Question

A block of mass \(m\) is moving on a horizontal frictionless surface with velocity \(v\). It hits a spring of constant \(k\) and compresses it by a distance \(x\). If the initial velocity is doubled, what will be the new compression?

• A. \(2x\)
• B. \(\sqrt{2}\,x\)
• C. \(4x\)
• D. \(x/2\)
• E.

Hint:

In spring–block energy problems, equate kinetic energy with spring potential energy. Since \(x^2 \propto v^2\), the compression \(x\) is directly proportional to the velocity \(v\). If velocity doubles, compression doubles; if velocity quadruples, compression quadruples.

Answer 1

The Correct Option is A

Concept: When a moving block compresses a spring on a frictionless surface, the kinetic energy of the block converts completely into the potential energy of the spring. \[ \text{Kinetic Energy} = \text{Spring Potential Energy} \] Thus, \[ \frac{1}{2}mv^2 = \frac{1}{2}kx^2 \]

Step 1: Relate compression with velocity. From the energy conservation equation: \[ \frac{1}{2}mv^2 = \frac{1}{2}kx^2 \implies x^2 = \frac{m}{k}v^2 \] Taking the square root on both sides: \[ x = v \sqrt{\frac{m}{k}} \] Hence, the compression of the spring \(x\) is directly proportional to the velocity \(v\) (\(x \propto v\)).

Step 2: Consider the velocity is doubled. If the new velocity \(v' = 2v\), the new compression \(x'\) will be: \[ x' \propto 2v \] Since \(x \propto v\), it follows that: \[ x' = 2x \]

Step 3: State the final result. The new compression of the spring becomes \[ \boxed{2x} \]