Chapter 17.4, Problem 1aTH

In the space provided, draw separate arrows representing the direction of the change in momentum vector of object A in the two experiments.
Direction of $\Delta {\stackrel{\to }{p}}_A$

Is the magnitude of the change in momentum of object A in experiment 1 greater than, less than, or equal to that in experiment 2? Explain.

To determine

The change in magnitude of the object $\text{A}$ of experiment $1$ and $2$ .

Explanation of Solution

Introduction:

Momentum has both magnitude as well as direction because it is a vector quantity. It is calculated by using the product of mass and velocity of object during motion. The total system’s momentum is equal to the vector summation of the individual momentum.

The change in momentum is given by product of mass and the change in velocity.

For experiment $1$:

The change in momentum is given by:

  ${\text{Δp = m}}_{\text{A}}\left({\text{V}}_{\text{Af}}{\text{-V}}_{\text{Ai}}\right)$

Thus, the direction of motion is along the velocity direction.

Therefore,

The change in momentum vector for experiment $1$ is given as ${\text{<mover accent="true"><mo>V</mo><mo>→</mo></mover>}}_{\text{Af}}+\left({\text{-<mover accent="true"><mo>V</mo><mo>→</mo></mover>}}_{\text{Ai}}\right)$

  

For experiment $2$:

  

Magnitude of velocity vectors from both resultant velocity vectors:

  $\begin{array}{l}{\text{V}}_{\text{R1}}\text{= 5 blocks}\\ {\text{V}}_{\text{R2}}\text{= 4 blocks}\end{array}$

Therefore, The change in object ‘s momentum for experiment $1$ is greater than experiment $2$ .

Conclusion:

The change in object’s momentum for experiment $1$ is greater than experiment $2$ .