A rocket-powered hockey puck has a thrust of 2.40 and a total mass of 1.40 . It is released from rest on a frictionless table, 2.90 from the edge of a 1.80 drop. The front of the rocket is pointed directly toward the edge. How far from the table does the puck land?
To determine how far from the table the puck lands, we need to find the horizontal displacement of the puck.
First, let's analyze the horizontal motion of the puck after it is released from rest.
The only horizontal force acting on the puck is the thrust force provided by the rocket, which can be calculated using Newton's second law:
F = ma
Where F is the force, m is the mass, and a is the acceleration.
In this case, the thrust force is equal to 2.40 N, and the mass of the puck is 1.40 kg.
So, the acceleration of the puck in the horizontal direction is:
a = F/m = 2.40 N / 1.40 kg = 1.71 m/s^2
Now, let's consider the vertical motion of the puck when it reaches the edge of the drop.
Since it is released from rest, we can use the kinematic equation to find the time it takes for the puck to reach the edge of the drop:
d = vt + (1/2)at^2
Where d is the distance, v is the initial velocity, a is the acceleration, and t is the time.
In this case, the distance is 2.90 m, the initial velocity is 0 m/s, and the acceleration is due to gravity (9.8 m/s^2).
Simplifying the equation for initial velocity gives us:
t = sqrt(2d/a)
t = sqrt(2 * 2.90 m / 9.8 m/s^2)
t ≈ 0.763 s
Now, let's find the horizontal distance traveled by the puck in this time.
The horizontal displacement can be calculated using the formula:
d = v * t
Where d is the distance, v is the initial velocity, and t is the time.
In this case, the initial velocity is the acceleration multiplied by the time:
v = a * t
v = 1.71 m/s^2 * 0.763 s ≈ 1.31 m/s
Now, we can calculate the horizontal displacement:
d = 1.31 m/s * 0.763 s ≈ 1.00 m
Therefore, the rocket-powered hockey puck will land approximately 1.00 meter from the edge of the table.
To find the distance from the table where the rocket-powered hockey puck lands, we need to analyze the motion of the puck.
Firstly, let's find the initial velocity of the puck just before it leaves the table. We can use the equation:
Acceleration = Force/Mass
Given that the thrust force of the rocket is 2.40 N and the total mass of the puck is 1.40 kg, the acceleration can be calculated as:
Acceleration = 2.40 N / 1.40 kg = 1.714 m/s²
Next, we can find the time it takes for the puck to fall from the table to the edge using kinematic equations. Let's use the equation:
Distance = Initial Velocity * Time + (0.5 * Acceleration * Time²)
At the edge of the table, the distance is 2.90 m, and the initial velocity is 0 m/s (the puck is released from rest). Plugging in these values, the equation becomes:
2.90 m = 0 * t + (0.5 * 1.714 m/s² * t²)
Simplifying the equation, we get:
1.714 m/s² * t² = 2.90 m
t² = 2.90 m / 1.714 m/s²
t² = 1.691 s²
Taking the square root of both sides, we find:
t = √(1.691 s²) ≈ 1.30 s
Therefore, it takes approximately 1.30 seconds for the puck to fall from the table to the edge.
Now, we can calculate the horizontal distance traveled by the puck during this time by using the equation:
Distance = Velocity * Time
The velocity in the horizontal direction remains constant at 0 m/s throughout the motion. Therefore, the distance is:
Distance = 0 m/s * 1.30 s = 0 m
Therefore, the rocket-powered hockey puck lands at a distance of 0 meters from the table.