Question

3.2.2 quiz: momentum
question 7 of 10
two hockey pucks, each with a mass of 0.15 kg, slide across the ice and collide. before the collision, puck 1 is moving 14 m/s to the west and puck 2 is moving 10 m/s to the east. after the collision, puck 1 is moving at 10 m/s to the east. what is the velocity of puck 2?
a. 14 m/s west
b. 14 m/s east
c. 10 m/s east
d. 10 m/s west

Explanation:

Step1: Define variables and direction

Let west - direction be positive. $m_1 = m_2=0.15$ kg, $v_{1i}=14$ m/s, $v_{2i}=- 10$ m/s, $v_{1f}=-10$ m/s.

Step2: Apply conservation of momentum

The conservation - of - momentum formula is $m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$.
Substitute the values: $0.15\times14 + 0.15\times(-10)=0.15\times(-10)+0.15\times v_{2f}$.
First, simplify the left - hand side: $0.15\times(14 - 10)=0.15\times4 = 0.6$.
The equation becomes $0.6=0.15\times(-10)+0.15\times v_{2f}$.
Then, simplify the right - hand side: $0.6=-1.5 + 0.15v_{2f}$.
Add 1.5 to both sides: $0.6 + 1.5=0.15v_{2f}$, so $2.1 = 0.15v_{2f}$.
Solve for $v_{2f}$: $v_{2f}=\frac{2.1}{0.15}=14$ m/s. Since the value is positive, the direction is west.

Answer:

A. 14 m/s west