# A ball with a mass of #5 kg# moving at #2 m/s# hits a still ball with a mass of #3 kg#. If the first ball stops moving, how fast is the second ball moving?

By using momentum conservation while the first ball is moving

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We use the law of conservation of momentum, which states that:

where:

So, we get:

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To find the final velocity of the second ball after the collision, you can use the principle of conservation of momentum:

[ m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f} ]

where ( m_1 ) and ( m_2 ) are the masses of the first and second balls, ( v_{1i} ) and ( v_{2i} ) are the initial velocities of the first and second balls, and ( v_{1f} ) and ( v_{2f} ) are the final velocities of the first and second balls.

Given that the first ball stops moving (( v_{1f} = 0 )), and the initial velocities are ( v_{1i} = 2 , \text{m/s} ) and ( v_{2i} = 0 ), and the masses are ( m_1 = 5 , \text{kg} ) and ( m_2 = 3 , \text{kg} ), we can solve for ( v_{2f} ):

[ 5 \cdot 2 + 3 \cdot 0 = 5 \cdot 0 + 3 \cdot v_{2f} ]

[ 10 = 3 \cdot v_{2f} ]

[ v_{2f} = \frac{10}{3} ]

Therefore, the final velocity of the second ball after the collision is ( \frac{10}{3} , \text{m/s} ).

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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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