Question

toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, taken as the x-y plane. The 4.00 kg puck has a velocity of 3.00 i m/s at one instant. Eight seconds later, its velocity is (8.00 i 10.00 j) m/s. Assuming the rocket engine exerts a constant force, find (a) the components of the force and (b) its magnitude.

Answer 1

Fx = 2.5 N

Fy = 5 N

|F| = 5.59 N

Step-by-step explanation:

Given:-

- The mass of puck, m = 4.0 kg

- The initial velocity of puck, u = 3.00 i m/s

- The final velocity of puck, v = ( 8.00 i + 10.00 j ) m/s

- The time interval for the duration of force, Δt = 8 seconds

Find:-

the components of the force and (b) its magnitude.

Solution:-

- We will set up a coordinate system ( x - y ) plane. With unit vectors i and j along x and y axes respectively.

- To model the situation we will seek help from Newton's second law of motion. Defined by the rate of change of linear momentum of the system.

`F_net = (m*( v - u ) )/(dt)`

Where,

Fnet: The net force that acts on the puck-rocket system

- Here we will assume that the mass of rocket is negligible compared to the mass of the puck. The only force ( F ) acting on the puck is due to the thrust produced of the rocket. The dry and air frictions are both neglected for the analysis.

- We will apply the newton's second law of motion in component forms. And determine the components of force F, as ( Fx ) and ( Fy ) as follows:

`F_x = (m* ( v_x - u_x))/(dt) \\\\F_x = (4* ( 8 - 3))/(8) \\\\F_x = 2.5 N\\\\F_y = (m* ( v_y - u_y))/(dt) \\\\F_y = (4* ( 10 - 0))/(8) \\\\F_y = 5 N\\\\`

- We will apply the Pythagorean theorem and determine the magnitude of the thrust force produced by the rocket with which the puck accelerated:

`| F | = √(( F_x)^2 + ( F_y)^2) \\\\| F | = √(( 2.5)^2 + ( 5)^2) \\\\| F | = √(31.25) \\\\| F | = 5.590`

Answer: the magnitude of the thrust force is F = 5.59 N