A0.500-kilogram cart traveling to the right on a horizontal, frictionless surface at 2.20 meters per second collides head on with a 0.800-kilogram cart moving to the left at 1.10 meters per second. what is the magnitude ofthe total momentum of the two-cart system after the collision?

Consider the motion towards right as positive and motion towards left as negative.

m₁ = mass of the cart moving to right = 0.500 kg

v₁ = initial velocity before collision of the cart moving towards right = 2.2 m/s

m₂ = mass of cart moving to left = 0.800 kg

v₂ = initial velocity before collision of the cart moving towards left = - 1.1 m/s

initial momentum of the system of carts before the collision is given as

P₁ = m₁ v₁ + m₂ v₂

P₁ = (0.500) (2.2) + (0.800) (- 1.1)

P₁ = 0.22 kgm/s

P₂ = momentum of system of carts after collision

As per conservation of momentum,

Momentum of system of carts after collision = Momentum of system of carts before collision

P₂ = P₁

P₂ = 0.22 kgm/s

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over