Question

1. a 1.30 kg object is dropped from a height of 6.5 m. how far did the object fall when its momentum is 6.0 kgm/s?

Explanation:

Step1: Find the velocity when momentum is \(6.0\space kg\cdot m/s\)

Using the formula for momentum \(p = mv\), where \(p = 6.0\space kg\cdot m/s\) and \(m = 1.30\space kg\).
We can solve for \(v\): \(v=\frac{p}{m}\)
Substituting the values: \(v=\frac{6.0}{1.30}\approx4.62\space m/s\)

Step2: Use the kinematic equation \(v^{2}=v_{0}^{2}+2gh\) to find the distance fallen \(h\)

Since the object is dropped, \(v_{0} = 0\space m/s\), \(g = 9.8\space m/s^{2}\) and \(v\approx4.62\space m/s\)
The kinematic equation simplifies to \(v^{2}=2gh\)
Solving for \(h\): \(h=\frac{v^{2}}{2g}\)
Substituting the values: \(h=\frac{(4.62)^{2}}{2\times9.8}=\frac{21.34}{19.6}\approx1.09\space m\)

Answer:

The object has fallen approximately \(1.09\space m\) when its momentum is \(6.0\space kg\cdot m/s\)