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College physics 1) A 600 g model rocket is on a cart that is rolling to...

College physics

1) A 600 g model rocket is on a cart that is rolling to the right at a speed of 4.0 m/s . The rocket engine, when it is fired, exerts a 9.0 N vertical thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is 25 mabove the launch point. At what horizontal distance left of the loop should you launch?

2) A m1 = 5.2 kg box is on a frictionless θ = 36 ∘ slope and is connected via a massless string over a massless, frictionless pulley to a hanging m2 = 1.7 kg weight. What is the tension in the string once the box begins to move?

3) A 300 g block on a 54.0 cm -long string swings in a circle on a horizontal, frictionless table at 60.0 rpm. What is the speed of the block? What is the tension in the string?

Solutions

Expert Solution

First problem -

(1) Mass of the rocket, m = 600 g = 0.600 kg

So, weight of the rocket, Fw = mg = (0.600 kg)(9.81 m/s²) = 5.89 N

The rocket engine exerts 9.0 N of force on the rocket. 5.89 N of that force will be used to counter the rocket's weight, leaving (9.0 - 5.87) = 3.11 N of force available to provide acceleration. We should calculate the rocket's upward acceleration:

Fnet = ma

Where Fnet is the net force (the force that remains after the rocket's weight is compensated), and a is the rocket's acceleration (in m/s²)

3.11 N = (0.600 kg) a

a = 5.18 m/s²

So now that we know the rocket's upward acceleration we can calculate how long it will take to rise 25 meters into the air.

Δy = vo*t + 1/2 at²

The rocket's original velocity, vo, is 0 m/s. We can remove the vo*t term from the equation, leaving:

Δy = 1/2 at²

25 m = 1/2(5.18 m/s²)t²

25 m = (2.59 m/s²)t²

9.65 s² = t²

t = 3.11 s

So the rocket will take 3.11 seconds to reach the hoop. It should be launched when the cart is 3.11 seconds away from being directly beneath the hoop. Given the cart's constant horizontal velocity of 4.0 m/s, we can calculate how far the cart will travel in 2.5 s

v = Δx / t

4.0 m/s = Δx / 3.11 s

Δx = 4.0 x 3.11 = 12.44 m

Therefore, the rocket should be launched when the cart is 12.44 m away from a point directly below the hoop. (Answer)


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