- #1

PirateFan308

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## Homework Statement

A 0.150 kg frame, when suspended from a coil spring, stretches the spring 0.070m. A 0.200kg lump of putty is dropped from rest onto the frame from a height of 30.0cm

Find the maximum distance the frame moves downward from its initial position

## Homework Equations

SE

_{p}+GE

_{p}+E

_{k}= SE

_{p}+GE

_{p}+E

_{k}

mgh+(0.5)(m)(v)

^{2}+ (1.5)(k)(x)

^{2}= mgh+(0.5)(m)(v)

^{2}+ (1.5)(k)(x)

^{2}

initial momentum = final momentum

## The Attempt at a Solution

To find velocity of the putty just before it hits the frame:

E

_{p}= E

_{k}

mgh = (0.5)mv

^{2}

(0.150)(9.81)(0.30) = (0.5)(0.150)(v)(v)

(v)(v) = 5.886

v = 2.426 m/s

To find the velocity of the putty and the frame after the putty hits:

initial momentum = final momentum

mv + mv = mv + mv

(0.200)(2.426) + 0 = (0.200+0.150)v

v = 1.386 m/s

To find k of the spring:

GE

_{p}= SE

_{p}

mgh = (0.5)kx

^{2}

(0.15)(9.81)(0.07) = (0.5)(k)(0.07)

^{2}

k = 42 N/m

To find the distance the frame moves downward:

mgh+(0.5)(m)(v)

^{2}+ (1.5)(k)(x)

^{2}= mgh+(0.5)(m)(v)

^{2}+ (1.5)(k)(x)

^{2}

(0.35)(9.81)(h-0.07) + (0.5)(0.35)(1.386)

^{2}+ (0.5)(42)(0.070)

^{2}= (0) + (0) + (1.5)(42)(h)

^{2}

(3.4335h - 0.240345) + (0.439) = 21h

^{2}

0 = 21h

^{2}- 3.4335h - 0.199

h = 0.20887 or -0.04537

h - 0.07 = 0.1389 or -0.1153

I figure the height (because of my reference) must be negative, so it must be -0.1153m

Is this correct? I got a bit confused about h on the last equation ...