Question

Select the correct answer.
A 0.100–kilogram white ball moving with a velocity of 5 meters/second along the positive x–axis hits a stationary blue ball of the same mass. Energy is not necessarily conserved, but momentum is. After the collision, the white ball moves off at an angle of 20° with respect to the positive x–axis and at a velocity 3.5 meters/second. The blue ball moves at an angle of –35° and a velocity of
2.1 meters/second. What is the final y–momentum of the system?

A.
0.0 kilogram·meters/second
B.
3.5 kilogram·meters/second
C.
4.1 kilogram·meters/second
D.
5.9 kilogram·meters/second

Answer 1

To find the final y-momentum of the system after the collision, we need to calculate the y-components of the momentum of both balls and then sum them.

1. **Calculate the y-momentum of the white ball:**
- Mass of the white ball, \( m_w = 0.100 \, \text{kg} \)
- Velocity of the white ball, \( v_w = 3.5 \, \text{m/s} \)
- Angle of the white ball, \( \theta_w = 20^\circ \)
- The y-component of the velocity of the white ball is given by:
\[
v_{wy} = v_w \cdot \sin(\theta_w) = 3.5 \cdot \sin(20^\circ)
\]
- Calculate \( \sin(20^\circ) \):
\[
\sin(20^\circ) \approx 0.342
\]
- Now calculate \( v_{wy} \):
\[
v_{wy} = 3.5 \cdot 0.342 \approx 1.197 \, \text{m/s}
\]
- Now, calculate the y-momentum of the white ball:
\[
p_{wy} = m_w \cdot v_{wy} = 0.100 \cdot 1.197 \approx 0.1197 \, \text{kg·m/s}
\]

2. **Calculate the y-momentum of the blue ball:**
- Mass of the blue ball, \( m_b = 0.100 \, \text{kg} \)
- Velocity of the blue ball, \( v_b = 2.1 \, \text{m/s} \)
- Angle of the blue ball, \( \theta_b = -35^\circ \)
- The y-component of the velocity of the blue ball is given by:
\[
v_{by} = v_b \cdot \sin(\theta_b) = 2.1 \cdot \sin(-35^\circ)
\]
- Calculate \( \sin(-35^\circ) \):
\[
\sin(-35^\circ) \approx -0.574
\]
- Now calculate \( v_{by} \):
\[
v_{by} = 2.1 \cdot (-0.574) \approx -1.206 \, \text{m/s}
\]
- Now, calculate the y-momentum of the blue ball:
\[
p_{by} = m_b \cdot v_{by} = 0.100 \cdot (-1.206) \approx -0.1206 \, \text{kg·m/s}
\]

3. **Sum the y-momenta:**
\[
p_{y, \text{total}} = p_{wy} + p_{by} \approx 0.1197 - 0.1206 \approx -0.0009 \, \text{kg·m/s}
\]

After rounding, the total y-momentum is essentially \( 0.0 \, \text{kg·m/s} \).

Thus, the answer is:

**A. 0.0 kilogram·meters/second**