To find the maximum compression of the spring when the collision is perfectly inelastic, we can follow these steps:

1. Conservation of Momentum in the Inelastic Collision

Since the collision is perfectly inelastic, the two masses stick together after the collision. Using the conservation of linear momentum, we have:

$m_1 v_0 + m_2 \cdot 0 = (m_1 + m_2) v_f$

Where:

• $m_1 = 3.0 \, \text{kg}$
• $v_0 = 15.0 \, \text{m/s}$
• $m_2 = 1.0 \, \text{kg}$

Substitute these values:

$3.0 \times 15.0 = (3.0 + 1.0) v_f$

$45 = 4 v_f$

$v_f = \frac{45}{4} = 11.25 \, \text{m/s}$

2. Energy Conservation to Find Maximum Compression

After the collision, the combined mass $(m_1 + m_2)$ moves with velocity $v_f$ and compresses the spring until the kinetic energy is fully converted to potential energy in the spring. The potential energy in the spring at maximum compression is given by:

$\frac{1}{2} k x^2$

And the kinetic energy of the combined mass just after the collision is:

$\frac{1}{2} (m_1 + m_2) v_f^2$

Setting these equal to each other due to energy conservation:

$\frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} k x^2$

Canceling $\frac{1}{2}$ from both sides and substituting the values:

$(3.0 + 1.0) \times (11.25)^2 = 100 \cdot x^2$

$4 \times 126.5625 = 100 \cdot x^2$

$506.25 = 100 \cdot x^2$

$x^2 = \frac{506.25}{100} = 5.0625$

$x = \sqrt{5.0625} \approx 2.25 \, \text{m}$

Answer

The maximum compression of the spring is approximately $\mathbf{2.25 \, \text{m}}$, which is not listed in the options provided. The solution should be double-checked in case there is an error in rounding or setup interpretation.

Do you need more details or have further questions?

1. What is a perfectly inelastic collision and how does it differ from elastic collisions?
2. How would the solution change if friction was present?
3. How does spring constant $k$ affect the maximum compression?
4. How would the solution change if the masses were swapped?
5. Can we solve this problem with an energy approach alone?

Tip: Remember that in energy conservation problems, converting units consistently can prevent common mistakes.