Question

two water balloons of mass 0.5 kg collide and bounce off of each other without breaking. before the collision, one water balloon moved east at a speed of 2 m/s, while the other moved west at a speed of 2.5 m/s. after the collision, one moves west at a speed of 1.5 m/s. what is the velocity of the other water balloon? a. 1.5 m/s east b. 1 m/s east c. 1.5 m/s west d. 1 m/s west

Explanation:

Step1: Define variables and direction

Let east - direction be positive. $m_1 = m_2=0.5$ kg, $v_{1i}=2$ m/s, $v_{2i}=- 2.5$ m/s, assume $v_{2f}=-1.5$ m/s, find $v_{1f}$.

Step2: Apply conservation of momentum

The law of conservation of momentum is $m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$.
Substitute the values: $0.5\times2 + 0.5\times(-2.5)=0.5v_{1f}+0.5\times(-1.5)$.

Step3: Simplify the equation

First, calculate the left - hand side: $0.5\times2+0.5\times(-2.5)=1 - 1.25=-0.25$.
The right - hand side is $0.5v_{1f}-0.75$.
So, $-0.25 = 0.5v_{1f}-0.75$.

Step4: Solve for $v_{1f}$

Add 0.75 to both sides: $-0.25 + 0.75=0.5v_{1f}$, which gives $0.5 = 0.5v_{1f}$.
Divide both sides by 0.5: $v_{1f}=1$ m/s. Since the value is positive, the direction is east.

Answer:

B. 1 m/s east