# A ball with a mass of #3 kg # and velocity of #7 m/s# collides with a second ball with a mass of #5 kg# and velocity of #- 4 m/s#. If #40%# of the kinetic energy is lost, what are the final velocities of the balls?

After a very lengthy solution, we find

This is a lengthy problem that asks us to solve for two unknowns (the two final velocities). To do this, we must generate two equations involving these two unknowns, and solve them simultaneously.

One equation will come from conservation of momentum, the other will come from the 40% kinetic energy lost condition.

First, cons. of momentum:

Inserting the values we know:

Now, the kinetic energy condition. The initial KE is:

The final KE is:

Since 40% of the KE is lost, the final KE equals 60% of the initial

With all that complete, our two equations are:

Substitute this value into the second equation:

Now, we must solve this equation. It will be simpler if we multiply every term by 18, to eliminate the denominators:

Use the quadratic formula to solve for v_(2f)

There are two answers:

We must reject the second answer in each case, as this would result in both balls continuing in their original directions. This could only happen if there was no collision , but somehow the balls lost 40% of their KE.

So, the only acceptable answers are

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To solve for the final velocities of the balls after the collision, we can use the principle of conservation of momentum and the equation for kinetic energy.

First, let's calculate the initial momentum of the system:

(p_{\text{initial}} = m_1 \cdot v_{1\text{initial}} + m_2 \cdot v_{2\text{initial}})

(p_{\text{initial}} = (3 , \text{kg}) \cdot (7 , \text{m/s}) + (5 , \text{kg}) \cdot (-4 , \text{m/s}))

(p_{\text{initial}} = 21 , \text{kg m/s} - 20 , \text{kg m/s})

(p_{\text{initial}} = 1 , \text{kg m/s})

Next, we calculate the total initial kinetic energy of the system:

(KE_{\text{initial}} = \frac{1}{2} m_1 \cdot v_{1\text{initial}}^2 + \frac{1}{2} m_2 \cdot v_{2\text{initial}}^2)

(KE_{\text{initial}} = \frac{1}{2} (3 , \text{kg}) \cdot (7 , \text{m/s})^2 + \frac{1}{2} (5 , \text{kg}) \cdot (-4 , \text{m/s})^2)

(KE_{\text{initial}} = \frac{1}{2} (3 , \text{kg}) \cdot 49 , \text{m}^2/\text{s}^2 + \frac{1}{2} (5 , \text{kg}) \cdot 16 , \text{m}^2/\text{s}^2)

(KE_{\text{initial}} = 73.5 , \text{J} + 40 , \text{J})

(KE_{\text{initial}} = 113.5 , \text{J})

Now, since 40% of the kinetic energy is lost during the collision, the final kinetic energy is 60% of the initial kinetic energy:

(KE_{\text{final}} = 0.6 \times KE_{\text{initial}})

(KE_{\text{final}} = 0.6 \times 113.5 , \text{J})

(KE_{\text{final}} = 68.1 , \text{J})

Using the principle of conservation of momentum, we can set the total initial momentum equal to the total final momentum:

(p_{\text{initial}} = p_{\text{final}})

(m_1 \cdot v_{1\text{initial}} + m_2 \cdot v_{2\text{initial}} = m_1 \cdot v_{1\text{final}} + m_2 \cdot v_{2\text{final}})

(3 , \text{kg} \cdot 7 , \text{m/s} + 5 , \text{kg} \cdot (-4 , \text{m/s}) = 3 , \text{kg} \cdot v_{1\text{final}} + 5 , \text{kg} \cdot v_{2\text{final}})

(21 , \text{kg m/s} - 20 , \text{kg m/s} = 3 , \text{kg} \cdot v_{1\text{final}} + 5 , \text{kg} \cdot v_{2\text{final}})

(1 , \text{kg m/s} = 3 , \text{kg} \cdot v_{1\text{final}} + 5 , \text{kg} \cdot v_{2\text{final}})

Now, using the fact that the kinetic energy is lost during the collision, we can write:

(KE_{\text{initial}} - KE_{\text{final}} = \frac{1}{2} m_1 \cdot (v_{1\text{initial}}^2 - v_{1\text{final}}^2) + \frac{1}{2} m_2 \cdot (v_{2\text{initial}}^2 - v_{2\text{final}}^2))

Substitute the known values:

(113.5 , \text{J} - 68.1 , \text{J} = \frac{1}{2} (3 , \text{kg}) \cdot (7 , \text{m/s})^2 - \frac{1}{2} (3 , \text{kg}) \cdot (v_{1\text{final}}^2) + \frac{1}{2} (5 , \text{kg}) \cdot (-4 , \text{m/s})^2 - \frac{1}{2} (5 , \text{kg}) \cdot (v_{2\text{final}}^2))

Solve for (v_{1\text{final}}) and (v_{2\text{final}}):

(v_{1\text{final}} = 2 , \text{m/s})

(v_{2\text{final}} = -1 , \text{m/s})

Therefore, the final velocities of the first and second balls are (2 , \text{m/s}) and (-1 , \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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