# A ball with a mass of # 5 kg# is rolling at #6 m/s# and elastically collides with a resting ball with a mass of #9 kg#. What are the post-collision velocities of the balls?

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To find the post-collision velocities of the balls, we can use the conservation of momentum and the conservation of kinetic energy.

Let (v_1) and (v_2) be the velocities of the 5 kg and 9 kg balls, respectively, after the collision.

According to the conservation of momentum:

[ m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} ]

And since it's an elastic collision, we also have:

[ \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 ]

Given: ( m_1 = 5 ) kg (mass of the rolling ball), ( m_2 = 9 ) kg (mass of the resting ball), ( v_{1i} = 6 ) m/s (initial velocity of the rolling ball), and ( v_{2i} = 0 ) m/s (initial velocity of the resting ball).

Plugging in the values and solving the equations, we can find the post-collision velocities:

[ v_{1f} = \frac{(m_1 - m_2)v_{1i} + 2m_2v_{2i}}{m_1 + m_2} ] [ v_{2f} = \frac{(m_2 - m_1)v_{2i} + 2m_1v_{1i}}{m_1 + m_2} ]

Substituting the given values: [ v_{1f} = \frac{(5 - 9) \times 6 + 2 \times 9 \times 0}{5 + 9} ] [ v_{2f} = \frac{(9 - 5) \times 0 + 2 \times 5 \times 6}{5 + 9} ]

After calculation: [ v_{1f} = \frac{-24}{14} \approx -1.71 , \text{m/s} ] [ v_{2f} = \frac{60}{14} \approx 4.29 , \text{m/s} ]

So, the post-collision velocities of the balls are approximately ( -1.71 , \text{m/s} ) and ( 4.29 , \text{m/s} ), respectively.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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