Few problems I've tried but still had issues on

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In summary, the conversation discusses the velocity of a bird after it swallows a bug and the distance a man can walk on a beam without exceeding the tension in the cable. The bird's weight is also mentioned.
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Nefarious
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#1 : A 300g bird is gliding through the air toward the east at 8m/s. A 10gram bug is buzzing along toward the west at 5ms direction toward the bird. The bird opens his mouth and swallows the bug. What is the velocity of the bird immediately after swallowing the bug?


Using conservation of momentum this is what I did.

300g * 8m/s = 2400
10g * 5m/s = 50

2400-50 = 300 * x

x=7.83m/s

However I'm aware that I'm wrong would love if someone could show me where I went wrong.

#2: A 10 foot beam has a mass of 200lbs. The cable at the end of the beam makes an angle of 40degrees with the horizontal. What distance x from the left end of the beam can a 150lb man walk out onto the beam if the tension in the cable is not to exceed 300lbs.
 
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  • #2
The bird has put on weight.

You realize this should have been posted in the homework forum?
 
  • #3



For the first problem, you are correct in using conservation of momentum to solve it. However, there is a small error in your calculation. The 300g bird should actually be 0.3kg, and the 10g bug should be 0.01kg. This will give you a final answer of 7.83 m/s, as you have calculated.

For the second problem, we can use the concept of torque to solve it. Torque is the product of force and the distance from the pivot point. In this case, the pivot point is the left end of the beam.

First, we need to find the tension in the cable. We can use trigonometry to find the vertical component of the tension: 300lbs * sin(40) = 193.9lbs. This means the horizontal component is 300lbs * cos(40) = 229.6lbs.

Next, we can calculate the torque exerted by the man on the beam. Torque = force * distance. In this case, the force is the weight of the man, which is 150lbs, and the distance is x. So the torque is 150lbs * x.

The total torque exerted on the beam must be balanced, otherwise the beam will rotate. So we can set up an equation: 193.9lbs * (10ft-x) = 150lbs * x. This is because the torque exerted by the tension in the cable must be equal and opposite to the torque exerted by the man.

Solving for x, we get x = 3.89ft. So the man can walk out 3.89ft from the left end of the beam before the tension in the cable exceeds 300lbs.
 

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