# A ball with a mass of #2 kg# is rolling at #12 m/s# and elastically collides with a resting ball with a mass of #1 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 principle of conservation of momentum and the principle of conservation of kinetic energy.

Let (v_1) and (v_2) be the velocities of the 2 kg ball and 1 kg ball respectively after the collision.

Using the conservation of momentum:

(m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2)

where: (m_1 = 2) kg (mass of the first ball) (m_2 = 1) kg (mass of the second ball) (u_1 = 12) m/s (initial velocity of the first ball) (u_2 = 0) m/s (initial velocity of the second ball)

Plugging in the values:

(2 \times 12 + 1 \times v_2 = 2 \times v_1 + 1 \times 0)

(24 + v_2 = 2v_1)

Now, using the conservation of kinetic energy:

(0.5m_1u_1^2 + 0.5m_2u_2^2 = 0.5m_1v_1^2 + 0.5m_2v_2^2)

Plugging in the values:

(0.5 \times 2 \times 12^2 + 0.5 \times 1 \times 0^2 = 0.5 \times 2 \times v_1^2 + 0.5 \times 1 \times v_2^2)

(144 = 2v_1^2 + 0.5v_2^2)

We now have a system of equations:

(24 + v_2 = 2v_1) ...(1)

(144 = 2v_1^2 + 0.5v_2^2) ...(2)

Solving these equations simultaneously will give us the values of (v_1) and (v_2).

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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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