Question

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.

Answer 1

Answer:

Explanation:

Given

Mass of block is $M$

spring constant $=k$

Amplitude is $A_1$

when putty is placed then amplitude decreases to $\frac{A_1}{2}$

Initially $\frac{1}{2}kA^2=\frac{1}{2}Mv^2\quad \ldots(i)$

Conserving momentum

$Mv_o=(m+M)v$

where $v_o$=initial velocity

$v=\frac{M}{M+m}v_o$

Now

$\frac{1}{2}k(\frac{A_1}{2})^2=\frac{1}{2}(M+m)v^2$

$\frac{1}{2}k(\frac{A_1}{2})^2=\frac{1}{2}(M+m)(\frac{M}{M+m}v_o)^2\quad \ldots(ii)$

divide (i) and (ii) we get

$\frac{4}{1}=\frac{M}{M+m}\times (\frac{m+M}{m})^2$

$4=\frac{m+M}{M}$

$m=3M$

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

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

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

So, $\frac{3}{4}$ fraction is converted into heat energy