Question

1. an average net force caused an 11.0 kg object to accelerate uniformly from rest. if this object travels 26.3 m west in 3.20 s, what is the change in momentum of the object?

Explanation:

Step1: Find the acceleration

Use the kinematic equation \(x = v_0t+\frac{1}{2}at^{2}\). Since \(v_0 = 0\) (starts from rest), the equation simplifies to \(x=\frac{1}{2}at^{2}\). Solving for \(a\):
\(a=\frac{2x}{t^{2}}\)
Substitute \(x = 26.3\space m\) and \(t = 3.20\space s\)
\(a=\frac{2\times26.3}{3.20^{2}}=\frac{52.6}{10.24}\approx5.14\space m/s^{2}\)

Step2: Find the final velocity

Use the equation \(v = v_0+at\). Since \(v_0 = 0\), \(v = at\)
Substitute \(a\approx5.14\space m/s^{2}\) and \(t = 3.20\space s\)
\(v=5.14\times3.20 = 16.45\space m/s\)

Step3: Calculate the change in momentum

The change in momentum \(\Delta p=m(v - v_0)\). Since \(v_0 = 0\), \(\Delta p=mv\)
Substitute \(m = 11.0\space kg\) and \(v = 16.45\space m/s\)
\(\Delta p=11.0\times16.45 = 181\space kg\cdot m/s\)

Answer:

\(181\space kg\cdot m/s\)