Question

a 0.330 - kg volleyball is thrown vertically downward with a speed of 0.150 m/s in a place where g = 9.81 m/s². it takes it 0.0655 s to reach the ground. what is the magnitude of its momentum just before it hits the ground?
a) 0.322 kg·m/s
b) 0.0216 kg·m/s
c) 0.262 kg·m/s
d) 0.163 kg·m/s
e) 0.0418 kg·m/s
f) 0.212 kg·m/s

Explanation:

Step1: Find the final velocity

Use the kinematic equation \(v = v_0+at\). Here, \(v_0 = 0.150\ m/s\), \(a = g=9.81\ m/s^{2}\), and \(t = 0.0655\ s\).

$$v=0.150 + 9.81\times0.0655$$

$$v=0.150+0.642555$$

$$v = 0.792555\ m/s$$

Step2: Calculate the momentum

Momentum is given by \(p = mv\). Given \(m = 0.330\ kg\) and \(v = 0.792555\ m/s\)

$$p=0.330\times0.792555$$

$$p = 0.261543\approx0.262\ kg\cdot m/s$$

Answer:

C. \(0.262\ kg\cdot m/s\)