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Chapter 7: Problem 9

An object of mass \(3.0 \mathrm{kg}\) is allowed to fall from rest under the force of gravity for 3.4 s. What is the change in its momentum? Ignore air resistance.

Short Answer

Expert verified
Answer: The change in momentum is approximately 99.96 kg路m/s.

Step by step solution

01

Calculate the final velocity of the object

Using the equation of motion \(v = u + at\), the final velocity can be calculated as follows: - \(u = 0 \mathrm{m/s}\) (initial velocity) - \(a = 9.8 \mathrm{m/s^2}\) (acceleration due to gravity) - \(t = 3.4 \mathrm{s}\) (time) Plugging in these values, we have \(v = 0 + 9.8 \times 3.4 = 33.32 \mathrm{m/s}\).
02

Calculate the change in momentum

Now, we can calculate the change in momentum using the formula \(\Delta p = m(v - u)\). Plugging in the values we have: - \(m = 3.0 \mathrm{kg}\) (mass) - \(v = 33.32 \mathrm{m/s}\) (final velocity) - \(u = 0 \mathrm{m/s}\) (initial velocity) So, \(\Delta p = 3.0 \times (33.32 - 0) = 99.96 \mathrm{kg \cdot m/s}\). The change in momentum of the object is approximately \(99.96 \mathrm{kg \cdot m/s}\).

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