A student throws a 140 g snowball at 7.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of the average force on the wall if the duration of the collision is 0.19 s ?
momentum = .140 * 7.5
force = change of momentum / time
= .140 * 7.5 / .19
Melissa - your method is correct but numbers are different.
To calculate the magnitude of the average force on the wall, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = ma).
First, let's calculate the acceleration of the snowball during the collision. To do this, we can use the formula for average acceleration:
acceleration = (final velocity - initial velocity) / time
Given:
Initial velocity (u) = 7.5 m/s
Final velocity (v) = 0 m/s (since the snowball sticks to the wall)
Time (t) = 0.19 s
Using the formula, we can calculate the acceleration:
acceleration = (0 - 7.5) / 0.19 = -39.47 m/s²
Since the snowball hits the wall and sticks, the direction of the acceleration is negative.
Next, we need to determine the force by multiplying the mass of the snowball by the calculated acceleration:
mass (m) = 140 g = 0.14 kg
acceleration (a) = -39.47 m/s²
force (F) = mass * acceleration
F = 0.14 kg * (-39.47 m/s²)
Calculating the result:
F = -5.526 N
Because we are interested in the magnitude of the force, we can ignore the negative sign. Thus, the magnitude of the average force on the wall is approximately 5.526 N.
F = (mv - 0 ) / t
F = (0.15 x 8.5 - 0 )/ 0.15 = 8.5 N
F = rate of change of momentum