A ball with a mass of #2 kg# is rolling at #6 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 solve this problem, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy.
Let ( m_1 ) be the mass of the first ball (2 kg), ( v_{1i} ) be the initial velocity of the first ball (6 m/s), ( m_2 ) be the mass of the second ball (1 kg), and ( v_{2i} ) be the initial velocity of the second ball (0 m/s).
Using the conservation of momentum:
[ m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f} ]
[ 2 \cdot 6 + 1 \cdot 0 = 2 \cdot v_{1f} + 1 \cdot v_{2f} ]
[ 12 = 2 \cdot v_{1f} + v_{2f} ]
Using the conservation of kinetic energy:
[ \frac{1}{2} m_1 \cdot (v_{1i})^2 + \frac{1}{2} m_2 \cdot (v_{2i})^2 = \frac{1}{2} m_1 \cdot (v_{1f})^2 + \frac{1}{2} m_2 \cdot (v_{2f})^2 ]
[ \frac{1}{2} \cdot 2 \cdot (6)^2 + \frac{1}{2} \cdot 1 \cdot (0)^2 = \frac{1}{2} \cdot 2 \cdot (v_{1f})^2 + \frac{1}{2} \cdot 1 \cdot (v_{2f})^2 ]
[ 36 + 0 = 2 \cdot (v_{1f})^2 + \frac{1}{2} \cdot (v_{2f})^2 ]
[ 36 = 2 \cdot (v_{1f})^2 + \frac{1}{2} \cdot (v_{2f})^2 ]
Now, we have a system of equations:
[ 12 = 2 \cdot v_{1f} + v_{2f} ] [ 36 = 2 \cdot (v_{1f})^2 + \frac{1}{2} \cdot (v_{2f})^2 ]
Solving this system of equations will give us the post-collision velocities of the balls.
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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.
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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