A 4-kg plate rests on spring A which is the larger of nested springs. The smalle

Question

A 4-kg plate rests on spring A which is the larger of nested springs. The smaller unstretched spring B is 75 mm below the location of the plate. An 8-kg plunger is released from rest in the position shown to impact the plate; the constant of the outer spring is kA = 5 kN/m and the constant of the inner spring is kB = 8 kN/m. If the plunger is released from the height h = 4 m. The coefficient of restitution between the plunger and the plate is 0.8. Determine:

1_ Maximum compression of each spring

2_ Energy lost on impact

Explanation / Answer

here

V1/V2=e therefore

mgh= 1/2mV1^2

so

V2= (sqrt(2gh))/e

hence max compression of both spring is 1/2m(V2)^2 = 1/2(Ka+Kb)x^2

THEREFORE x= V2 sqrt (m/(ka+kb))

compression =sqrt(2ghm/(ka+kb)

energy lost in collision = 1/2mV2^2 - 1/2m(V1)^2

=1/2m ( (2gh/e)^2 - (2gh)^2)

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