Inelastic Collision, only masses known

[BATBLady]

Inelastic Collision, only masses known need help

Homework Statement

A projectile (mass = 0.28 kg) is fired at and embeds itself in a target (mass = 2.35 kg). The target (with the projectile in it) flies off after being struck. What percentage of the projectile's incident kinetic energy does the target (with the projectile in it) carry off after being struck?

Homework Equations

m₁V1_i+m₂V2_i=m₁V1_f+m₂V2_f

The Attempt at a Solution

I just don't know how to solve without velocities. I know the KE+PE must equal the final KE+PE. No idea how to get percentages with this information.

 

[Cyosis]

Kinetic energy isn't conserved in an inelastic collision and there isn't really any potential energy to speak of.

What you do know is that the kinetic energy of the bullet is 1/2 mv^2 and the kinetic energy of the block+bullet 1/2 M u^2. You can solve u as a function of v through conservation of momentum. From that you can calculate the percentage.

So the first step is to write down the equation for conservation of momentum and solve for u.

 

[BATBLady]

Am I assuming the KE is 0 or that it is equal to the KE of just the bullet? If KE=0 then u=1.1467. If not,

u= sqrt(KEb+B/(.5*(.28+2.35)))

 

[Cyosis]

You can't calculate the velocities explicitly. Also Kinetic energy isn't conserved, the kinetic energy of the bullet will not be equal to the kinetic energy of the bullet+block after impact so you cannot use that equation. Momentum however is conserved so try to answer these two questions.

What is the momentum before the impact in terms of the mass of the bullet and velocity of the bullet?

What is the momentum of the bullet+block after the collision in terms of mass of the bullet, mass of the block and speed of the composite system?

 

[BATBLady]

Ok, so conservation of momentum would mean that
0.28V= before
(2.35+0.28)Vf= after
thus 0.28V should = (2.35+0.28)Vf

 

[Cyosis]

Yep that is correct. Now solve that equation for Vf. You can then plug vf into the equation for kinetic energy. This way you will have both kinetic energies, for bullet and composite as a function of v. Calculating the percentage will then cause v to drop out.

 

[BATBLady]

So, if I'm doing this right

0.5*0.28V=0.28/2.63
V=0.76

plugged into the original KE, KE=0.081

Do I then plug that number in for KEb+B then use that number to compute the percentage of the original KE that the larger mass has?

 

[Cyosis]

You cannot calculate v nor is it necessary to do so. Two questions need to be answered.

1) What is the kinetic energy of the bullet as a function of its mass and v
2) What is the kinetic energy of the bullet+block composite as a function of the mass of the bullet the mass of the block and the speed vf.

 

[BATBLady]

KEb=0.14V^2
KEb+B= 1.315Vf^2

the answer for the problem should be (KEb+B/KEb)*100 the V's cannot cancel because they are not equal. If they were, it'd be over 100% of the KEb. I took the liberty of multiplying out the 0.5

 

[Cyosis]

The v's are not equal that is correct, but you have already calculated vf as a function of v. This will make the "v"s equal.

[quote='BATBLady]Ok, so conservation of momentum would mean that
0.28V= before
(2.35+0.28)Vf= after
thus 0.28V should = (2.35+0.28)Vf
[/quote]

 

[BATBLady]

plugging into equation:

0.5(0.28)V^2=0.28V/2.36 ...V's cancel

0.14^2 (does not)= 0.28/2.36

0.0196 (does not)= 0.1186

That's where I'm at. I think I'm putting the equations into the wrong things...but I'm not sure. Comparing these give over 100% of the KE from the original.

 

[BATBLady]

I'm sorry this is taking so long. I missed two lectures because I was sick, which is probably why I'm having this issue.

 

[Cyosis]

0.5(0.28)V^2=0.28V/2.36 ...V's cancel

These v's don't cancel there is a square on one side and a first power on the other side. This means you will be left with one v.

Use the momentum equation and the momentum equation alone to calculate vf as a function of v.

Edit: There is no need to apologize we will get to the answer!

 

[BATBLady]

Ok, that's

0.28V/2.63=Vf

 

[Cyosis]

KEb=0.14V^2
KEb+B= 1.315Vf^2

the answer for the problem should be (KEb+B/KEb)*100 the V's cannot cancel because they are not equal. If they were, it'd be over 100% of the KEb. I took the liberty of multiplying out the 0.5

That is correct now plug it into the kEb+B equation.

 

[BATBLady]

=0.14V

which is the 0.5(0.28)V, excluding a ^2

 

[Cyosis]

I am not sure what this excluding a square means.

We have KEb+B=1.1315 vf^2 and vf=0.28 v/2.63, combining these yields KEb+b=1.1315 (0.28v/2.63)^2. Now you have a v^2 in both kinetic energy equations. Can you see how to proceed from here?