What Is the Final Velocity of the Wood After Bullets Impact?

[omeromer]

Homework Statement

A piece of wood mass of 10 kilograms Placed on a smooth surface.
15 bullets were fired towards the piece and stabilized in .
The mass of one bullet is 40 grams and its velocity is 1000 meters in second.
Find the velocity of the piece of wood after the 15 bullets were stabilized in .

Homework Equations

The Attempt at a Solution

 

[maniacp08]

Although I am new, even I know this will not be answered because no one is going to do the homework for you.

 

[omeromer]

Thank you mania,

Actually this is not a homework question, and I am not studying a physics course. I am mathematician not physician. so why did I post the question ? The answer is that i was asked to convent my sisters daughter in the solution of her teacher.

Her teacher answered the problem as follows


Linear Momentum(before) = Linear Momentum(after)
10 X zero = (10 + 0.6 X 1000) X V where V is the wanted velocity



My sisters daughter wants the solution to be as follows

Linear Momentum(before) = Linear Momentum(after)
10 X zero + 0.6 X 1000 = (10 + 0.6 X 1000) X V, (V still the wanted velocity)


She is claiming that the Linear Momentum(before) depinds on the motion of the bullets as it depends on the motion of the wood piece


What I am suugesting is


Linear Momentum(initial) = Linear Momentum(after the 1st bullet stabilized in the wood piece ) = Linear Momentum(after the 2nd bullet stabilized in the wood piece ) = ......= Linear Momentum(after the 15th bullet stabilized in the wood piece )



This yeils to the following

10 X zero + 0.040 X 1000 = (10 + 0.040 ) X V_1 = (10+ 0.040 + 0.040) X V_2 = ......= (10 + 15 X 0.040) X V_15


(V_n is the velocity after the nth bullet stabilized in the wood piece , n= 1,2,...,15 )

Note that v_15 is the wanted velocity


So V = 40 / 10.6 meters /second


Is this true ? or which is?
Thank you all

 

[LowlyPion]

Thank you mania,

Actually this is not a homework question, and I am not studying a physics course. I am mathematician not physician. so why did I post the question ? The answer is that i was asked to convent my sisters daughter in the solution of her teacher.

Her teacher answered the problem as follows Linear Momentum(before) = Linear Momentum(after)
10 X zero = (10 + 0.6 X 1000) X V where V is the wanted velocity
My sisters daughter wants the solution to be as follows

Linear Momentum(before) = Linear Momentum(after)
10 X zero + 0.6 X 1000 = (10 + 0.6 X 1000) X V, (V still the wanted velocity)She is claiming that the Linear Momentum(before) depinds on the motion of the bullets as it depends on the motion of the wood pieceWhat I am suugesting is Linear Momentum(initial) = Linear Momentum(after the 1st bullet stabilized in the wood piece ) = Linear Momentum(after the 2nd bullet stabilized in the wood piece ) = ......= Linear Momentum(after the 15th bullet stabilized in the wood piece )
This yeils to the following

10 X zero + 0.040 X 1000 = (10 + 0.040 ) X V_1 = (10+ 0.040 + 0.040) X V_2 = ......= (10 + 15 X 0.040) X V_15 (V_n is the velocity after the nth bullet stabilized in the wood piece , n= 1,2,...,15 )

Note that v_15 is the wanted velocitySo V = 40 / 10.6 meters /second


Is this true ? or which is?
Thank you all

Welcome to PF.

Actually the problem doesn't state as it is worded whether they are fired serially or simultaneously.

But happily it doesn't really matter. 15*bullet mass * speed = (mass of block+ 15 bullets)*final speed.

After the first bullet, if you were doing it serially, you have the momentum of the first in the block and it would be moving forward subsequently with the momentum imparted by the first and it would then add to the momentum of the second to become the new momentum of block and 2 bullets. And so on. And so on to the 15th that would be insinuating itself into a block that already had the momentum of the first 14.

10 X zero + 0.6 X 1000 = (10 + 0.6 X 1000) X V

Should be
mb*vb = (B+m)*V
.6 *1000 = 10.6 *V
V= 600/10.6