A baseball is thrown vertically upward and feels no air resistance. As it is rising:
2 answers:
Answer:
C) Its momentum is not conserved, but its mechanical energy is conserved.
Explanation:
There is no air resistance acting on the ball: this means that there are only conservative forces acting on the ball, therefore the total mechanical energy of the ball is conserved.
The total mechanical energy is defined as
where
U is the gravitational potential energy
K is the kinetic energy
The gravitational potential energy is given by:
where m is the mass, g the acceleration of gravity, h the heigth - since the height change, U changes as well, so the potential energy is not conserved.
The kinetic energy is given by
where v is the speed of the ball: as the ball rises, its speed decreases, so the kinetic energy is not conserved as well.
The momentum of the ball is given by:
As we stated above, the speed of the ball is decreasing, so v is decreasing, and this means that the momentum is also not conserved.
So the only correct choice is
C) Its momentum is not conserved, but its mechanical energy is conserved.
A baseball is thrown vertically upward and feels no air resistance. As it is rising: C. its momentum is not conserved, but its mechanical energy is conserved.
<h3>Explanation: </h3>Baseball is a sport played on a field by two teams that against each other. A player on one team throws a small round ball at a player on the other team who tries to hit it with a bat. Momentum is the quantity of motion that an object has. Momentum can be defined as "mass in motion." Whereas mechanical energy is the sum of kinetic and potential energy in an object that used to do work. Mechanical energy is energy in an object due to its motion or position, or both.
Momentum is the mass m multiplied by velocity v so momentum may be conserved if a change of mass is complemented by a change of velocity. Energy and momentum are always conserved. Momentum can be conserved if mechanical energy is not conserved although In a perfect scenario mechanical energy is always conserved, therefore the kinetic energy is not conserved in an inelastic collision because it is converted to another form of energy (heat, etc.). The sum of all types of energy including kinetic is the same before and after the collision
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Answer:
a) The takeoff speed is 10 m/s.
b) The maximum height above the ground is 1.2 m.
Explanation:
The position of the kangaroo and its velocity at any given time "t" can be calculated by the following equations:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v =(v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time "t".
x0 = initial horizontal position.
v0 = initial velocity.
α = jumping angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
v = velocity vector at time "t"
a) Please see the attached figure for a better understanding of the problem. In red is depicted the position vector at the final time (r final). The components of r final are known:
r final = (8.7 m, 0 m)
Then at final time:
8.7 m = x0 + v0 · t · cos α
0 m = y0 + v0 · t · sin α + 1/2 · g · t²
(notice in the figure that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0). Then:
8.7 m = v0 · t · cos α
Solving for "v0":
8.7 m /(t · cos α) = v0
Replacing v0 in the equation of the y-component, we can obtain the final time:
0 m = 8.7 m · tan 29° - 1/2 · 9.8 m/s² · t² (remember: sin α / cos α = tan α)
- 8.7 m · tan 29° / -4.9 m/s² = t²
t = 0.99 s
Now, we can calculate the initial speed:
8.7 m /t · cos α = v0
v0 = 8.7 m / (0.99 s · cos 29°)
<u>v0 = 10 m/s</u>
The takeoff speed is 10 m/s
b) When the kangaroo is at its maximum height, the velocity vector is horizontal (see figure). That means that the y-component of the velocity at that time is 0:
0 = v0 · sin α + g · t
Solving for "t":
-v0 · sin α / g = t
t = - 10 m/s · sin 29° / 9.8 m/s²
t = 0.49 s
Notice that we could have halved the final time (0.99 s, calculated above) to obtain the time at which the kangaroo is at its maximum height. That´s because the trajectory is parabolic.
Now, let´s find the height of the kangaroo at that time:
y = y0 + v0 · t · sin α + 1/2 · g · t²
y = 10 m/s · 0.49 s · sin 29° - 1/2 · 9.8 m/s² · (0.49 s)²
<u>y = 1.2 m</u>
The maximum height above the ground is 1.2 m.
