Momentum of a Child and Bike: What is the Total and Individual Momentum?

[Ronaldo21]

1. (part 1 of 3) A 25 kg child is riding a 6.4 kg bike with a velocity of 4.2 m/s to the northwest. a)What is the total momentum of the child and the bike together?
Answer in units of kg · m/s.

2. (Part 2 of 3) b) What is the momentum of the child? Answer in units of kg · m/s.

3. (part 3 of 3) c) What is the momentum of the bike? Answer in units of kg · m/s.

THANK YOU!

 

[hitmeoff]

ok, like the last question: p = mv, but we are talking about a system here. so we have m of the bike and the m of the boy so

p = (m_boy + m_bike)*v

p = (25+6.4)*4.2 = 31.4*4.2 = about 132Ns

Now if you were to calculate just the boy, and have him still travel at the same speed the its just 25*4.2, just the bike at the same speed its 6.4*4.2.

If you add both these momentums together, they should equal the momentum of the system we found in part 1

 

[Ronaldo21]

thank you soo much!

and what about this one, would you do the same??

Two 1 kg balls move away from each other, one traveling 8 m/s to the right, the other
9 m/s to the left.What is the magnitude of the total momen-
tum of the system?
Answer in units of kg · m/s.

i did 1 times 8 + 1 times -9 and got -1. is that correct?

 

[hitmeoff]

Yes, though Id state your answer as 1kgm/s to the left.

Its just me, but I generally like to state things in physics in terms of direction instead of negatives and positives.

 

[Ronaldo21]

and do you know how to find the magnitude of the normal force
exerted on the block by the plane.

because it says "Assume you are on a planet similar to Earth where the acceleration of gravity g ≈10 m/s2 . A plane 39 m in length is inclined at an angle of 22.6 ◦ as shown. A block of weight 390 N is placed at the top of the plane and allowed to slide down.

and on part one it said find the mass so i just did 390/10 and got 39.

 

[hitmeoff]

so normal force is the force exerted on the object by the plane, and it is perpendicular to the surface of the plane.

So if the plane is how I imagine it and the block is at an angle of 22.6 degrees, then the Normal force is equal to the y component of the weight of the block. I should say that by saying "y-component" I don't mean the component that goes straight down, that's just simply equal to the weight and it would be equal to the normal IF the block was horizontal.

But the "y-component" is actually equal to mg*cos 22.6 degs. I wish I had a scanner handy so I could show you how that works out.

Basically, if you "twist" your coordinates so that Normal point directly up, the weight vector gets "rotated" backwards 22.6 degs. So now the weight vector points 22.6 degrees down and to the left. So the y-component of this vector is mg x cos 22.6

 

[hitmeoff]

If you want to know why we would even want to bother with "rotating" or "twisting" axis' its because you want the y and/or the x-axis to be along the plane of motion. The math works out a lot nicer if you try to orient as much as you can along the plane of motion.

 

[Ronaldo21]

that makes perfect sense, your really good at this!

okay so if i want to calculate the work done on the block by the gravitational force during the 39 m slide down the plane in J, would i just do (mg*cos(22.6)) which is 352.850945 times 39 since work =fd??

 

[hitmeoff]

that makes perfect sense, your really good at this!

okay so if i want to calculate the work done on the block by the gravitational force during the 39 m slide down the plane in J, would i just do (mg*cos(22.6)) which is 352.850945 times 39 since work =fd??

Actually, work would be change in kinetic energy. SO at the top of the slide there is no kinetic energy, but the potential energy would be mgh. Total Energy = Kinetic + Potential. Since energy is conserved, right before the moment of impact there is no more potential energy, but now you have all kinetic energy. Again since energy is conserved, in this case, KE = mgh

So $$\Delta$$KE = KE_f - KE_i = mgh - 0

h is the height of the ball, which is 39m x SIN 22.6

 

[hitmeoff]

remember this is work done by gravity. So any motion in the x-component is not work done by gravity. Gravity (in the case of block-planet) is in the vertical only.

W does equal F*d but the d in this case is h.

 

[Ronaldo21]

ohhh! i see!

and for this one "A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant force of 44 N. How far must the student be pushed, starting from rest, so that her final kinetic energy is 350 J ?
Answer in units of m."

i did KE=1/2mv^2 and i did 44/9.8 i get the mass, then i just solved for v. is that correct?

 

[Ronaldo21]

nevermind, i understand! thank you sooo much!