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NEET PHYSICSWORK, ENERGY AND POWERMedium

Question

A mass of 0.5 kg0.5 \text{ kg} moving with a speed of 1.5 m/s1.5 \text{ m/s} on a horizontal smooth surface, collides with a nearly weightless spring of force constant k=50 N/mk = 50 \text{ N/m}. The maximum compression of the spring would be:

A

0.15 m0.15 \text{ m}

B

0.12 m0.12 \text{ m}

C

1.5 m1.5 \text{ m}

D

0.5 m0.5 \text{ m}

Step-by-Step Solution

According to the law of conservation of mechanical energy, the kinetic energy of the moving mass is completely converted into the elastic potential energy of the spring at the point of maximum compression . Therefore, Kf+Vf=Ki+ViK_f + V_f = K_i + V_i 0+12kxm2=12mv2+00 + \frac{1}{2}kx_m^2 = \frac{1}{2}mv^2 + 0 12kxm2=12mv2\frac{1}{2}kx_m^2 = \frac{1}{2}mv^2 Where m=0.5 kgm = 0.5 \text{ kg}, v=1.5 m/sv = 1.5 \text{ m/s}, and k=50 N/mk = 50 \text{ N/m}. Solving for xmx_m (maximum compression): xm2=mkv2x_m^2 = \frac{m}{k}v^2 xm2=0.550×(1.5)2x_m^2 = \frac{0.5}{50} \times (1.5)^2 xm2=1100×2.25=0.0225x_m^2 = \frac{1}{100} \times 2.25 = 0.0225 xm=0.0225=0.15 mx_m = \sqrt{0.0225} = 0.15 \text{ m} Thus, the maximum compression of the spring is 0.15 m0.15 \text{ m}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWORK, ENERGY AND POWERmovinghorizontalsmoothsurfacecollides

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