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? 8. an average net force of 16.0 n acts on a box for 3.33×10^{-3} min causing it to accelerate from rest to 3.50 m/s. what is the mass of the object?

Explanation:

Step1: Calculate the velocity

Use the formula for momentum \(p = mv\). Given \(p = 6.0\space kg\cdot m/s\) and \(m=1.30\space kg\), we can find \(v\) by rearranging the formula to \(v=\frac{p}{m}\).

$$v=\frac{6.0}{1.30}\approx 4.62\space m/s$$

Step2: Use the kinematic equation

The kinematic equation \(v^{2}=v_{0}^{2}+2gh\) (where \(v_{0} = 0\space m/s\) as the object is dropped). We want to find \(h\), so rearranging the equation gives \(h=\frac{v^{2}}{2g}\). Taking \(g = 9.8\space m/s^{2}\)

$$h=\frac{(4.62)^{2}}{2\times9.8}=\frac{21.34}{19.6}\approx1.09\space m$$

Answer:

The object has fallen approximately \(1.09\space m\)