Please Please Please help on these 3 questions :) 20 POINTS!! - NO LINKS PLEASE
1 answer:
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Answer:
<em>Second option</em>
Explanation:
<u>Linear Momentum</u>
The linear momentum of an object of mass m and speed v is
P=mv
If two or more objects are interacting in the same axis, the total momentum is
Where the speeds must be signed according to a fixed reference
The images show a cart of mass 2m moves to the left with speed v since our reference is positive to the right
The second cart of mass m goes to the right at a speed v
The total momentum before the impact is
The total momentum after the collision is negative, both carts will join and go to the left side
The first option shows both carts with the same momentum before the collision and therefore, zero momentum after. It's not correct as we have already proven
The third option shows the 2m cart has a positive greater momentum than the other one. We have proven the 2m car has negative momentum. This option is not correct either
The fourth option shows the two carts keep separated after the collision, which contradicts the condition of the question regarding "they hook together".
The second option is the correct one because the mass has a negative momentum and then the sum of both masses keeps being negative
Answer:
The arrow will hit 112.07 m from the point of release.
Explanation:
The equation for the position of an object in a parabolic movement is as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal position
v0 = initial velocity
α = launching angle
y0 = initial vertical position
t = time
g = acceleration due to gravity
We know that at the final time the y-component of the vector "r" (see figure") is -1.88 m. The x-component of that vector will be the horizontal distance traveled by the arrow. Using the equation of the y-component of "r", we can obtain the final time and with that time we can calculate the value of the x-component (horizontal distance).
Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
Since the origin of the frame of reference is located at the point where the arrow is released, y0 = 0. Notice that the angle α = 26° + 15° = 41° ( see figure)
-1.88 m = 33 m/s · sin 41° · t - 1/2 · 9.8 m/s² · t² (g is downward)
0 = -4.9 m/s² · t² + 33 m/s · sin 41° · t + 1.88 m
Solving the quadratic equation:
t = 4.5 s ( the negative value is discarded)
Now, with this time we can calculate the horizontal distance:
x = x0 + v0 · t · cos α (x0 = 0, the same as y0)
x = 33 m/s · 4.5 s · cos 41° = 112.07 m
