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 positive direction

Let east be the positive direction.

Step2: Write the momentum conservation equation

The momentum conservation equation is \(m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}\). Given \(m_1 = m_2=m = 0.2\space kg\), \(v_{1i}=16\space m/s\), \(v_{2i}=- 11\space m/s\), \(v_{1f}=-11\space m/s\). Substitute these values into the equation: \(m\times16+m\times(-11)=m\times(-11)+m\times v_{2f}\).

Step3: Simplify the equation

Divide both sides of the equation by \(m\) (since \(m
eq0\)): \(16 - 11=-11 + v_{2f}\).

Step4: Solve for \(v_{2f}\)

\(v_{2f}=16\space m/s\) (positive value means east direction).

Answer:

D. \(16\space m/s\) east