A body of mass m falls from a height `h_1` and rises to a height `h_2`. The magnitude of the change in momentum during the impact with the ground.

Options

A. `mg (h_1 + h_2)`
B. `m (sqrt(2 gh_(1)) + sqrt(2 gh_(2)) )`
C. `m(sqrt(2 gh_(1)) - sqrt(2 gh_(2)))`
D. Zero

Answer

To find the magnitude of the change in momentum during the impact of a body of mass \( m \) that falls from a height \( h_1 \) and rises to a height \( h_2 \), we can follow these steps:

### Step 1: Calculate the initial velocity just before impact
When the body falls from height \( h_1 \), its potential energy is converted to kinetic energy just before it hits the ground. The potential energy at height \( h_1 \) is given by:
\[ PE = mgh_1 \]
This potential energy converts into kinetic energy (KE) just before impact:
\[ KE = \frac{1}{2} mv^2 \]
Setting the potential energy equal to the kinetic energy:
\[ mgh_1 = \frac{1}{2} mv_i^2 \]
From this, we can solve for the initial velocity \( v_i \) just before impact:
\[ v_i = \sqrt{2gh_1} \]

### Step 2: Calculate the final velocity just after rebound
After hitting the ground, the body rebounds to a height \( h_2 \). At this height, the kinetic energy just after impact converts back to potential energy:
\[ PE = mgh_2 \]
Setting the kinetic energy equal to the potential energy at height \( h_2 \):
\[ \frac{1}{2} mv_f^2 = mgh_2 \]
From this, we can solve for the final velocity \( v_f \) just after rebound:
\[ v_f = \sqrt{2gh_2} \]

### Step 3: Determine the change in momentum
The change in momentum \( \Delta p \) is given by the mass times the change in velocity:
\[ \Delta p = m(v_f - v_i) \]
Substituting the expressions for \( v_f \) and \( v_i \):
\[ \Delta p = m\left(\sqrt{2gh_2} - \sqrt{2gh_1}\right) \]

### Step 4: Calculate the magnitude of change in momentum
Since we are interested in the magnitude of the change in momentum, we take the absolute value:
\[ |\Delta p| = m\left(\sqrt{2gh_2} + \sqrt{2gh_1}\right) \]

### Final Result
Thus, the magnitude of the change in momentum during the impact with the ground is:
\[ |\Delta p| = m\left(\sqrt{2gh_2} + \sqrt{2gh_1}\right) \]