Question

Two hockey players strike a puck of mass 0.159 kg with their sticks simultaneously, exerting forces of 1.22 103 N, directed west, and 9.90 102 N, directed 30.0° east of north. Find the instantaneous acceleration of the puck. magnitude

Answer 1

Concept - Take positive X axis along east and positive Y axis along north. Express each force along the coordinate axes and add them to find the resultant force. Then use Newton’s law of motion to find the acceleration magnitude.

***********************************************************************************************
This concludes the answers. If there is any mistake or omission, let me know immediately and I will fix it....

Answer 2

To find the instantaneous acceleration of the puck, we need to calculate the net force acting on the puck and then use Newton's second law of motion, which states that acceleration is equal to the net force divided by the mass of the object.

Let's break down the forces acting on the puck:

1. The force exerted by the first hockey player (directed west) is 1.22 * 10^3 N.

2. The force exerted by the second hockey player (30.0° east of north) is 9.90 * 10^2 N.

Since the forces are not in the same direction, we need to calculate their horizontal and vertical components separately.

Horizontal components:

• The force directed west remains unchanged.

• The force directed east of north can be broken down into its horizontal component: F_horizontal = 9.90 * 10^2 N * cos(30°).

Vertical components:

• The force directed east of north can also be broken down into its vertical component: F_vertical = 9.90 * 10^2 N * sin(30°).

Now, let's find the net force in both the horizontal and vertical directions:

Net horizontal force = F_horizontal - F_horizontal (west) = 0 - 9.90 * 10^2 N * cos(30°). Net vertical force = F_vertical + F_vertical (west) = 9.90 * 10^2 N * sin(30°) + 1.22 * 10^3 N.

Next, we can calculate the net force's magnitude:

Net force = sqrt((Net horizontal force)^2 + (Net vertical force)^2).

Finally, we can use Newton's second law to find the instantaneous acceleration:

acceleration = Net force / mass.

Given that the mass of the puck is 0.159 kg, we can substitute the values and calculate the magnitude of the instantaneous acceleration.

Answer 3

To find the instantaneous acceleration of the puck, we need to calculate the net force acting on the puck and then use Newton's second law of motion, which states that the net force (F_net) on an object is equal to its mass (m) multiplied by its acceleration (a), i.e., F_net = m * a.

1. Resolve the forces into their horizontal (x) and vertical (y) components:

Force1 (directed west): Magnitude = 1.22 * 10^3 N x-component: -1.22 * 10^3 N (since it is directed west) y-component: 0 N (there is no vertical component)

Force2 (directed 30.0° east of north): Magnitude = 9.90 * 10^2 N x-component: 9.90 * 10^2 N * cos(30°) ≈ 8.58 * 10^2 N (since cos(30°) = √3/2) y-component: 9.90 * 10^2 N * sin(30°) = 4.95 * 10^2 N (since sin(30°) = 1/2)

2. Calculate the net force in the x and y directions:

Net force in the x-direction (F_net,x) = -1.22 * 10^3 N + 8.58 * 10^2 N ≈ -3.66 * 10^2 N Net force in the y-direction (F_net,y) = 0 N + 4.95 * 10^2 N = 4.95 * 10^2 N

3. Find the magnitude of the net force (F_net):

F_net = √(F_net,x^2 + F_net,y^2) F_net = √((-3.66 * 10^2 N)^2 + (4.95 * 10^2 N)^2) F_net ≈ √(133956 N^2 + 245025 N^2) F_net ≈ √378981 N^2 F_net ≈ 615.8 N

4. Calculate the acceleration (a) of the puck using Newton's second law:

F_net = m * a 615.8 N = 0.159 kg * a

5. Solve for the magnitude of the acceleration (a):

a = 615.8 N / 0.159 kg ≈ 3872.95 m/s²

The magnitude of the instantaneous acceleration of the puck is approximately 3872.95 m/s².