Final answer:
The professor is 0 meters from the wall when he stops at the opposite end of the skateboard from where he started.
Step-by-step explanation:
To solve this problem, we can apply the principle of conservation of momentum. The initial momentum of the system, consisting of the professor and the skateboard, is equal to the final momentum of the system. Initially, both the professor and the skateboard are at rest, so the initial momentum is zero.
When the professor walks towards the wall, he exerts a force on the ground, causing an equal and opposite force on him (according to Newton's third law). This force propels the skateboard forward, resulting in a change in momentum of the professor-skateboard system. The total momentum is conserved during this process.
When the professor reaches the opposite end of the skateboard, he comes to a stop. At this point, the skateboard would have moved a certain distance, which we'll call x. If the professor and the board have the same mass, the professor would have moved a distance equal to (d + x) from the wall, while the board would have moved a distance equal to x from the wall.
From the conservation of momentum, we can write:
(0) + (m)(v) = (m)(v) + (M)(V)
Here, m represents the mass of the professor, v represents his initial velocity, M represents the mass of the skateboard, and V represents its final velocity.
Since the professor starts from rest, his initial velocity v is zero. The final velocity V of the skateboard can be calculated using the equation:
(m)(v) = (M)(V)
From the given information, we know that the professor's mass is equal to the skateboard's mass, so m = M. Plugging this into the equation, we get:
(M)(0) = (M)(V)
This simplifies to:
0 = V
Since the final velocity V of the skateboard is zero, we can conclude that the professor comes to a stop at the opposite end of the skateboard. Therefore, he would be a distance x from the wall. To find the value of x, we need to analyze the forces and motion of the system.
When the professor exerts a force on the ground, causing an equal and opposite force on him, the system experiences a net force. According to Newton's second law, the net force on the system is equal to the product of the mass of the system and its acceleration.
Let's define the positive direction as the direction the professor is moving towards the wall. The net force on the system in the positive direction is:
F_net = F_applied - F_opposing
Where F_applied is the force exerted by the professor on the ground, and F_opposing is the sum of all the opposing forces, such as friction.
Since the skateboard rolls with negligible friction, the opposing forces can be considered negligible. Therefore, we have:
F_net = F_applied
From Newton's second law, we can write:
F_net = (M + m)a
Where a is the acceleration of the system.
Plugging in the given information, we have:
150 N = (2M)a
Solving for a gives:
a = 75 N / M
Now, let's consider the motion of the skateboard. Since there is no friction, the only horizontal force acting on the skateboard is the force exerted by the professor on the ground. This force causes the skateboard to accelerate in the positive direction.
The distance x is related to the acceleration a and the time taken t to reach the opposite end of the skateboard. We can use the kinematic equation:
x = (1/2)at^2
Since the professor is initially at rest and comes to a stop at the opposite end, his final velocity vf is zero. We can use the equation:
vf = vi + at
Where vi is the initial velocity of the professor, and a is the acceleration.
Since the professor is initially at rest, his initial velocity vi is zero. Plugging in the known values, we have:
0 = 0 + (75 N / M)t
This simplifies to:
t = 0
Since the time t is zero in this case, the professor reaches the opposite end of the skateboard instantaneously. Therefore, the distance x is also zero.
In conclusion, the professor is 0 meters from the wall when he stops at the opposite end of the skateboard from where he started.