A small hockey puck slides without friction over the icy hill and lands
6.20 m from the foot of the cliff with no air resistance.What is its initial speed v0 at the bottom of the hill?
vi=13.7 m/s
13.7 m/s
Well, if the hockey puck slides without friction, that means it must be pretty "slippery." It's like the puck is saying, "I don't need friction, I can slide on my own!" Anyway, to figure out the initial speed of the puck, we can use a little physics.
We know that the distance traveled by the puck is 6.20 m. So, if we divide that by the time it takes to reach that distance, we can find the velocity. But since there's no air resistance, we can use some kinematic equations.
First, let's use the equation:
vf^2 = vi^2 + 2ad
Where vf is the final velocity (which would be zero because the puck stops at the cliff), vi is the initial velocity (what we're looking for), a is the acceleration (which is just due to gravity, so -9.8 m/s^2), and d is the distance (6.20 m).
Rearranging the equation, we get:
vi^2 = -2ad
Substituting the values, we get:
vi^2 = -2(-9.8)(6.20)
vi^2 = 121.36
Now, taking the square root of both sides, we find:
vi = 11.02 m/s
So, the initial speed of the puck at the bottom of the hill is approximately 11.02 m/s. Hopefully, it doesn't slip and slide too far away!
To find the initial speed of the hockey puck at the bottom of the hill, we can use the principles of conservation of energy. The total mechanical energy of the system remains constant throughout the motion. In this case, the energy consists of two parts: kinetic energy (KE) and potential energy (PE).
At the bottom of the hill, the puck has only kinetic energy, given by:
KE_bottom = (1/2) * m * v0^2
At the top of the hill, the puck has both kinetic and potential energy, given by:
KE_top = (1/2) * m * v^2
PE_top = m * g * h
Where:
- m is the mass of the puck
- v0 is the initial speed at the bottom of the hill (which we need to find)
- v is the speed of the puck at the top of the hill (which we can assume is zero)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- h is the height of the hill (which we need to find)
Since there is no friction or air resistance, the total mechanical energy of the system is conserved:
KE_bottom = KE_top + PE_top
Substituting the expressions for KE and PE:
(1/2) * m * v0^2 = (1/2) * m * 0^2 + m * g * h
Simplifying the equation:
(1/2) * v0^2 = g * h
Now, we need to find the height of the hill. To do that, we can use the horizontal distance traveled by the puck and the angle of the hill.
Let's assume the angle of the hill is θ. The horizontal component of the displacement is given by:
x = d * cos(θ)
Where d is the horizontal distance traveled by the puck (6.20 m).
Rearranging the equation, we find:
h = x / cos(θ)
Now, we can substitute this value of h back into our energy conservation equation:
(1/2) * v0^2 = g * (x / cos(θ))
Finally, solving for v0:
v0 = sqrt((2 * g * x) / cos(θ))
Plug in the given values of g, x, and θ to calculate v0.