(6\right)+\left(1.0\right)\left(-12\right)=\left(0.5\right)\left(-14\right)+\left(1.0\right)\left(v_{2f}\right)\\ v_{2f}=\left(3-12\text{+7}\right)\text{m/s}\\ v_{2f}=-2\text{m/s}\end{array}$

Negative sign of the above velocity denotes that after the collision the second ball moves in the opposite direction of 1^st ball’s initial motion.

Conclusion:

Thus, the final speed of the $1.00\text{kg}$ ball is $2\text{m/s}$ in the opposite direction of 1^st ball’s initial motion.