Physics Final-Multiple Problems-Help

[Matt H]

Physics Final---Multiple Problems---Help

Hey everyone, I have my physics final on Monday and I'm working through a practice test right now that is giving me a ton of trouble. If you can help with any/all of the problems i'd appreciate it.



1)

Homework Statement

1)A 74 kg bungee jumper is tied to a 39 m cord (unstretched). She leaps (from rest) off a
tall bridge (much taller than 39 m), so that she falls freely for 39 m before the cord begins
to stretch. The stretched cord stops her briefly above the water before yanking her back
upward. What is the magnitude of the impulse exerted on the bungee jumper while the
cord stretches during the downward motion?

Answer: 2046 kg m^2




2)

Homework Statement

10)A weight of Mg = 50 N is attached to the free end of a light string wrapped around a
pulley of radius R = 0.25 m and mass m = 3 kg. The pulley is free to rotate frictionlessly
in a vertical plane about the horizontal axis passing through its center, but remember that
the pulley is a solid disk and is not massless. The weight is released from rest 5 m above
the floor. The speed of the weight when it reaches the floor is:

Answer 8.7 m/s


I tried using the formula T = F_perpd = (I)(alpha) I wanted to find alpha, change it to linear acceleration and then plug that acceleration into a standard kinematic to find V_final. No such luck.



3)

Homework Statement

11) A car of mass 1000 kg moves on a circular track of radius 100 m with a speed of 50
m/s. What is the magnitude of its angular momentum (in kg m2/s) relative to the center of the race track?

Answer: 5 x 10^6

I tried using the formula Angular Momentum = (I)(angular velcoity) and manipulating some of the variables around. I couldn't make it work.



4)

Homework Statement

12) Two 1 kg particles are connected by a string 1 m long. They are set spinning about
their center of mass with an angular velocity of 2 rad/s in outer space, where external
forces are negligible. The tension in the string is:

Answer: 2N



5)

Homework Statement

13) A disk of radius R and mass M has a moment of inertia given by I = ½ MR2 for
rotations about an axis through the center of the disk. If a second disk of twice the mass
and half the radius is added as shown in the figure, what is the new moment of inertia of
the combined disks?

Answer: 3/4 MR^2



6)

Homework Statement

A spherical asteroid of mass Mast = 1016kg and radius Rast = 10 km (moment of inertia
Iast = 4×1023kg m2) is slowly rotating in free space at an angular velocity of ωi =
+6.28×10−4rad/s (one revolution every 10,000 seconds) around an axis through its center
of mass. A small comet of mass mc = 1011kg and traveling at vi = 100,000 m/s hits the
asteroid at a glancing blow, in a direction directly opposing the motion of that part of the
asteroid, as described in the figure. If the comet sticks to the edge of the asteroid (embeds
itself in it), what is the final angular velocity of the composite object?

Answer: +3.78 × 10−4 rad/s (same rotation direction)




7)

Homework Statement

A mass m = 0.1 kg is attached to a massless spring of constant k = 10 kg/s2. The other
end of the spring is attached to a pivot point about which the spring is free to rotate
frictionlessly on a tabletop. The mass also moves frictionlessly on the tabletop. The
unstretched spring length is r = 20 cm. The mass is set to rotate at a constant angular
velocity, which stretches the spring to a new length R = 25 cm, but the spring does not
oscillate. What is the angular momentum of the mass about the pivot point?

Answer: 0.028 kg m2/s



8)

Homework Statement

A 3 kg air-track frictionless glider is attached to each end of the track by two coil
springs. It takes a horizontal force of 0.9 N to displace the glider to a new equilibrium
position, x = 0.3 m. What is the period of oscillations about the equilibrium point?

Answer: 6.28 s



9)

Homework Statement

The Earth has a mass mE = 6×1024 kg, and orbits around the Sun (mass mS=2×1030kg)
along an approximately circular orbit of radius R=1.5×1011m. What is the total
mechanical energy (K+U) of the Earth, in reference to the Sun? Take U=0 at infinity.
(Remember: 1 year = 3.16×107s, G = 6.67×10-11 N m2/kg2; Hint: the circumference of a
circle is 2πR)

Answer: −2.7×1033 J

 

[Ofey]

Perhaps too late, but better late than never.

In the first assignment. What is the momentum of the jumper when when the rope begins to tighten? Since you know that the jumper eventually stops moving at the bottom, you can solve the impulse.

In the second assignment you will save a lot of work by simply using the rule of conservation of energy. What forms of energy are transformed into what?

Work on those, and see where you end up.

 

[nlightner]

10)

Homework Statement

When a 2 kg mass is hung from a vertical spring, the spring stretches 0.2 m. What is the
spring constant of this spring?

Answer: 20 N/m



1) To find the magnitude of the impulse exerted during the downward motion, we can use the equation Impulse = Change in Momentum. The bungee jumper's initial momentum is 0, as she is at rest. When she reaches the end of the unstretched cord, her momentum is still 0, as she is momentarily stopped before being pulled back up. Therefore, the change in momentum is 0. The impulse is equal to the force applied multiplied by the time it is applied for. In this case, the force is the weight of the bungee jumper (mg) and the time is the time it takes for the cord to stretch. We can calculate the time using the formula t = √(2h/g), where h is the height she falls freely (39m) and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, we get t = 2.83 seconds. Therefore, the magnitude of the impulse is (74 kg)(9.8 m/s^2)(2.83 s) = 2046 kg m^2.

2) To find the speed of the weight when it reaches the floor, we can use conservation of energy. At the start, the weight has potential energy (mgh) and no kinetic energy. At the bottom, the weight has no potential energy and only kinetic energy. Therefore, mgh = 1/2mv^2. Plugging in the values, we get v = √(2gh) = √(2(9.8 m/s^2)(5m)) = 8.7 m/s.

3) To find the angular momentum, we can use the formula L = Iω, where I is the moment of inertia and ω is the angular velocity. In this case, the moment of inertia is equal to the mass times the radius squared (I = MR^2). Plugging in the values, we get L = (1000 kg)(100 m)^2(50 m/s) = 5 x 10^6 kg m^2/s.

4) To find the tension in the string, we can use Newton's Second Law