Momentum,impulse, force question

  • Thread starter Thread starter study earth
  • Start date Start date
  • Tags Tags
    Force
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
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: $$ \frac {m_A v_{iA}^2} {2} + \frac {m_B v_{iB}^2} {2} = \frac {m_A v_{fA}^2} {2} + \frac {m_B v_{fB}^2} {2} $$