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Answer:
Stunt Car A experiences a large force over a short period of time. Stunt Car B experiences a small force over a long period of time. Because of the force experienced by Stunt Car A, it will sustain more damage than Stunt Car B.
Explanation:
Both cars have the same mass and velocity, therefore they have the same momentum. During the collision, the total momentum of the car A and brick wall is conserved as well as the total momentum of the car B and the pile of leaves.
However, if we are to investigate the damage on each car, we should look at the cars not the whole system. So, the momentum difference between the cars gives us the impulse that the car felt.
Since the Car A will crash the wall quicker than the other car crashes through the pile of leaves,
which gives us
Answer:
distance stop 1.52m,
velocity 4.0 m/s y^
Explanation:
The movement of the particle is two-dimensional since it has acceleration in the x and y axes, the way to solve it is by working each axis independently.
a) At the point where the particle begins to return its velocity must be zero (Vfx = 0)
Vfₓ = V₀ₓ + aₓ t
t = - V₀ₓ/aₓ
t = - 2.4/(-1.9)
t= 1.26 s
At this time the particle stops, let's find his position
X1 = V₀ₓ t + ½ aₓ t²
X1= 2.4 1.26 + ½ (-1.9) 1.26²
X1= 1.52 m
At this point the particle begins its return
b) The velocity has component x and y
As a section, the X axis x Vₓ = 0 m/s is stopped, but has a speed on the y axis
Vfy= Voy + ay t
Vfy= 0 + 3.2 1.26
Vfy = 4.0 m/s
the velocity is
V = (0 x^ + 4.0 y^) m/s
c) In order to make the graph we create a table of the position x and y for each time, let's start by writing the equations
X = V₀ₓ t+ ½ aₓ t²
Y = Voy t + ½ ay t²
X= 2.4 t + ½ (-1.9) t²
Y= 0 + ½ 3.2 t²
X= 2.4 t – 0.95 t²
Y= 1.6 t²
With these equations we build the table to graph, for clarity we are going to make two distance graph with time, one for the x axis and another for the y axis
Chart to graph
Time (s) x(m) y(m)
0 0 0
0.5 0.960 0.4
1 1.45 1.6
1.50 1.46 3.6
2.00 1.00 6.4
