Consider two frictionless inclined planes with the same vertical height. Plane 1 makes an angle of 25.0 degree with the horizontal, and plane 2 makes an angle of 55.0 degree with the horizontal. Mass m_1 is placed at the top of plane 1, and mass m_2 is placed at the top of plane 2. Both masses are released at the same time.
At the bottom, which mass is going faster?
A. m_1
B. m_2
C. Neither; they both have the same speed at the bottom.
They both lose the same amount of gpe.
Oh, what a gravity-ious question! Gravity sure knows how to make things interesting. So, we have two frictionless inclined planes with the same vertical height. Plane 1 makes an angle of 25.0 degrees, and Plane 2 makes an angle of 55.0 degrees. Mass m_1 is placed at the top of Plane 1, and mass m_2 is placed at the top of Plane 2. Both masses are released at the same time.
Now, brace yourself for the punchline! At the bottom, which mass is going faster?
The answer is...C. Neither! They both have the same speed at the bottom. Why? Because the speed at the bottom only depends on the vertical height and not the angle of the incline. Gravity treats everyone equally, my friend!
So, whether you're rolling down a gentle slope like Plane 1 or a steeper slope like Plane 2, you'll end up at the bottom with the same speed. Isn't gravity just hilarious? It doesn't discriminate, it just pulls everyone down with the same comedic force!
Keep those physics questions coming, and I'll keep spinning out the laughs!
To determine which mass is going faster at the bottom of the inclined planes, we can use the principles of conservation of mechanical energy. Since the inclined planes are frictionless, there is no energy lost due to friction.
The potential energy at the top of each inclined plane can be calculated using the formula:
Potential energy (PE) = mass (m) * gravity (g) * height (h)
The initial potential energy for both masses can be calculated as:
PE1 = m1 * g * h
PE2 = m2 * g * h
where m1 and m2 are the masses of objects placed at the top of each inclined plane, g is the acceleration due to gravity, and h is the vertical height of the inclined planes.
At the bottom of each inclined plane, all of the initial potential energy is converted into kinetic energy (KE), given by the formula:
Kinetic energy (KE) = (1/2) * mass * velocity^2
Since both masses start from rest and are released at the same time, they both have zero initial velocity (v). Therefore, the final kinetic energy for both masses is zero.
Equating the initial potential energy to the final kinetic energy gives:
PE1 = KE1
PE2 = KE2
Since the potential energy is the same as the initial potential energy:
PE1 = m1 * g * h
PE2 = m2 * g * h
The masses cancel out in both equations, so we are left with:
g * h = (1/2) * v1^2
g * h = (1/2) * v2^2
It can be observed that the final velocity (v) will depend on the square root of the height (h) only.
Comparing the angles of the inclined planes, we see that the height (h) is the same for both. Therefore, the square root of the height is also the same.
As a result, v1 = v2.
This means that both masses m1 and m2 will have the same speed at the bottom of their respective inclined planes.
Therefore, the correct answer is C. Neither; they both have the same speed at the bottom.
To determine which mass is going faster at the bottom, we need to analyze the motion of the objects on the inclined plane.
When an object moves down an inclined plane, its motion can be divided into two components: one parallel to the plane (down the slope) and one perpendicular to the plane (along the vertical direction).
First, let's break down the forces acting on each mass:
For mass m₁ on plane 1:
- There is a component of gravity acting down the slope, which is mg sin(25°).
- There is a component of gravity acting perpendicular to the slope, which is mg cos(25°).
For mass m₂ on plane 2:
- There is a component of gravity acting down the slope, which is mg sin(55°).
- There is a component of gravity acting perpendicular to the slope, which is mg cos(55°).
Since the inclined planes are frictionless, there is no friction force opposing the motion.
Now, let's consider the acceleration of each mass:
For mass m₁ on plane 1:
- The acceleration along the slope is a₁ = g sin(25°).
For mass m₂ on plane 2:
- The acceleration along the slope is a₂ = g sin(55°).
Since both masses are released at the same time and have the same acceleration due to gravity (g), their motion down the inclined planes is determined by the values of sin(25°) and sin(55°).
Now, let's compare sin(25°) and sin(55°):
- sin(25°) ≈ 0.423
- sin(55°) ≈ 0.819
Therefore, sin(55°) > sin(25°), which means that the acceleration of mass m₂ on plane 2 is greater than the acceleration of mass m₁ on plane 1.
Since the acceleration of mass m₂ is greater, it will reach the bottom of the inclined plane faster than mass m₁. Hence, the correct answer is:
B. m₂ (mass on plane 2) is going faster at the bottom.