A block of mass $M$ lies at rest connected to a massless spring of spring constant $k$ on a frictionless surface. A bullet of mass $m$ hits the block horizontally with speed $v$ as shown in the figure and is completely stuck to the block. What is the maximum compression in the spring resulting from this impact (assuming that at this point the spring is still not fully compressed)?
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For a completely inelastic collision followed by spring compression, you can write the conservation equations directly.
The final mechanical energy stored in the spring is equal to the kinetic energy just after the collision, which is $K = \frac{p^2}{2(M+m)}$ where $p = mv$.
Equating $\frac{(mv)^2}{2(M+m)} = \frac{1}{2}kx_{\max}^2$ immediately yields the answer.
Step 1: Understanding the Question:
This problem involves a two-stage physical process: first, a completely inelastic collision between a bullet and a stationary block, followed by the compression of a spring attached to the combined mass. Step 2: Key Formulas and Approach:
1. Conservation of Linear Momentum during the collision (since the impact is instantaneous and external spring force is negligible during this very short time interval):
\[ p_{\text{initial}} = p_{\text{final}} \implies m v = (M + m) V \]
2. Conservation of Mechanical Energy during the subsequent spring compression:
\[ \frac{1}{2} (M + m) V^2 = \frac{1}{2} k x_{\max}^2 \] Step 3: Detailed Explanation:
• Let us first find the velocity $V$ of the combined mass $(M + m)$ immediately after the completely inelastic collision. Using conservation of linear momentum:
\[ m v = (M + m) V \implies V = \frac{m v}{M + m} \]
• Once the bullet is embedded in the block, the combined system acts as a single mass $(M + m)$ with initial kinetic energy:
\[ K = \frac{1}{2} (M + m) V^2 \]
• As the combined mass moves to the left, it compresses the spring of spring constant $k$. Since the surface is frictionless, mechanical energy is conserved during the compression process.
• The kinetic energy of the combined mass is entirely converted into elastic potential energy of the spring at the point of maximum compression $x_{\max}$:
\[ \frac{1}{2} (M + m) V^2 = \frac{1}{2} k x_{\max}^2 \]
• Substitute the expression for $V$ into the energy conservation equation:
\[ (M + m) \left( \frac{m v}{M + m} \right)^2 = k x_{\max}^2 \]
\[ \frac{m^2 v^2}{M + m} = k x_{\max}^2 \]
• Solve for $x_{\max}$:
\[ x_{\max}^2 = \frac{m^2 v^2}{k(M + m)} \implies x_{\max} = \sqrt{\frac{m^2 v^2}{k(M + m)}} \] Step 4: Final Answer:
The maximum compression in the spring is $\sqrt{\frac{m^2v^2}{k(M + m)}}$, which corresponds to Option (A).
