A 1100- kg car collides with a 1300- kg car that was initially at rest at the origin of an x-y coordinate system. After the collision, the lighter car moves at 20.0 km/h in a direction of 30 o with respect to the positive x axis. The heavier car moves at 23 km/h at -46 o with respect to the positive x axis. What was the initial speed of the lighter car (in km/h)

37.45 km/hr

Explanation:

To solve this, we use the law of conservation of momentum in two directions (x, and y).

in x direction

1100 * v * cosθ = 1100 * 20 * Cos30 + 1300 * 23 * cos46

1100 * vcosθ = 22000 * 0.866 + 29900 * 0.695

1100 * vcosθ = 19052 + 20780.5

1100 * vcosθ = 39832.5

vcosθ = 39832.5 / 1100

vcosθ = 36.21

In the y direction

1100 * v * Sinθ = 1100 * 20 sin30 - 1300 * 23 sin46

1100 * vsinθ = 22000 * 0.5 - 29900 * 0.719

1100 * vsinθ = 11000 - 21498.1

1100 * vsinθ = -10498.1

vsinθ = -10498.1 / 1100

vsinθ = -9.54

Since we are looking for v, then

v² = vcos²θ + vsin²θ

v² = 36.21² + (-9.54²)

v² = 1311.16 + 91.01

v² = 1402.17

v = √1402.17

v = 37.45 km/hr

Thus, the initial speed of the lighter car is 37.45 km/hr

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.