Question:
A block of mass 'm' moving on a frictionless horizontal surface collides with a spring of spring constant 'K' and compresses it through a distance 'x'. The maximum momentum of the block after collision is
A block of mass 'm' moving on a frictionless horizontal surface collides with a spring of spring constant 'K' and compresses it through a distance 'x'. The maximum momentum of the block after collision is
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The block's kinetic energy before contact equals the spring energy at maximum compression.
Updated On: Oct 1, 2026
- zero
- \(\sqrt{\text{km}} x\)
- \(\text{mx}^2/\text{K}\)
- \(\text{Kx}^2/2\text{m}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
On a frictionless surface the total mechanical energy is conserved. All the kinetic energy of the block goes into the spring at maximum compression \(x\).
Step 2: Energy balance:
\[ \frac12mv^2 = \frac12Kx^2 \Rightarrow v = x\sqrt{\frac Km} \]
Step 3: Momentum:
\[ p = mv = m\,x\sqrt{\frac Km} = x\sqrt{Km} \]
This is the largest momentum the block carries, since its speed only falls once the spring is compressed.
Step 4: Check the options:
Option (B) \(\sqrt{km}\,x\) matches. Option (A) is the momentum at the instant of maximum compression. Options (C) and (D) have units that do not match momentum: \(mx^2/K\) and \(Kx^2/2m\) are not kg m/s.
Final Answer:
Momentum is m times x times root of K over m. \[ \boxed{\text{(B) }\sqrt{Km}\,x} \]
On a frictionless surface the total mechanical energy is conserved. All the kinetic energy of the block goes into the spring at maximum compression \(x\).
Step 2: Energy balance:
\[ \frac12mv^2 = \frac12Kx^2 \Rightarrow v = x\sqrt{\frac Km} \]
Step 3: Momentum:
\[ p = mv = m\,x\sqrt{\frac Km} = x\sqrt{Km} \]
This is the largest momentum the block carries, since its speed only falls once the spring is compressed.
Step 4: Check the options:
Option (B) \(\sqrt{km}\,x\) matches. Option (A) is the momentum at the instant of maximum compression. Options (C) and (D) have units that do not match momentum: \(mx^2/K\) and \(Kx^2/2m\) are not kg m/s.
Final Answer:
Momentum is m times x times root of K over m. \[ \boxed{\text{(B) }\sqrt{Km}\,x} \]
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