Answer:
Incomplete question: "Each block has a mass of 0.2 kg"
The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s
Explanation:
Given data:
θ = angle of the surface = 37°
m = mass of each block = 0.2 kg
v = speed = 0.35 m/s
t = time to collision = 0.5 s
Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?
Change in momentum:
It is neccesary calculate the force:
Here, g = gravity = 9.8 m/s²
Answer:
32000 N
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Mass (m) of car = 400 Kg
Force (F) =?
Next, we shall determine the acceleration of the the car. This can be obtained as follow:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Acceleration (a) =?
v² = u² + 2as
0² = 40² + (2 × a × 10)
0 = 1600 + 20a
Collect like terms
0 – 1600 = 20a
–1600 = 20a
Divide both side by –1600
a = –1600 / 20
a = –80 m/s²
The negative sign indicate that the car is decelerating i.e coming to rest.
Finally, we shall determine the force needed to stop the car. This can be obtained as follow:
Mass (m) of car = 400 Kg
Acceleration (a) = –80 m/s²
Force (F) =?
F = ma
F = 400 × –80
F = – 32000 N
NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.
The y-component of the stone's velocity when it is 8 m below the hand is 14.86 m / s
v² = u² + 2 a s
s = Displacement
u = Initial velocity
a = Acceleration
u = 8 m / s
s = 8 m
v² = 8² + 2 * 9.8 * 8
v² = 64 + 156.8
v = √ 220.8
v = 14.86 m / s
The equation used to solve the problem is an equation of motion. These equations are designed to locate an object in motion using components such as velocity, displacement, acceleration and time.
Therefore, the y-component of the stone's velocity is 14.86 m / s
To know more about Equations of motion
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Answer:
a) m=20000Kg
b) v=0.214m/s
Explanation:
We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.
For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, 
, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as 
, and the mass of the first and second coals as 
and 
respectively
We start with the transition between parts A and B, so we have:
Which means
And since we want the mass of the first coal thrown () we do:
Substituting values we obtain
For the transition between parts B and C, we can write:
Which means
Since we want the new final speed of the car (
) we do:
Substituting values we obtain
