Snow Day Blog #2
Conservation of Momentum
By: Lauren Kern
In this blog, I found the conversation of momentum when pushing a box full of rocks into a candle. I put the box of rocks on one end of a table, and the candle about a foot away, and pushed the box towards the candle. Once the box hit the candle, the candle moved a little farther away than its original position, but the box stopped moving. This is how I found the conservation of momentum and the kinetic energy lost during the collision:
Next I needed to find the amount of kinetic energy lost during the collision. To do this I found the kinetic energy before the collision, the kinetic energy after the collision, and then I had to subtract the two to find the total KE lost.
KE BEFORE:
= 1/2mv^2 + 1/2mv^2
= 1/2(1.36)(-2.507)^2 + 1/2(1.36)(0)^2
= (0.68)(6.285) + 0
= 4.27 J
KE AFTER:
= 1/2mv^2 + 1/2mv^2
= 1/2(1.36)(0)^2 + 1/2(1.36)(-0.335)^2
= 0 + (0.68)(0.112)
= 0.076 J
KE LOST:
4.27 J - 0.076 J = 4.19 J
The momentum in this experiment is conserved but not shown. It is conserved because the answers would be equal if there was no loss of friction, but the numbers are not equal since one number is -1.147 and the other number it -0.4556. Due to lost momentum in friction, the numbers are not equal.
In this experiment not only was the conservation of momentum found, but the kinetic energy lost during the collision, 4.19 J, was also found.
Conservation of Momentum
By: Lauren Kern
In this blog, I found the conversation of momentum when pushing a box full of rocks into a candle. I put the box of rocks on one end of a table, and the candle about a foot away, and pushed the box towards the candle. Once the box hit the candle, the candle moved a little farther away than its original position, but the box stopped moving. This is how I found the conservation of momentum and the kinetic energy lost during the collision:
CANDLE (m1) = 1.36kg
BOX OF ROCKS (m2) = 1.36 kg
m1v1 + m2v2 = m1v1f + m2v2f
1.36(-2.507) + 1.36(0) = 1.36(0) + 1.36(-0.335)
1.36(-2.507) = 1.36(-0.335)
-1.147 = -0.4556
In this equation, two of the velocities are zero. One velocity that is zero is the initial velocity of m2, and that is because originally m2 is not moving and it is just sitting on the table, so since it is at rest the velocity is zero. The other zero velocity is the final velocity of m1 and that is because after the collision, m1 stops moving. Since it is no longer moving and is at rest on the table, its final velocity is zero.Next I needed to find the amount of kinetic energy lost during the collision. To do this I found the kinetic energy before the collision, the kinetic energy after the collision, and then I had to subtract the two to find the total KE lost.
KE BEFORE:
= 1/2mv^2 + 1/2mv^2
= 1/2(1.36)(-2.507)^2 + 1/2(1.36)(0)^2
= (0.68)(6.285) + 0
= 4.27 J
KE AFTER:
= 1/2mv^2 + 1/2mv^2
= 1/2(1.36)(0)^2 + 1/2(1.36)(-0.335)^2
= 0 + (0.68)(0.112)
= 0.076 J
KE LOST:
4.27 J - 0.076 J = 4.19 J
The momentum in this experiment is conserved but not shown. It is conserved because the answers would be equal if there was no loss of friction, but the numbers are not equal since one number is -1.147 and the other number it -0.4556. Due to lost momentum in friction, the numbers are not equal.
In this experiment not only was the conservation of momentum found, but the kinetic energy lost during the collision, 4.19 J, was also found.
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