Chapter 9, Problem 9STP

To determine

The impulse on the rock if its velocity at the moment it strikes the ground is $20\text{m/s}$ downward.

Answer to Problem 9STP

  $240\text{kg}\cdot \text{m/s}$

Explanation of Solution

Given:

Mass of the rock, $m=12.0\text{kg}$

Final speed of the rock at the moment it strikes the ground, $v_f=20.0\text{m/s}$

Formula used:

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

  $p=mv$

And, Impulse on an object is given by the change in momentum of the object,

  $\text{Impulse = }{p}_f-p_i$

Calculation:

Since the rock falls on ground, thus its initial velocity will be zero,

  $v_i=0\text{m/s}$

Using the given values in above mentioned formulas, the Impulse on the rock will be,

  $\begin{array}{c}\text{Impulse = }{p}_f-p_i\\ =m\left(v_f-v_i\right)\\ =12\left(20-0\right)\\ =240\text{kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the impulse on the rock is $240\text{kg}\cdot \text{m/s}$ .