A block with mass M attached to a horizontal spring with force constant k is moving with simple harmonic motion having amplitude A1. At the instant when the block passes through its equilibrium position, a lump of putty with mass m is dropped vertically onto the block from a very small height and sticks to it. Part APart complete What should be the value of the putty mass m so that the amplitude after the collision is one-half the original amplitude? Express your answer in terms of the variables M, A1, and k. m = 3M Previous Answers Correct Part B For this value of m, what fraction of the original mechanical energy is converted into heat? Express your answer in terms of the variables M, A1, and k.

Respuesta :

Answer:

Explanation:

Given

Mass of block is [tex]M[/tex]

spring constant [tex]=k[/tex]

Amplitude is [tex]A_1[/tex]

when putty is placed then amplitude decreases to [tex]\frac{A_1}{2}[/tex]

Initially [tex]\frac{1}{2}kA^2=\frac{1}{2}Mv^2\quad \ldots(i)[/tex]

Conserving momentum

[tex]Mv_o=(m+M)v[/tex]

where [tex]v_o[/tex]=initial velocity

[tex]v=\frac{M}{M+m}v_o[/tex]

Now

[tex]\frac{1}{2}k(\frac{A_1}{2})^2=\frac{1}{2}(M+m)v^2[/tex]

[tex]\frac{1}{2}k(\frac{A_1}{2})^2=\frac{1}{2}(M+m)(\frac{M}{M+m}v_o)^2\quad \ldots(ii)[/tex]

divide (i) and (ii) we get

[tex]\frac{4}{1}=\frac{M}{M+m}\times (\frac{m+M}{m})^2[/tex]

[tex]4=\frac{m+M}{M}[/tex]

[tex]m=3M[/tex]

Fraction of energy converted into heat[tex]=\frac{1}{2}kA_1^2-\frac{1}{2}k(\frac{A_1}{2})^2[/tex]

[tex]=\frac{1}{2}kA_1^2[1-\frac{1}{4}][/tex]

[tex]=\frac{1}{2}kA_1^2[0.75][/tex]

So, [tex]\frac{3}{4}[/tex] fraction is converted into heat energy