# A ball with a mass of # 5 kg# is rolling at #8 m/s# and elastically collides with a resting ball with a mass of #4 kg#. What are the post-collision velocities of the balls?

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

Let ( m_1 = 5 , \text{kg} ) be the mass of the first ball rolling at ( v_1 = 8 , \text{m/s} ), and ( m_2 = 4 , \text{kg} ) be the mass of the second ball initially at rest.

By conservation of momentum: [ m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2 ] [ 5(8) + 4(0) = 5(u_1) + 4(u_2) ] [ 40 = 5u_1 + 4u_2 ]

By conservation of kinetic energy in an elastic collision: [ \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 ] [ \frac{1}{2}(5)(8)^2 + \frac{1}{2}(4)(0)^2 = \frac{1}{2}(5)(u_1)^2 + \frac{1}{2}(4)(u_2)^2 ] [ 160 = \frac{5}{2}(u_1)^2 + \frac{4}{2}(u_2)^2 ]

Solving the system of equations: [ 40 = 5u_1 + 4u_2 ] [ 160 = \frac{5}{2}(u_1)^2 + \frac{4}{2}(u_2)^2 ]

We find ( u_1 = 2 , \text{m/s} ) and ( u_2 = 10 , \text{m/s} ). Therefore, the post-collision velocities of the balls are ( u_1 = 2 , \text{m/s} ) and ( u_2 = 10 , \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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