Questions

Two identical blocks A and B, each of mass m resting on smooth floor are connected by a light spring of natural length L and spring constant k with the spring at its natural length. A third identical block C (mass m) moving with a speed v along the line joining A and B collides with A elastically , the maximum compression in the spring is

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a

v m2k

b

m v2k

c

mvk

d

mv2k

detailed solution

Correct option is A

After striking with A, the block C comes to rest and block A moves with velocity v. When compression in spring ismaximum, both A and B will be moving with common velocity v.From law of conservation of linear momentum, mv=(m+m)V⇒V=v2 …(i)From law of conservation of energy, KE of block C = KE of system + PE of system 12mv2=12(2m)v2+12kx2⇒ 12mv2=12(2m)v22+12kx2 [from Eq. (i)]⇒ kx2=12mv2⇒x=vm2k

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Three particles each of mass m are located at vertices of an equilateral triangle ABC. They start moving with equal speeds each along the meridian of the triangle and collide at its center G. If after collisions A comes to rest and B returns its path along GB, then C

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