Question

two hockey pucks, each with a mass of 0.2 kg, slide across the ice and collide. before the collision, puck 1 is moving 16 m/s to the east and puck 2 is moving 11 m/s to the west. after the collision, puck 1 is moving at 11 m/s to the west. what is the velocity of puck 2?
a. 11 m/s west
b. 11 m/s east
c. 16 m/s west
d. 16 m/s east

Explanation:

Step1: Define Variables and Directions

Let east be positive and west be negative. Mass of each puck, \( m_1 = m_2 = 0.2 \, \text{kg} \). Initial velocity of puck 1, \( v_{1i} = 16 \, \text{m/s} \) (east, positive). Initial velocity of puck 2, \( v_{2i} = -11 \, \text{m/s} \) (west, negative). Final velocity of puck 1, \( v_{1f} = -11 \, \text{m/s} \) (west, negative). Let final velocity of puck 2 be \( v_{2f} \).

Step2: Apply Conservation of Momentum

Conservation of momentum: \( m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \). Since \( m_1 = m_2 \), we can divide both sides by \( m \) (mass):
\( v_{1i} + v_{2i} = v_{1f} + v_{2f} \).

Step3: Substitute Values and Solve

Substitute values: \( 16 + (-11) = -11 + v_{2f} \).
Simplify left side: \( 5 = -11 + v_{2f} \).
Add 11 to both sides: \( v_{2f} = 5 + 11 = 16 \, \text{m/s} \). Positive means east.

Answer:

D. 16 m/s east