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NEET PHYSICSWORK, ENERGY AND POWERMedium

Question

A bullet from a gun is fired on a rectangular wooden block with velocity uu. When the bullet travels 24 cm24\text{ cm} through the block along its length horizontally, velocity of bullet becomes u/3u/3. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is:

A

30 cm

B

27 cm

C

24 cm

D

28 cm

Step-by-Step Solution

  1. Identify the Principle: The resistive force assumed constant causes a constant deceleration (aa). We can use the Work-Energy Theorem or the kinematic equation v2u2=2asv^2 - u^2 = 2as [Class 11 Physics, Ch 5, Sec 5.6].
  2. Analyze First Part of Motion:
  • Initial velocity vi=uv_i = u
  • Final velocity vf=u/3v_f = u/3
  • Displacement s1=24 cms_1 = 24\text{ cm} (u/3)2u2=2as1(u/3)^2 - u^2 = 2as_1 u29u2=2a(24)\frac{u^2}{9} - u^2 = 2a(24) 8u29=48a    a=u254-\frac{8u^2}{9} = 48a \implies a = -\frac{u^2}{54}
  1. Analyze Total Motion: Let the total length of the block be LL. The bullet enters with uu and stops (v=0v=0) after traversing distance LL. 02u2=2aL0^2 - u^2 = 2aL u2=2(u254)L-u^2 = 2\left(-\frac{u^2}{54}\right)L 1=2L54    L=27 cm1 = \frac{2L}{54} \implies L = 27\text{ cm}
  2. Alternative Method: Calculate the remaining distance s2s_2 from velocity u/3u/3 to 00. 02(u/3)2=2as20^2 - (u/3)^2 = 2as_2 u29=2(u254)s2    s2=3 cm-\frac{u^2}{9} = 2\left(-\frac{u^2}{54}\right)s_2 \implies s_2 = 3\text{ cm} Total length L=24+3=27 cmL = 24 + 3 = 27\text{ cm}.

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 POWERbulletrectangularwoodenvelocitybullet

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