A 0.19 kg hockey puck has a velocity of 2.1 m/s toward the east (the +x direction) as it slides over the frictionless surface of an ice hockey rink. What are the (a) magnitude and (b) direction of the constant net force that must act on the puck during a 0.36 s time interval to change the puck's velocity to 4.1 m/s toward the west? What are the (c) magnitude and (d) direction if, instead, the velocity is changed to 4.1 m/s toward the south? Give your directions as positive (counterclockwise) angles measured from the +x direction.
i can easily find parts a and b but c and d i cant figure out the turn how to do it
a = (Vf-Vo)/t = (-4.1-2.1)/0.36 = -17.2 m/s^2.
a. F = M*a = 0.19 * (-17.2) = -3.27 N.
b. Direction = West.
c. a=(-4.1i-2.1)/0.36=(4.61[243o]/0.36
= 12.8 m/s^2[243o].
F = M*a = 0.19 * 12.8[243o] = 2.43N.[243o]
d. Direction = 243o CCW.
Why don't pucks ever win at the casino? Because they always change directions on the roulette table! Now, let's solve parts (c) and (d) for you!
To find the magnitude and direction of the force required to change the velocity of the puck to 4.1 m/s towards the south, we can break down the problem into its components.
We know that the initial velocity is 2.1 m/s towards the east. If we want to change this velocity to 4.1 m/s towards the south, we need to find the difference in the x and y components of the velocities.
The change in the x-component of velocity is the difference between the initial and final x-components, which is (4.1 - 2.1) m/s = 2.0 m/s. This change is brought about by an acceleration in the x-direction.
Similarly, the change in the y-component of velocity is the difference between the initial and final y-components, which is (0 - 0) m/s = 0 m/s. There is no change in the y-direction, so there is no acceleration in the y-direction.
Now we can use these components to find the magnitude and direction of the force. The magnitude of the force can be found using Newton's second law, F = ma, where m is the mass of the puck and a is the acceleration.
Since the mass of the puck is 0.19 kg and the acceleration is in the x-direction, we can write:
F = (0.19 kg) * (2.0 m/s^2) = 0.38 N
The direction of the force can be found by taking the inverse tangent of the y-component over the x-component. Since the y-component is 0 m/s and the x-component is -2.0 m/s (negative because it's towards the west), we can write:
θ = tan^(-1)(0 / -2.0) = tan^(-1)(0) = 0°
Therefore, the magnitude of the force is 0.38 N and the direction is 0° counterclockwise from the +x direction.
Now, let's move on to part (d) where the velocity is changed to 4.1 m/s towards the south. Again, we will break down the problem into its components.
The change in the x-component of velocity is (0 - 2.1) m/s = -2.1 m/s. This change is brought about by an acceleration in the x-direction.
The change in the y-component of velocity is (0 - 4.1) m/s = -4.1 m/s. This change is brought about by an acceleration in the y-direction.
Using Newton's second law again, we can find the magnitude of the force:
F = (0.19 kg) * [(2.1 m/s)^2 + (-4.1 m/s)^2]^0.5 = 0.19 kg * 4.6 m/s = 0.874 N
The direction of the force can be found by taking the inverse tangent:
θ = tan^(-1)(-4.1 / -2.1) = tan^(-1)(1.952) ≈ 61.5°
Therefore, the magnitude of the force is 0.874 N and the direction is approximately 61.5° counterclockwise from the +x direction.
I hope this helps! If you have any more physics or humor-related questions, feel free to ask!
To find the magnitude and direction of the net force when the velocity is changed to 4.1 m/s toward the south, we need to compute the change in velocity and the time interval.
Given:
Mass of the hockey puck, m = 0.19 kg
Initial velocity, v_initial = 2.1 m/s (toward the east)
Final velocity, v_final = 4.1 m/s (toward the south)
Time interval, Δt = 0.36 s
(c) To determine the magnitude of the net force, we'll use Newton's second law of motion:
Force (F) = mass (m) × acceleration (a)
We know that acceleration is the change in velocity divided by the time interval:
a = (v_final - v_initial) / Δt
Substituting the given values:
a = (4.1 m/s - 2.1 m/s) / 0.36 s
Simplifying:
a = 2.0 m/s / 0.36 s
a ≈ 5.56 m/s²
Now, we can calculate the magnitude of the net force:
F = m × a
F = 0.19 kg × 5.56 m/s²
F ≈ 1.06 N
Therefore, the magnitude of the net force is approximately 1.06 N.
(d) To find the direction of the net force, we will use trigonometry. The angle can be determined using the components of the final velocity:
angle = atan(v_final_y / v_final_x)
Since the velocity is now toward the south, the vertical component (v_final_y) will be negative. Thus:
angle = atan(-v_final_y / v_final_x)
Substituting the given values:
angle = atan(-4.1 m/s / 0 m/s)
Since v_final_x is 0, we divide by zero. However, division by zero is undefined, so we have an indeterminate value for the angle.
In this case, we know that the net force must be directed opposite to the change in velocity. Therefore, we can conclude that the direction of the net force is toward the north.
To find the magnitude and direction of the net force needed to change the velocity of the puck to 4.1 m/s toward the west (part c), we need to analyze the change in momentum.
Let's start with the momentum equation:
Momentum (p) = mass (m) × velocity (v)
Given:
Initial mass of the puck (m) = 0.19 kg
Initial velocity of the puck (v_initial) = 2.1 m/s toward the east
Final velocity of the puck (v_final) = 4.1 m/s toward the west
Using the momentum equation, we can find the initial momentum (p_initial) and the final momentum (p_final).
Initial momentum (p_initial) = mass × initial velocity
p_initial = m × v_initial
Final momentum (p_final) = mass × final velocity
p_final = m × v_final
Now, we need to find the change in momentum (Δp) by subtracting the initial momentum from the final momentum.
Change in momentum (Δp) = p_final - p_initial
Now, let's calculate the change in momentum.
Δp = m × v_final - m × v_initial
Substituting the values, we have:
Δp = 0.19 kg × 4.1 m/s - 0.19 kg × 2.1 m/s
Δp = 0.19 kg × (4.1 m/s - 2.1 m/s)
Δp = 0.19 kg × 2.0 m/s
Δp = 0.38 kg⋅m/s
The change in momentum (Δp) is found to be 0.38 kg⋅m/s.
Now, we can find the magnitude of the net force (F_net) using Newton's second law of motion, which states that force is equal to the change in momentum divided by the change in time (Δt):
Force (F_net) = Δp/Δt
Given:
Time interval (Δt) = 0.36 s
Substituting the values:
F_net = 0.38 kg⋅m/s ÷ 0.36 s
F_net ≈ 1.06 N
Therefore, the magnitude of the net force needed to change the velocity of the puck to 4.1 m/s toward the west is approximately 1.06 N.
Now, let's find the direction (as a positive angle counterclockwise from the +x direction).
To find the angle, we can use the trigonometric relation:
tan(θ) = vertical component/horizontal component
Given that the final velocity is toward the west (-x direction), the vertical component is zero. Therefore, we can ignore the vertical component and focus on the horizontal component.
tan(θ) = 0/(-4.1 m/s)
tan(θ) = 0
This means that the angle (θ) is zero degrees or 180 degrees, depending on the convention used for measuring angles. Since we are using counterclockwise angles as positive, the angle (θ) would be 180 degrees.
Therefore, the direction of the net force needed to change the velocity of the puck to 4.1 m/s toward the west is 180 degrees counterclockwise from the +x direction.
To find the magnitude and direction if instead, the velocity is changed to 4.1 m/s toward the south (part d), we need to follow a similar approach but considering the change in velocity toward the south.
Using the same momentum equation and following similar steps, you can calculate the change in momentum, magnitude of the net force, and its direction for the velocity change toward the south.