Does an Inelastic Collision Affect Conservation of Mechanical Energy?

[eagles12]

Homework Statement

a .440 kg block of wood hangs from he ceiling by a string, and a 7.3x10^-2 kg wad of putty is thrown straight upward, striking the bottom of the block with a speed of 5.50 m/s. The wad of putty sticks to the block.

is the mechanical energy of this system conserved?
how high does the putty-block system rise above the original position of the block?

Homework Equations

pi=pf
E1=E2

The Attempt at a Solution

yes the mechanical energy is conserved because it is an inelastic collision

pi=pf
mvi=(m1+m2)v
v=mvi/m1+m2
v=(7.3x10^-2)(5.5)/(7.3x10^-2)+.440
v=.4015/.513
v=.7827

E1=E2
K1U1=K2U2
1/2 (m1+m2)v^2=1/2 Kx^2
.2565(.7827^2)=1/2 kx^2

is this right so far?

 

[HallsofIvy]

No, energy is conserved in a perfectly elastic collision, not an inelastic collision.

Use conservation of momentum instead.

 

[emailanmol]

Hey,


You can think of the putty acting like the way bullet does.

The putty(like the bullet) collides and gets stuck on the block.
During the collision the forces acting on block are the force of tension, gravity and the force exerted by putty,(lets call it N).

(NOTE:It's a STRING.Not a SPRING. You are thinking of it as a SPRING).

The forces acting on putty is the force of gravity and the reaction force exerted by block (-N).

Strictly speaking external forces of tension and gravity act on system (block +putty) so momentum is not conserved. However, the impulse caused by these forces is v small in comparison to the impulse caused by N, so from practical point of view momentum is nearly conserved .(The difference in result is of negligible order).

Now, the putty gets stuck on block.What is the new velocity of system containing block and putty.

Was this collision elastic, partially elastic or completely inelastic?

What happens to mechanical energy after this type of collision ? Does it increase, decrease or remain constant?
(Your view that energy is conserved in inelastic collision is wrong.It's the other way round as the above post suggests)


What is the the new kientic energy of the system?

Remember, as soon as the block amd putty start moving upwards (after the collision) tension in string vanishes (WHY?)

So you can conserve mechanical energy during the course of journey after the collision.(Why?)


This will help you calculate the maximum height attained.

Hope this helps.

 

[eagles12]

the system was completely inelastic but i don't understand what you mean by the spring thing. I though i was using conservation of momentum.

 

[emailanmol]

You have used the conservation of momentum correctly. :-)


Now energy before collision is not equal to energy after collision since its an inelastic collision .(Right?)

The new energy of system after collision is (M+m)V^2/2 which you have again used correctly.

After the collision, the system starts moving up losing kinetic energy and gaining gravitational potential energy.During this journey energy is conserved as no net external force acts on system (Tension is 0 as STRING is not stretched).


However, in your energy equation you have used the term kX^2/2 which is for the potential energy of a spring

Here, its string. So no such term enters :-).

a .440 kg block of wood hangs from he ceiling by a STRING

-----
E1=E2
K1U1=K2U2
1/2 (m1+m2)v^2=1/2 Kx^2/2

The letters in bold are important

 

[eagles12]

so i use mgH instead of kx^2/2

 

[emailanmol]

Yes. :-)

 

[eagles12]

in mgH m would be the block+the putty correct?

 

[emailanmol]

Yes.

See,

After the collision the block + putty system starts moving up.

Only force which acts on the system is the gravity which acts in the direction opposite to system's velocity thus slowing it down.

The system moves up till its velocity becomes 0.
This is the point where it attains its maximum height.
After this it starts moving down again.

Simce its velovity is 0 at highest point,
all its kinetic energy which it possesed right after the collision,
has been converted into potential energy.

This potential energy change is mgh, where (m is mass of system as entire system is moving up, not just the block.Putty is struck to the block),

h is distance between
the highest point and the point where collision took place.