Answer:
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
Answer:
Given
acceleration (a) =1.5ms2
Force(F) =2100N
R. t. c mass (m) =?
Form
F=ma(divided by m both sides)
m=F/a
m=2100/105
m=1400kg
mass of car =1400kg
