WPE CYU 8

Solution to Question 8

Problem Analysis:

A bar of mass $M$ carrying a block of mass $m$ strikes a spring with velocity $u$. We need to find the range of the spring constant $k$ such that two conditions are met:

1. The block $m$ does not slip on $M$ due to the deceleration.
2. The maximum compression does not exceed $l$ (since inter-turn spacing vanishes at $l$, implying the spring becomes rigid).

Step 1: Dynamics of the Collision

Assuming no slip occurs, the system $(M+m)$ moves together as a single oscillator. The maximum compression $x_{max}$ is found using conservation of energy:

$$ \frac{1}{2}(M+m)u^2 = \frac{1}{2}k x_{max}^2 \implies x_{max} = u \sqrt{\frac{M+m}{k}} $$

The maximum restoring force is $F_{max} = k x_{max}$. The maximum deceleration $a_{max}$ of the system is:

$$ a_{max} = \frac{F_{max}}{M+m} = \frac{k}{M+m} \left( u \sqrt{\frac{M+m}{k}} \right) = u \sqrt{\frac{k}{M+m}} $$

Step 2: Condition for No Slipping

The block $m$ decelerates due to the static friction force $f$. The required force to decelerate $m$ at rate $a_{max}$ is $m a_{max}$. This must not exceed the maximum static friction:

$$ m a_{max} \le \mu m g \implies a_{max} \le \mu g $$

Substituting $a_{max}$:

$$ u \sqrt{\frac{k}{M+m}} \le \mu g $$

Squaring both sides:

$$ \frac{k u^2}{M+m} \le \mu^2 g^2 \implies k \le \frac{(M+m)\mu^2 g^2}{u^2} $$

Step 3: Condition for Spring Compression

The spring becomes solid at compression $l$. Therefore, the physical compression $x_{max}$ must not exceed $l$:

$$ x_{max} \le l $$ $$ u \sqrt{\frac{M+m}{k}} \le l $$ $$ \frac{M+m}{k} u^2 \le l^2 \implies k \ge \frac{(M+m)u^2}{l^2} $$

Step 4: Combining Conditions

For a valid solution to exist, the lower bound for $k$ must be less than or equal to the upper bound:

$$ \frac{(M+m)u^2}{l^2} \le k \le \frac{(M+m)\mu^2 g^2}{u^2} $$

This implies a condition on velocity $u$:

$$ \frac{(M+m)u^2}{l^2} \le \frac{(M+m)\mu^2 g^2}{u^2} \implies u^4 \le \mu^2 g^2 l^2 \implies u \le \sqrt{\mu g l} $$