Two identical masses, m, can glide along a frictionless surface, as shown above.

Question

Two identical masses, m, can glide along a frictionless surface, as shown above. The left-hand mass is pushed against a spring-plunger with spring constant k, and the plunger is compressed through a distance 2d. Conversely, the right-hand mass is pushed against a spring-plunger with stronger spring constant 2k, but it is only compressed through a shorter distance d. Both masses are released from rest at the same time. When they eventually collide, they stick together. For both parts of this question, Show all steps of your derivation. Simplify, each final answer to its most compact algebraic form. Find an expression for the final velocity of the stuck-together pair of masses after collision. Specify whether it is to the right or to the left. Express your final answer ONLY in terms of m, k, d, and any necessary mathematical constants. Find the fraction of initial mechanical energy that is lost in the collision. You may give your final answer as a simplified, pure mathematical expression OR as a decimal value (expressed to 3 sig.figs.) OR as a percentage (expressed to 3 sig.figs.).

Explanation / Answer

for left mass 1/2mV^2=1/2k(2d)^2

V=2d sqrt(k/m)

similarly second mass

v=-d sqrt(2k/m) -ve for opposite direction

let u be combined velocity

conservation of momentum

2mu=mV+mv

2u=V+v

=2d sqrt(k/m)-d sqrt (2k/m)

u =d/2 sqrt(k/m)(2-sqrt2) towards right

lost energy =1/2mV^2 +1/2 mv^2 -(1/2) 2m u^2

=(1/2)m (4d^2 k/m+d^2 *2k /m-2d^2 k/4m * (4+2-4sqrt2))

=.5md^2 k/m(4+2-4-2+4sqrt2)

=2d^2k( sqrt2)

initial energy=1/2 k(2d)^2 +1/2 *2k d^2

=3kd^2

lost fraction=2sqrt2 /3=.94

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