 # a system of masses m and 2m connected together by a tensionless spring moves with constant v on a frictionless surface. after an elastic collision with a wall what is the maximum compression of the spring observed

10 years ago

Dear Jishal,

Let k be the spring constant and x be the maximum compression.

kinetic energy before collision=potential energy at maximum compression.

mv2+(mv2/2)=(k/2)x2

solve for x

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10 years ago

according to work enrgy theorem ΔK.E is associated with the increase in potential energy which lead to the compression in the spring

according to law of conservation of momentum (m+3m)vfinal=-2mv+mv

thus vfinal=v/3

now since it is elastic collision energy conservation too comes under consern

thus change in kinetic energy is the max potential energy stored in the spring.

so,

ΔK.E= 1/2(kx2)  i.e final K.e-initial K.E =1/2(3mv2)-1/2(3mvfinal2)    ...................>( vfinal = v/3)

so 3mv2/2(1-1/9) = 1/2(kx2)

thus solving for x you ill finally reach the answer x= √2m/√3k

this is the correct answer. ive got it just recently . and i wanted to share it with you 