To find the maximum compression of the spring in this perfectly inelastic collision, we can use the principles of conservation of momentum and conservation of energy.

Step-by-Step Solution

1. Conservation of Momentum: Since the collision is perfectly inelastic, the masses $m_1$ and $m_2$ stick together after the collision and move with a common velocity $v'$.

   Given:

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

   Initial momentum of the system: $p_{\text{initial}} = m_1 \cdot v_0 = 3.0 \times 15.0 = 45.0 \, \text{kg} \cdot \text{m/s}$

   After the collision, the combined mass $m_1 + m_2$ moves with velocity $v'$. Thus: $p_{\text{final}} = (m_1 + m_2) \cdot v'$

   Setting $p_{\text{initial}} = p_{\text{final}}$: $45.0 = (3.0 + 1.0) \cdot v'$ $v' = \frac{45.0}{4.0} = 11.25 \, \text{m/s}$

2. Conservation of Energy: After the collision, the kinetic energy of the combined masses is transformed into the potential energy stored in the spring at maximum compression.

   The initial kinetic energy after collision: $KE = \frac{1}{2} (m_1 + m_2) v'^2$ Substituting values: $KE = \frac{1}{2} \times 4.0 \times (11.25)^2$ $KE = 2.0 \times 126.5625 = 253.125 \, \text{J}$

3. Spring Compression: At maximum compression, all the kinetic energy is converted to spring potential energy, given by $PE = \frac{1}{2} k x^2$.

   Equating $KE = PE$: $\frac{1}{2} k x^2 = 253.125$ Given $k = 100 \, \text{N/m}$: $\frac{1}{2} \times 100 \times x^2 = 253.125$ $50 x^2 = 253.125$ $x^2 = \frac{253.125}{50} = 5.0625$ $x = \sqrt{5.0625} \approx 2.25 \, \text{m}$

Answer:

The maximum compression of the spring is approximately $2.25 \, \text{m}$. None of the options provided in the image match this value exactly, so it might be a rounding or calculation check question. However, given our calculations, the closest option would be 2.25 m if it were listed.

Would you like a detailed explanation of any particular step or have further questions?

Further Questions

1. How does conservation of momentum apply to inelastic collisions in general?
2. Why is kinetic energy not conserved in inelastic collisions?
3. How would the answer change if the collision were elastic?
4. What factors determine the stiffness of a spring in physics problems?
5. How would friction impact the maximum compression in this scenario?

Tip

When solving physics problems involving springs, always check whether energy is conserved or if you need to account for inelastic losses.