StreakPeaked· Practice

ExamsJEE AdvancedPhysics

Three identical blocks each of mass m lie on a smooth horizontal line. Block 1 moves with speed v and undergoes a perfectly inelastic collision with block 2, which is connected to block 3 by a light spring; block 2 and 3 are initially at rest with the spring relaxed. Find: (i) the velocity of the combined (1+2) just after the collision, (ii) the maximum kinetic energy attained by block 3, (iii) the minimum kinetic energy of block 2 during the motion, and (iv) the maximum compression of the spring (spring constant k).

  1. v1 = v/2; KE3_max = 2mv²/9; KE2_min = mv²/72; x_max = v*sqrt(m/(6k))
  2. v1 = v/3; KE3_max = mv²/2; KE2_min = mv²/8; x_max = v*sqrt(m/k)
  3. v1 = v; KE3_max = mv²/2; KE2_min = mv²/4; x_max = (v/4)*sqrt(m/k)
  4. v1 = v/2; KE3_max = mv²/4; KE2_min = 0; x_max = (v/2)*sqrt(m/k)

Correct answer: v1 = v/2; KE3_max = 2mv²/9; KE2_min = mv²/72; x_max = v*sqrt(m/(6k))

Solution

Just after impact the combined block moves at v/2. Treating the spring interaction as elastic, block 3 reaches maximum speed 2v/3 (max KE 2mv²/9), the combined block slows to v/6 (min KE mv²/72), and maximum compression occurs at the common velocity v/3.

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →