Q103 / 150 · MHT CET 2026 (16 April Shift 1) · Physics
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
- Azero
- Correct answerB$\sqrt{\text{km}}\ x$
- C$\text{mx}^2/\text{K}$
- D$\text{Kx}^2/2\text{m}$
Solution
By the principle of conservation of mechanical energy, the maximum kinetic energy of the block is equal to the maximum potential energy stored in the spring at maximum compression.
$\dfrac{p^2}{2m} = \dfrac{1}{2} K x^2$
$p^2 = m K x^2$
$p = \sqrt{m K} x$
The maximum momentum of the block after the collision is equal in magnitude to its initial momentum, which is $\sqrt{m K} x$.
Answer: $\sqrt{\text{km}}\ x$
$\dfrac{p^2}{2m} = \dfrac{1}{2} K x^2$
$p^2 = m K x^2$
$p = \sqrt{m K} x$
The maximum momentum of the block after the collision is equal in magnitude to its initial momentum, which is $\sqrt{m K} x$.
Answer: $\sqrt{\text{km}}\ x$
Hence, the correct answer is option: 'B'.