Question

Two blocks A and B are attached to the two ends of a spring having length L and force constant k on a horizontal surface. Initially the system is in equilibrium. Now a third block having same mass m, moving with velocity v collides with block A. In this situation ……….

(A) During maximum contraction of the spring, the kinetic energy of the system A–B will be zero. 

(B) During maximum contraction of the spring, the kinetic energy of the system A–B will be mV²/4 

(C) Maximum contraction of the spring is V√(m/k) 

(D) Maximum contraction of the spring is V√(2m/k)

Answer 1

The correct option (B) During maximum contraction of the spring, the kinetic energy of the system A–B will be mV²/4

Explanation:

initially system is in equilibrium. Spring is mass less.

when C collides with A, both A & B will gain equal momentum.

also A & B have equal mass, both will have same velocity.

Let this velocity be u.

Hence by conservation of momentum

mV   = mu + mu

∴ u = (V/2)   (1)

Let compression produced in the spring is x.

by law of energy conservation,

(1/2) mV²  = (1/2) mu² + (1/2) mu² + (1/2) kx²

∴ V² = 2u² + (kx²/m)

∴ V²    = {2V² / 4) + (kx²/m) --- from (1)

 ∴ (kx²/m) = (2V²/4)

∴ x  = V √(m/2K)   (2)

Now block A & B will have equal kinetic energy.

∴ (1/2) Kx²  = (1/2) mu² + (1/2) mu²

∴ (1/2) Kx²  = mu²

∴ from (2)

(1/2) K × V² × (m/2K) = mu²

∴ {mV²/4} = mu²

∴ During maximum contraction, Kinetic energy of system A – B is mu² = (mV²/4)