On an icy road, a 1100 kg car moving at 55 km/h strikes a 480 kg truck moving in the same direction at 37 km/h . The pair is soo
2 answers:
Answer:
the speed of the wreckage is 48.39km/hr
Explanation:
Given that ,
Mass of first car, m₁ =1100kg
Speed of the first car, v₁ = 55km/hr
Mass of second car, m₂ = 480kg
Speed of second car, v₂ = 37km/hr
Mass of third car, m₃ =1300kg
Speed of third car, v₃ = 47km/hr
In the first case,
In the second case
where v is the velocity of the combined system
Thus , the speed of the wreckage is 48.39km/hr
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Answer:
Diagrams in pictures
Explanation:
Using energy I can get
m g h = 1/2 m v^2
So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.
(H= 1,3m sin25)
V=2,32m/s
V= a t
And
X= v t +1/2 a t^2
Knowing v=2,32 m/s and x= 1,3 m
I can get
a= 6,21m/s2
F= m a
I can get the force of the box when it collides with the spring
F= 12, 425 N
The force the spring makes on the box then is
F = -12,425N = -k d
Then the spring's constant is k= 51,75N/m
To make the two diagrams I need the functions of time when the box slows down
I use the same two equations
V= a t
And
X= v t + 1/2 a t^2
Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.
I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).
Then,
V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec
X(t)= 2,32 m/s t + 1/2 33,67 m/s2 t^2.
for time between 0 and 0,689 sec
Diagrams and equations are in the pictures
