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## Homework Statement

A block (m1 = 2.5 kg) sits at rest on a horizontal frictionless surface, connected to an unstretched spring (k = 190 N/m) whose other end is fixed to a wall. Another block (m2 = 1.0 kg) whose speed is 4.0 m/s collides head-on with it.

http://www.webassign.net/hrw/10_44alt.gif

a.) If the two blocks stick together after the collision,

of the initial mechanical energy of the (m1, m2 & spring) system:

how much is lost to the environment during the collision? _____ %

how much is stored in the spring at maximum compression? _____ %

what is that maximum compression of the spring (when the blocks momentarily stop)?

0.1551 m (i got this one right)

b.) If the two blocks instead collide elastically,

what will be the speed and direction of each block immediately after the collision?

block 1: _____m/s,

block 2: _____m/s, left( travel to the left is the right answer)

of the initial mechanical energy of the system:

_____how much is lost to the environment during the collision? %

_____how much is stored in the spring at maximum compression? %

(Where is the rest of the energy?)

what will be the maximum compression of the spring afterwards?

_____m ( i put 0meters since this is elastic collision, because the spring will compressed and return to its initial state, but the answer is wrong)

## Homework Equations

This is about momentum and conservation of energy

formula:

(i=initial. f=final)

momentum conservation

pi=pf

(m1v1)i+(m2v2)i=(m1v1)f+(m2v2)f

energy conservation (KE=kinetic energy, PE=potential energy)

Ei=Ef

KEi+PEi=KEf+PEf

(1/2mv^2)i+(mgh)i=(1/2mv^2)f+(mgh)f

## The Attempt at a Solution

heres how i find the maximum compression of the spring

first i use momentum conservation equation

(m1v1)i+(m2v2)i=(m1v1)f+(m2v2)f

(2.5*0)+(1*4)=(2.5*vf)+(1*vf)

4=(2.5+1)vf

4/(2.5+1)=vf=1.1428m/s

now i need to find the maximum distance the spring compressed

so i already know the velocity of (m1+m2) when they start compress the spring.

i set up the equation as

KEblock=KEspring

1/2mv^2=1/2kx^2 (where k is spring constant of 190N, x is distance)

we are looking for x, so i rearrange the equation

((1/2mv^2)*(2/k))^(1/2)=x

((mv^2)/k)^(1/2)=x

((3.5*1.1428^2)/190)^(1.2)= x =0.1551m

how much is lost to the environment during the collision? _____ % (inelastic)

i didnt get this one right, but this is how i do

KEi=KEf

KE(block2)=KE(block1+2)+PEspring

(1/2m2v^2)=(1/2(m1+m2)v^2)+(1/2kx^2)

(1/2*1*4*2)=(1/2(2.5+1)*1.1428^2)+(1/2(190)*0.1551^2)

8=2.28548+2.2843

8=4.5707

to find percent lost

((KEfinal-KEinitial)/KEinitial)*100)=((4.5707-8)/8)*100)=42.86% ( either this or -42.86% is wrong)

I have tried 2-3 times on some problems but i cant get it right. If anyone can explain and show me what equation should i use for all of them, i am really grateful because i am a person that learn through visual not listen. Again, thank you for your help and i really appreciate it