Question

Two blocks of masses m and M are moving with speeds V1
and V2 (V1 > V2) in the same direction on the frictionless
surface respectively, M being ahead of m.

An ideal spring of force constant k is attached to the backside of M (as shown). The maximum
compression of the spring when the block collides is :
(A) v1 V
Womac
(B) V2
(C) (V1 - V2) ImM
(D) None of above is correct.
V(M+ m)K​

Answer 1

Answer:

Explanation:Check the attachment this must be a correct solution of given problem

Answer 2

applying law of conservation of linear momentum,

as external force, F = 0,

so, initial momentum = final momentum

mv1 + Mv2 = (m + M)v

or, v = (mv1 + Mv2)/(m + M)......(1)

now according to law of conservation of energy,

initial energy of system = final energy of system

or, 1/2 mv1² + 1/2Mv2² = 1/2 (m + M)v² + 1/2 kx²

or mv1² + Mv2² = (m + M)v² + kx²

or, m1² + Mv2² - (m + M)v² = kx²

from equation (1),

or, mv1² + Mv2² - (m + M){(mv1 + Mv2)/(m + M)}² = kx²

or, {(m + M)(mv1² + Mv2²) - (mv1 + Mv2)²}/(m + M) = kx²

or, (mMv2² + mMv1² - 2mMv1v2)/(m + M) = kx²

or, mM(v1² + v2² - 2v1.v2)/(m + M) = x²

or, {mM/(m + M)}(v1 - v2)² = kx²

or, x = (v1 - v2) √{mM/(m + M)k}

hence, maximum compression of the spring when the block collides is (v1 - v2) √{mM/(m + M)k}