Answers Collision derivation problem. If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the initial displacement is 2.25 m, and the change in momentum is 0.58 kg*m/s, how far will it coast back up the ramp before changing directions?

4.8967m

Explanation:

Given the following data;

M = 0.2kg

∆p = 0.58kgm/s

S(i) = 2.25m

Ratio h/w = 12/75

Firstly, we use conservation of momentum to find the velocity

Therefore, ∆p = MV

0.58kgm/s = 0.2V

V = 0.58/2

V = 2.9m/s

Then, we can use the conservation of energy to solve for maximum height the car can go

E(i) = E(f)

1/2mV² = mgh

Mass cancels out

1/2V² = gh

h = 1/2V²/g = V²/2g

h = (2.9)²/2(9.8)

h = 8.41/19.6 = 0.429m

Since we have gotten the heigh, the next thing is to solve for actual slant of the ramp and initial displacement using similar triangles.

h/w = 0.429/x

X = 0.429×75/12

X = 2.6815

Therefore, by Pythagoreans rule

S(ramp) = √2.68125²+0.429²

S(ramp) = 2.64671

Finally, S(t) = S(ramp) + S(i)

= 2.64671+2.25

= 4.8967m

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.