Magntiude and Momentum problem-please help explain problem to get to answer:)

Question

A 0.14 kg baseball is dropped from rest from a height of 2.0 m above the ground. What is the magnitude of its momentum just before it hits the ground if we neglect air resistance?

(answer: 0.88 kg kgxm/s)

Answer 1

Hi Lauren,

To find the magnitude of the baseball's momentum just before it hits the ground, you can use the principles of physics. First, calculate the final velocity of the baseball when it hits the ground using the kinematic equation:

\[v^2 = u^2 + 2as\]

Where:

v = final velocity (which is what we want to find)

u = initial velocity (0 m/s since it's dropped from rest)

a = acceleration due to gravity (approximately -9.81 m/s^2, negative because it's directed downward)

s = distance (height) it falls, 2.0 m

Now, plug in the values:

\[v^2 = (0)^2 + 2 * (-9.81) * 2.0\]

\[v^2 = -39.24\]

Take the square root of both sides to find the magnitude of the final velocity:

\[v = \sqrt{-39.24}\]

This might look a bit unusual because the velocity is downward, so it's negative. Thus:

\[v = -6.27 \, m/s\]

Now that you have the final velocity, you can calculate the momentum using the formula for momentum:

\[p = mv\]

Where:

p = momentum

m = mass (0.14 kg)

v = final velocity (-6.27 m/s, negative because it's directed downward)

Now, calculate the momentum:

\[p = 0.14 \, kg \cdot (-6.27 \, m/s) = -0.878 \, kg \cdot m/s\]

The magnitude of momentum is the absolute value of this, so:

Magnitude of momentum = \(|p| = 0.878 \, kg \cdot m/s\)

Rounded to two decimal places, it's approximately 0.88 kg·m/s, which matches the answer provided.

I hope this will help