Question

Ball A has a mass of 12.0 kg and is moving due west with a velocity of 13.0 m/s while ball B has a mass of 3.5 kg and is moving west with a velocity of 1.0 m/s. Ball A catches up to ball B and collides with it from behind. After the collision, ball A is moving westward with a velocity of 6.0 m/s. What is the velocity of ball B after the collision?

Answer 1

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum before the collision must equal the total momentum after the collision.

Let's denote the variables:

• \( m_A = 12.0 , \text{kg} \) (mass of ball A)
• \( v_{A_i} = 13.0 , \text{m/s} \) (initial velocity of ball A, moving west)
• \( m_B = 3.5 , \text{kg} \) (mass of ball B)
• \( v_{B_i} = 1.0 , \text{m/s} \) (initial velocity of ball B, moving west)
• \( v_{A_f} = 6.0 , \text{m/s} \) (final velocity of ball A after collision)
• \( v_{B_f} \) (final velocity of ball B after collision, unknown)

Using the law of conservation of momentum:

\[ \text{Total momentum before collision} = \text{Total momentum after collision} \]

Calculating the total momentum before the collision:

\[ \text{Initial Momentum} = (m_A \cdot v_{A_i}) + (m_B \cdot v_{B_i}) \]

Substituting in the values:

\[ \text{Initial Momentum} = (12.0 , \text{kg} \cdot 13.0 , \text{m/s}) + (3.5 , \text{kg} \cdot 1.0 , \text{m/s}) \]

\[ \text{Initial Momentum} = 156.0 , \text{kg} \cdot \text{m/s} + 3.5 , \text{kg} \cdot \text{m/s} = 159.5 , \text{kg} \cdot \text{m/s} \]

Now calculating the total momentum after the collision:

\[ \text{Final Momentum} = (m_A \cdot v_{A_f}) + (m_B \cdot v_{B_f}) \]

Substituting in the known values:

\[ \text{Final Momentum} = (12.0 , \text{kg} \cdot 6.0 , \text{m/s}) + (3.5 , \text{kg} \cdot v_{B_f}) \]

\[ \text{Final Momentum} = 72.0 , \text{kg} \cdot \text{m/s} + (3.5 , \text{kg} \cdot v_{B_f}) \]

Setting the initial momentum equal to the final momentum:

\[ 159.5 , \text{kg} \cdot \text{m/s} = 72.0 , \text{kg} \cdot \text{m/s} + (3.5 , \text{kg} \cdot v_{B_f}) \]

Now, we can solve for \( v_{B_f} \):

1. Rearrange the equation to isolate \( v_{B_f} \):

\[ 159.5 , \text{kg} \cdot \text{m/s} - 72.0 , \text{kg} \cdot \text{m/s} = 3.5 , \text{kg} \cdot v_{B_f} \]

\[ 87.5 , \text{kg} \cdot \text{m/s} = 3.5 , \text{kg} \cdot v_{B_f} \]

2. Divide both sides by \( 3.5 , \text{kg} \):

\[ v_{B_f} = \frac{87.5 , \text{kg} \cdot \text{m/s}}{3.5 , \text{kg}} \approx 25.0 , \text{m/s} \]

Thus, the velocity of ball B after the collision is approximately \( 25.0 , \text{m/s} \) moving west.