Sailing with a Trolley and a Bag of Sugar

  • Thread starter Thread starter kateman
  • Start date Start date
  • Tags Tags
    Flat
AI Thread Summary
Mieko, a girl with a trolley and a bag of sugar, throws the bag forward while coasting at 0.5 m/s. The momentum equations reveal that the initial momentum is 11.5 kg·m/s, and the bag's final velocity is 4.5 m/s after being thrown. The confusion arises from calculating the final velocity of Mieko and the trolley, as initial misinterpretations lead to incorrect values. Ultimately, the correct final speed of Mieko after throwing the bag is approximately 0.1 m/s backward. The discussion highlights the importance of careful reading and understanding of momentum principles in physics.
kateman
Messages
113
Reaction score
0

Homework Statement


While coasting along a street at .5m/s, Mieko, a 15kg girl in a 5kg trolly, sees a vicious dog in front of her. She has with her only a 3kg bag of sugar which she is bringing home from the store, and throws it with forward velocity of 4m/s relative to her original motion. How fast is she moving after she throws the bag of sugar? whit is her final direction of motion?


Homework Equations


p = mv


The Attempt at a Solution


right so i did p=mv of the trolley (worked out to be 11.5) and then of the bag being thrown (worked out to be 12)

the next step confused me. if i divide 11.5 by 12, i get the right answer of .9582 backwards. but i cannot see how that is mathematically correct.

now if i use the m1v1i + m2v2i = m1v1f + m2v2f formula, i only get a value half of the correct answer!
can someone please explain this to me?
 
Physics news on Phys.org
Are you sure that .9582 is the correct answer? I'm not getting that.

You can consider Mieko and the trolley together as one system of 20kg. so m1 is 20kg. m2 is 3kg.

use:

m1*v_{trolleyinitial} + m2*v_{baginitial} = m1*v_{trolleyfinal} + m2*v_{bagfinal}

You're right that initial total momentum is 11.5. The velocity of the bag is 4m/s relative to the initial motion, so the final velocity of the bag is 4.5m/s.
 
well the correct answer is .1m/s backwards
i just realized what i had done. i misread .9 and though it was .09 so that it could be rounded up to .1

and then i didnt think that the bag would also have the two velocities (of bag and trolley added together)

haha i have to stop doing physics late at night :P

thanks learningphysics, its always helpful to have someone else to see what i have overlooked :)
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Conservation of momentum of astronaut in her space suit
Replies
12
Views
22K
Conservatino of Momentum: Bowling Ball and Earth
Replies
1
Views
6K

Hot Threads

Recent Insights

Back
Top