Momentum,impulse, force question

  • Thread starter Thread starter study earth
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
In an elastic collision between two bumper cars, momentum is conserved, meaning the total momentum before the collision equals the total momentum after. The change in momentum for each car can be calculated by subtracting the initial momentum from the final momentum. For car A, the change in momentum is determined by its final momentum of 270 kg·m/s west minus its initial momentum of 540 kg·m/s east. To calculate impulse, the average force can be derived from the change in momentum divided by the contact time of 2 seconds. Understanding these principles helps clarify the dynamics of the collision and the forces involved.
study earth
Messages
3
Reaction score
0
Hi,

i have a question iv been working on for 4 hours and I am soo confused.

My example is a bumper car in a elastic collision.

bumper car A travels at 3ms-1 east with a mass of 180kg
Bumper car B travels at 4ms-1 west witha mass of 200kgs.
the momentum before the collision is 1340kgms-1..this will be equal to momentum after the collision.

Car A travels at 1.5ms-1 after the collision int he opposite direction west
while Car B travels at 5.35ms-1 to the east.

what i need help with is change in momentum.

what is the change in momentum for both cars.

Δmomentum= F* T ( how am i supposed to get the average force for both cars)
the contact time was 2 secs but what is the change in momentum?

i don't get if its change in momentum for the whole system or just 1 car?

please help me out..

feel free to substitue any values for F or T etc..keep it simple as well (im only 16)

Thank you
 
Physics news on Phys.org
If there is no other force acting on cars A and B during the collision, then the total momentum of the A&B system does not change. This is "conservation of momentum". What may and usually does change is the momentum of each car.
 
hi thanks for that.
yes momentum for car a and b does change..
so can i get change in momentum for car A?
will it be final momentum (270Kgms-1 to west) - Inital momentum ( 540kgms-1 to the east)
 
what i don't get is how to calculate impulse when both Bumper car A and b collide...?
what force do i use?
what change in momentum do i use?
I know the Time?Thank you
 
In a fully elastic collision you have conservation of momentum and you have conservation of energy. Together they are a system of two equations for two unknowns that you then solve. Let ## m_A ## and ## m_B ## be the masses of the cars, and ## v_{iA}, \ v_{iB}, \ v_{fA}, \ v_{fB} ## be their initial and final velocities. Conservation of momentum: $$ m_A v_{iA} + m_B v_{iB} = m_A v_{fA} + m_B v_{fB} $$ Conservation of energy: $$ \frac {m_A v_{iA}^2} {2} + \frac {m_B v_{iB}^2} {2} = \frac {m_A v_{fA}^2} {2} + \frac {m_B v_{fB}^2} {2} $$
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top