Question

two water balloons of mass 0.35 kg collide and bounce off of each other without breaking. before the collision, one water balloon moved at a velocity of 2.5 m/s east, while the other moved at a velocity of 2.25 m/s west. after the collision, one balloon moves at a velocity of 1.75 m/s west. what is the velocity of the other water balloon?

a. 2 m/s west
b. 2 m/s east
c. 0.25 m/s west
d. 0.25 m/s east

Explanation:

Step1: Define the positive - direction

Let the east - direction be positive. So the initial velocity of the first balloon $v_{1i}=2.5\ m/s$ and the initial velocity of the second balloon $v_{2i}=- 2.25\ m/s$ (negative because it moves west). The mass of each balloon $m = 0.35\ kg$. After the collision, let the velocity of the first balloon be $v_{1f}=-1.75\ m/s$ (negative as it moves west), and the velocity of the second balloon be $v_{2f}$ (unknown).

Step2: Apply the law of conservation of momentum

The law of conservation of momentum is $m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$. Since $m_1 = m_2=m$, we can simplify the equation to $m(v_{1i}+v_{2i})=m(v_{1f}+v_{2f})$. Canceling out the mass $m$ on both sides, we get $v_{1i}+v_{2i}=v_{1f}+v_{2f}$.

Step3: Solve for $v_{2f}$

Substitute the known values into the equation: $2.5+( - 2.25)=-1.75 + v_{2f}$. First, simplify the left - hand side: $2.5−2.25 = 0.25$. Then, solve for $v_{2f}$: $v_{2f}=0.25 + 1.75=2\ m/s$. Since the result is positive, the direction is east.

Answer:

B. 2 m/s east