Question

an object of mass ( m = 0.5 \text{ kg} ) is moving in the ( xy )-plane with velocity components of ( v_x = 3.0 \text{ m/s} ) and ( v_y = - 4.0 \text{ m/s} ). what is the magnitude of the objects momentum?
a ( 0.5 \text{ kg}cdot\text{m/s} )
b ( 2.5 \text{ kg}cdot\text{m/s} )
c ( 3.5 \text{ kg}cdot\text{m/s} )
d ( 5.0 \text{ kg}cdot\text{m/s} )

Explanation:

Step1: Find the magnitude of velocity

The velocity has components \( v_x = 3.0 \, \text{m/s} \) and \( v_y = - 4.0 \, \text{m/s} \). The magnitude of velocity \( v \) is given by the Pythagorean theorem:
\( v=\sqrt{v_x^2 + v_y^2} \)
Substitute \( v_x = 3.0 \) and \( v_y = - 4.0 \):
\( v=\sqrt{(3.0)^2+(-4.0)^2}=\sqrt{9 + 16}=\sqrt{25}=5.0 \, \text{m/s} \)

Step2: Calculate momentum magnitude

Momentum \( p \) is defined as \( p = mv \), where \( m = 0.5 \, \text{kg} \) and \( v = 5.0 \, \text{m/s} \).
Substitute the values:
\( p=(0.5 \, \text{kg})(5.0 \, \text{m/s}) = 2.5 \, \text{kg·m/s} \) (Wait, no—wait, recalculating: \( 0.5 \times 5 = 2.5 \)? Wait, no, wait: \( 0.5 \times 5 = 2.5 \)? Wait, no, the velocity magnitude is 5, mass is 0.5, so \( 0.5 \times 5 = 2.5 \)? Wait, but let's check again. Wait, \( v_x = 3 \), \( v_y = -4 \), so \( v = \sqrt{3^2 + 4^2} = 5 \). Then \( p = m v = 0.5 \times 5 = 2.5 \). So the correct answer is B.

Answer:

B. \( 2.5 \, \text{kg·m/s} \)