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mmoadi
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Homework Statement
A wooden block with a length of 10 cm and a mass of 1 kg lies on an icy plane. A projectile with mass of 2 g hits the wooden block with velocity of 300 m/s and breaks through its center of gravity. How much are the final velocities of the wooden block and the projectile? While moving through the wooden block, the projectile worked on it with a force of 500 N. This is a frictionless system.
Homework Equations
KE= ½ mv²
p=mv
The Attempt at a Solution
½ m(1)v(1-initial)²= ½ m(1)v(1-final)² + ½m(2)v(2-final)²+F*dAnd
m(1)v(1-initial)= m(1)v(1-final) + m(2)v(2-final)
Correct formula thanks to help from "kuruman"- PF Contributor and Homework Helper Homework Helper Thank you again!
Then I tried to express v(2-final) with the components of the second formula (conservation of the momentum) and insert it into the first formula (KE) but it just got very complicated and weird.
Then I did some more research and find an easier approach of solving my weirdness:
First, I applied:
x= v(1-final)/ / v(1-initial)
y= v(2-final) / v(1-initial)
and
m(2)/m(1)= 1kg/ 0.002kg= 500
F*d= 50J
After I plugged in the above, I obtained:
1. For the conservation of the moment: 1= x + 500y → x= 1- 500y
2. For the KE: 1= x² + 500y² + 50
Now, I plugged x value from the first equation into the second equation and after arranging I got:
0= 20y (250.5y – 1)
We now that velocity cannot be 0, so the correct solution is y= 1/ 250.5
Last step:
y= v(2-final) / v(1-initial) → v(2-final)= y* v(1-initial)= 1.19 m/s
v(2-final)= 1.19 m/s → velocity of the wooden block after the bullet passes through
x= 1- 500y= -0.996
x= v(1-final) / v(1-initial) → v(1-final)= x* v(1-initial)= -298.8 m/s
v(1-final)= 298.8 m/s → velocity of the bullet after it passes through the block
Please tell me that this is correct?!
Thank you!