SOLUTION: 6 2 convervation of momentum lecture worksheet - Studypool - Free Printable

Let's solve the Elastic Collision problem from your image step by step.

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Problem Statement (Elastic Collision Example):

A ball of mass $ m_1 = 3.0 \, \text{kg} $ is moving at $ v_1 = 5.0 \, \text{m/s} $ and strikes a second ball of mass $ m_2 = 4.0 \, \text{kg} $ that is initially at rest ($ v_2 = 0 \, \text{m/s} $). After the collision, the first ball stops immediately. We are to find the velocity of the second ball after the collision.

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Step 1: Use Conservation of Momentum

In elastic collisions, both momentum and kinetic energy are conserved. However, since we're told the first ball stops after the collision, we can use conservation of momentum directly.

The formula for conservation of momentum is:

$$
m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}
$$

Where:
- $ v_{1i} = 5.0 \, \text{m/s} $
- $ v_{2i} = 0 \, \text{m/s} $
- $ v_{1f} = 0 \, \text{m/s} $ (given: first ball stops)
- $ v_{2f} = ? $ (what we want to find)

Plug in the values:

$$
(3.0 \, \text{kg})(5.0 \, \text{m/s}) + (4.0 \, \text{kg})(0) = (3.0 \, \text{kg})(0) + (4.0 \, \text{kg})(v_{2f})
$$

$$
15.0 = 4.0 \cdot v_{2f}
$$

Now solve for $ v_{2f} $:

$$
v_{2f} = \frac{15.0}{4.0} = 3.75 \, \text{m/s}
$$

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✔ Answer: The velocity of the second ball after the collision is $ \boxed{3.75 \, \text{m/s}} $.

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Check: Is this consistent with an elastic collision?

Let’s verify if kinetic energy is also conserved (as it should be in elastic collisions).

#### Initial Kinetic Energy:

$$
KE_i = \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2}(3.0)(5.0)^2 + 0 = \frac{1}{2}(3.0)(25) = 37.5 \, \text{J}
$$

#### Final Kinetic Energy:

$$
KE_f = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 = 0 + \frac{1}{2}(4.0)(3.75)^2
$$

$$
= 2.0 \times 14.0625 = 28.125 \, \text{J}
$$

Wait! This is not equal to 37.5 J → Kinetic energy is not conserved, which contradicts the assumption of an elastic collision.

But the problem says "elastic collision" and gives that the first ball stops. So what's going on?

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🔍 Important Insight:

In real elastic collisions, if a lighter object hits a heavier one at rest, it does not stop completely unless the masses are equal or special conditions apply.

Let’s check whether it's possible for the first ball to stop in an elastic collision.

There is a known result in physics:

> In a one-dimensional elastic collision, if a moving object of mass $ m_1 $ hits a stationary object of mass $ m_2 $, and $ m_1 < m_2 $, then $ m_1 $ will bounce back (reverse direction), but cannot stop completely unless $ m_1 = m_2 $.

Here, $ m_1 = 3.0 \, \text{kg} $, $ m_2 = 4.0 \, \text{kg} $, so $ m_1 < m_2 $. It's impossible for the first ball to stop completely in an elastic collision — it would have to bounce back.

So the scenario described (first ball stops) cannot happen in an elastic collision unless additional forces or constraints are involved.

However, the problem assumes that the first ball stops, and asks us to find the velocity of the second ball.

This suggests that the problem may be testing conservation of momentum only, even if the physical situation is unrealistic for a true elastic collision.

Alternatively, perhaps the problem intends for us to ignore the inconsistency and just apply momentum conservation.

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🟩 Final Answer (based on momentum conservation):

Even though the scenario violates the laws of elastic collisions, if we accept the given condition (first ball stops), then:

$$
\boxed{v_{2f} = 3.75 \, \text{m/s}}
$$

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⚠️ Note:

For a true elastic collision, the final velocities are given by:

$$
v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i}
$$
$$
v_{2f} = \frac{2m_1}{m_1 + m_2} v_{1i}
$$

Let’s compute those:

$$
v_{1f} = \frac{3.0 - 4.0}{3.0 + 4.0} \cdot 5.0 = \frac{-1.0}{7.0} \cdot 5.0 = -0.714 \, \text{m/s}
$$
$$
v_{2f} = \frac{2 \cdot 3.0}{7.0} \cdot 5.0 = \frac{6.0}{7.0} \cdot 5.0 = 4.286 \, \text{m/s}
$$

So in reality, the first ball bounces back at ~0.714 m/s, and the second ball moves forward at ~4.29 m/s.

But the problem says the first ball stops, so it's either:
- A hypothetical scenario,
- Or it's testing momentum conservation without requiring full elastic behavior.

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✔ Conclusion:

Given the problem statement, and assuming the first ball stops, the velocity of the second ball is:

$$
\boxed{3.75 \, \text{m/s}}
$$

But note: this is not physically possible in a true elastic collision. The scenario likely assumes idealized momentum conservation only.

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Let me know if you'd like help solving the inelastic collision example (Granny and grandson)!