Car 1 approaches an intersection with a momentum of p1i=(11,000 kg*m/s)x + (-370 kg*m/s)y. Car 2 approaches the same intersection with momentum p2i= (4,000 kg*m/s)x + (2470 kg*m/s)y. The cars collide and immediately the momentum of Car2 is p2f= (2100 kg*m/s)y. What is the final momentum of Car 1??
The final momentum of car two is whatever it takes both cars to equal the total momentum of both cars before the collision.
p1i + p2i = p2f + p1f
Solve for pif
A 40-kg cart moving at 120 km/h collides head on with an approaching 50-kg cart, moving at 20 km/h ( in the opposite direction). if the two carts stick together what will be their speed?
To find the final momentum of Car 1 after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Let's denote the final momentum of Car 1 as p1f = (p1fx, p1fy). Since momentum is a vector quantity, we need to find the x- and y-components of the final momentum separately.
Before the collision:
p1i = (11,000 kg*m/s)x + (-370 kg*m/s)y
p2i = (4,000 kg*m/s)x + (2,470 kg*m/s)y
After the collision:
p2f = (0 kg*m/s)x + (2,100 kg*m/s)y (assuming Car 2 is brought to rest in the x-direction)
Using the conservation of momentum, we can write the equation:
p1i + p2i = p1f + p2f
Now, let's substitute the given values into the equation:
(11,000 kg*m/s)x + (-370 kg*m/s)y + (4,000 kg*m/s)x + (2,470 kg*m/s)y = p1fx + p1fy + (0 kg*m/s)x + (2,100 kg*m/s)y
Combining the like terms for the x- and y-components, we get:
15,000 kg*m/sx + 2,100 kg*m/sy = p1fx + p1fy
Therefore, the final momentum of Car 1 is p1f = (15,000 kg*m/s)x + (2,100 kg*m/s)y.