Question

A golf ball of mass 0.0500 kg rolls off a horizontal cliff with an initial speed of 12.3 m/s. The ball falls a vertical distance of 14.5 m into a lake below. What is the magnitue of the golf ball's momentum just before it strikes the water? (Answer must be in kg.m/s)

Answer 1

To solve for the magnitude of the golf ball's momentum just before it strikes the water, we need to calculate its final velocity using the principles of physics. We will use the kinematic equation for free falling objects and then apply the definition of momentum.

The momentum of the golf ball just before it strikes the water is approximately 0.133 kg.m/s.

Step-by-step explanation:

The momentum of an object is calculated by multiplying its mass by its velocity. In this case, the mass of the golf ball is 0.0500 kg and its velocity just before it strikes the water can be found using the principle of conservation of energy.

First, we can find the final velocity of the ball just before striking the water using the equation:

mgh = ½mv²

Where m is the mass of the ball, g is the acceleration due to gravity, h is the height it falls, and v is the final velocity. Rearranging the equation to solve for v, we have:

v = sqrt(½gh)

Substituting the known values, we have:

v = sqrt(½ * 9.8 m/s² * 14.5 m)

v = sqrt(7.14)

v ≈ 2.67 m/s

Now that we have the final velocity, we can calculate the momentum using the equation:

p = mv

Substituting the known values, we have:

p = 0.0500 kg * 2.67 m/s

p ≈ 0.133 kg.m/s