Troubleshooting Physics Problems: Understanding Force and Velocity Equations

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SUMMARY

This discussion focuses on troubleshooting physics problems related to force and velocity equations. Key issues include calculating buoyant force, impact force, and the conservation of momentum in collisions. The correct application of vector direction is crucial, particularly in problems involving opposing forces. Participants emphasized the importance of using negative signs for velocities in opposite directions to achieve accurate results.

PREREQUISITES
  • Understanding of buoyant force calculations using Archimedes' principle
  • Familiarity with Newton's second law (F=ma) for impact force calculations
  • Knowledge of conservation of momentum in collision scenarios
  • Basic vector analysis, including the significance of direction in velocity
NEXT STEPS
  • Study Archimedes' principle for buoyant force calculations
  • Learn about Newton's laws of motion, specifically focusing on impact force scenarios
  • Explore conservation of momentum in elastic and inelastic collisions
  • Review vector addition and subtraction, particularly in the context of physics problems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone looking to deepen their understanding of force and velocity equations in mechanics.

vitaly
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I have a few questions that I got wrong, and I'm trying to find out what I did. All help is appreciated.

1. A cube of aluminum has side length .10 m. It is dropped into a deep swimming pool of water withd ensity 1.00x10^3 kg/m^3. The cube comes gently to rest at the bottom of the pool. If the density of aluminum is 2.7x10^3 kg/m^3, the magnitude of the buoyant force acting on the aluminum is ________.
The answer is 9.8 N, but I keep getting 2.7 by multiplying (1.00x10^3) by(2.7x10^3). Is there a certain formula I must use?

2. A cement truck of mass 16,000 kg moving at 15 m/s slams into a cement wall and comes to a halt. What is the force of impact on the truck?
Is this question just asking me for the force? F=ma, so I got F=(16,000 kg)(15 m/s). But, that was wrong on my paper... What exactly is it asking me for? Those are the only two variables stated in there, so what else could I use?

3. A 40-kg footbal player leaps through the air to colide with and tackle a 60-kg player heading toward him, also in the air. If the 40-kg player is heading to the right at 7 .0 m/s, and the 60-kg player is heading toward the left at 3.0 m/s, what is the speed and direction of the tangled players?
The vectors would look like -------> <---. The equation I use was:
(40 kg)(7 m/s) + (60 kg)(3 m/s) = (40 + 60 kg)(final velocity) =
final velocity = 4.6 meters per second. That was counted wrong, and I don't know "what direction" the players are in... How would you know?

And lastly,

4. A 5.0-kg blob of clay moving horizontally at 2.0 m/s has a head-on collision with a 6.0-kg blob of clay that moves toward it at 1.0 m/s. What is the speed of the two blobs stuck together immediately after the collision?
I used the same equation as above and got 1.5 m/s. But, it's also wrong.

Are there two types of equations, one for where one of the forces is not moving, and another for where both forces are moving? Is that what is causing the problems? If so, what is the other equation? Oh, and thanks for all help in advance.
 
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alright. The problem is that velocity is magnitude and direction (a vector) That means the velocity can tell you speed as well as direction.

For example... #3...
(40 kg)(7 m/s) + (60 kg)(-3 m/s) = (40 + 60 kg)(final velocity)
v final=1m/s

in this case, the 60kg football player is heading in the opposite direction of the 40 kg player so it has a negative velocity. Positive velocity usually means going to the right. Negative usually means going to the left. You can apply this to many of your problems.

for #4:
(5*2)+(6*-1)=11x
x=0.63m/s
 
Oh, thanks. I forgot to use the negative sign for force heading in the opposite direction. Thanks again for that help.
 

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