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Arejang
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[SOLVED] A Ballistic Pendulum
A bullet of mass 0.01 kg moving horizontally strikes a block of wood of mass 1.5 kg which is suspended as a pendulum. The bullet lodges in the wood, and together they swing upward a distance of 0.40 m. What was the velocity of the bullet just before it struck the wooden block? The length of the string is 2 meters.
Momentum Conservation:
m_{a}v_{a1}+m_{b}v_{b1}=(m_{a}+m_{b})*v_{2}
Energy Conservation:
1/2mv^{2}=mgy
Since the block is at rest before the bullet hit, if we use the momentum conservation formula, we only have to deal with the initial speed of the bullet. The resultant formula is
m_{b}v_{b}=(m_{w}+m_{b})*v_{2}
Once the bullet is embedded in the block, it will have a potential energy of zero and a kinetic energy of
K=1/2(m_{b}+m_{w})v_{2}^{2}
and the block with the bullet travels up a height .40m, and comes to a rest. At this point, the block/bullet unit has a kinetic energy of zero, and a potential energy of
U=(m_{b}+m_{w})gy
Using energy conservation we get:
1/2(m_{b}+m_{w})v_{2}^{2}=(m_{b}+m_{w})gy
we can solve for velocity here and get the speed after the bullet hit the block. The masses should cancel out, leaving:
v_{2}=\sqrt{2gy}
We sub in this expression for v back into the first momentum formula, getting:
m_{b}v_{b}=(m_{b}+m_{w})*\sqrt{2gy}
solving for the initial velocity v_{b}, we get v_{b}=(m_{b}+m_{w})/m_{b}*\sqrt{2gy}
At this point I just plugged and chugged, using the given values in the problem and came up with 423 m/s, but it turned out to be the wrong answer. Can anyone help me figure out what I did wrong? Much thanks in advance!
Homework Statement
A bullet of mass 0.01 kg moving horizontally strikes a block of wood of mass 1.5 kg which is suspended as a pendulum. The bullet lodges in the wood, and together they swing upward a distance of 0.40 m. What was the velocity of the bullet just before it struck the wooden block? The length of the string is 2 meters.
Homework Equations
Momentum Conservation:
m_{a}v_{a1}+m_{b}v_{b1}=(m_{a}+m_{b})*v_{2}
Energy Conservation:
1/2mv^{2}=mgy
The Attempt at a Solution
Since the block is at rest before the bullet hit, if we use the momentum conservation formula, we only have to deal with the initial speed of the bullet. The resultant formula is
m_{b}v_{b}=(m_{w}+m_{b})*v_{2}
Once the bullet is embedded in the block, it will have a potential energy of zero and a kinetic energy of
K=1/2(m_{b}+m_{w})v_{2}^{2}
and the block with the bullet travels up a height .40m, and comes to a rest. At this point, the block/bullet unit has a kinetic energy of zero, and a potential energy of
U=(m_{b}+m_{w})gy
Using energy conservation we get:
1/2(m_{b}+m_{w})v_{2}^{2}=(m_{b}+m_{w})gy
we can solve for velocity here and get the speed after the bullet hit the block. The masses should cancel out, leaving:
v_{2}=\sqrt{2gy}
We sub in this expression for v back into the first momentum formula, getting:
m_{b}v_{b}=(m_{b}+m_{w})*\sqrt{2gy}
solving for the initial velocity v_{b}, we get v_{b}=(m_{b}+m_{w})/m_{b}*\sqrt{2gy}
At this point I just plugged and chugged, using the given values in the problem and came up with 423 m/s, but it turned out to be the wrong answer. Can anyone help me figure out what I did wrong? Much thanks in advance!
