a hockey puck rebounds from a board. the puck is in contact with the board for 2.5 ms. dertermine the acceleration of the puck over the time interval.
V(i) = 26 m/s...at angle of 22 degrees.
V(f) = 21 m/s...at angle of 22 degrees.
a=?
To determine the acceleration of the puck over the time interval, we need to calculate the change in velocity.
Step 1: Analyze the initial velocity (V(i)) and final velocity (V(f)):
- V(i) = 26 m/s at an angle of 22 degrees.
- V(f) = 21 m/s at an angle of 22 degrees.
Step 2: Separate the velocities into horizontal and vertical components:
- V(i)x = V(i) * cos(theta)
- V(i)y = V(i) * sin(theta)
- V(f)x = V(f) * cos(theta)
- V(f)y = V(f) * sin(theta)
Given that the angle (theta) is the same for both velocities, we can calculate the components:
- V(i)x = 26 m/s * cos(22 degrees)
- V(i)y = 26 m/s * sin(22 degrees)
- V(f)x = 21 m/s * cos(22 degrees)
- V(f)y = 21 m/s * sin(22 degrees)
Step 3: Calculate the change in velocity for both horizontal and vertical directions:
- ∆Vx = V(f)x - V(i)x
- ∆Vy = V(f)y - V(i)y
Step 4: Determine the time interval (t) during which the puck is in contact with the board:
- t = 2.5 ms = 2.5 × 10^(-3) s
Step 5: Calculate the acceleration (a) using the formula:
- a = (∆Vx + ∆Vy) / t
Plug in the values calculated in Step 2, 3, and 4 to find the acceleration:
- ∆Vx = 21 m/s * cos(22 degrees) - 26 m/s * cos(22 degrees)
- ∆Vy = 21 m/s * sin(22 degrees) - 26 m/s * sin(22 degrees)
Substitute the values into the formula:
- a = (∆Vx + ∆Vy) / t
- a = (21 m/s * cos(22 degrees) - 26 m/s * cos(22 degrees) + 21 m/s * sin(22 degrees) - 26 m/s * sin(22 degrees)) / (2.5 × 10^(-3) s)
Calculate the numerator:
- (∆Vx + ∆Vy) = (21 m/s - 26 m/s) * cos(22 degrees) + (21 m/s - 26 m/s) * sin(22 degrees)
Finally, divide the numerator by the time interval (t) to obtain the acceleration (a) of the puck over the given time interval.
To determine the acceleration of the puck, we need to find the change in velocity or ΔV and divide it by the time interval Δt.
1. We begin by breaking down the given initial and final velocities into their horizontal and vertical components.
Given:
Initial velocity (V(i)) = 26 m/s at an angle of 22 degrees.
Final velocity (V(f)) = 21 m/s at an angle of 22 degrees.
To find the horizontal component of velocity:
V(i)x = V(i) * cosθ(i)
V(f)x = V(f) * cosθ(f)
To find the vertical component of velocity:
V(i)y = V(i) * sinθ(i)
V(f)y = V(f) * sinθ(f)
2. The change in horizontal velocity (ΔVx) is the difference between the final and initial horizontal velocities:
ΔVx = V(f)x - V(i)x
3. The change in vertical velocity (ΔVy) is the difference between the final and initial vertical velocities:
ΔVy = V(f)y - V(i)y
4. The total change in velocity (ΔV) can be found using the Pythagorean theorem:
ΔV = √(ΔVx^2 + ΔVy^2)
5. Next, we need to calculate the time interval (Δt). Given that the puck is in contact with the board for 2.5 ms, we'll convert the time to seconds:
Δt = 2.5 ms = 2.5 * 10^(-3) s
6. Finally, we can find the acceleration (a) using the equation:
a = ΔV / Δt
Now let's plug in the values and calculate the acceleration:
Step 1:
V(i)x = 26 m/s * cos(22°)
V(f)x = 21 m/s * cos(22°)
V(i)y = 26 m/s * sin(22°)
V(f)y = 21 m/s * sin(22°)
Step 2:
ΔVx = V(f)x - V(i)x
Step 3:
ΔVy = V(f)y - V(i)y
Step 4:
ΔV = √(ΔVx^2 + ΔVy^2)
Step 5:
Δt = 2.5 * 10^(-3) s
Step 6:
a = ΔV / Δt
By following these steps and substituting the values provided, you can determine the acceleration of the puck over the given time interval.
Is the 22 degrees measured from the normal to the board?
Whatever the answer to that question is, determine the change in the component of velocity normal to the board
It will be either sin 21*47 m/s or cos21*47 m/s.
Divide that by the contact tiome for the acceleration normal to the board.
There will also be a deceleration along the direction of the board (since the velocity drops), but it will be much smaller.