10A spar buoy is a cylindrical object that floats with its axis in the vertical

Question

10A spar buoy is a cylindrical object that floats with its axis in the vertical direction. When it is disturbed a distance h from its equilibrium position, the magnitude of the restoring force is equal to pghs, where p is the density of the water and S is the cross-sectional area of the A certain spar buoy has a mass of 100 kg a cross-sectional area of 1.25 m2, and is ting in seawater, p- 1025 kg/m3, What is the angular frequency of its oscillations if it is displaced from its equilibrium position? A) 11.2 rad/s B) 1.78 rad/s C) 12.8 rad/s D) 1.31 rad/s I A 420-g bullet embeds itself in a 20.0-kg block, which is attached to a horizontal spring with a force constant of 960 N/m. The maximum speed of the bullet just before it hits the block? compression of the spring is 2.20 am. What is the A) 726 m/s B) 700 m/s C) 672 m/s D) 952 m/s A 16.2-g bullet with an initial speed of 850 m/s embeds itself in a 40.0-kg block, which is attached to a horizontal spring with a force constant of 1000 N/m. What is the maximum compression of the spring? A) 5.57 a B) 3.42 cm C)1.26 am D) 6.88 am 13 Amass of 500 g is resting on a vertical spring with a force constant of 55.0 N/m. A mass of 2S0 gîs dropped from a height of ?2.0 cm onto the larger mass and sticks to it. What is the amplitude of the resulting oscillations? A) 4.71 cm . B) 9.90 c C) 11.4 cm D) 5.97 cm An astronaut has landed on an asteroid and conducts an experiment to determine the acceleration of gravity on that asteroid. She uses a pendulum that has a period of oscillation of 2.00 s on Earth and finds that on the asteroid the period is 11.3 s. What is the acceleration of gravity on that asteroid? A) 0.307 m/s2 B) 1.74 m/s2 C) 5.51 m/s D) 0.0544 m/s2 14

Explanation / Answer

10) f = (1/2pi) x sqrt (A p g/m)

f = (1/ 2 pi) sqrt (1.25 x 1025 x 9.81 /100)

f = 1.78 rad/sec

Option (b)

11) Let V1 be the initial velocity of bullet and V2 be the combined velocity of bullet & block

KE of block+bullet = 1/2(m1+m2)v^2

work done on spring 1/2kd^2

Using law of conservation of energy,

1/2*960*(0.022)^2 = 1/2(20.0042)v2^2

V2 = 0.152 m/sec

Now conservation of momentum we get (Mass bullet)*(Vel. bullet) = (mass bullet + block)*(combined Vel.)

so V1 = 726 m/s

Option (a)

12) Using law of conservation of energy

mV = (m+M)U

U is the vel. of block and bullet after collision. The kinetic energy of bullet and block will then compress spring and be converted to elastic PE;

(1/2)(m+M)U^2 = (1/2)kX^2

elliminate U and solve for X;

X = mV/ sqrt{k(m+M)} = (0.0162x850)/sqrt (1000(40 + 0.0162)) = 6.88 cm

Option (d)

13) PE = (0.25)(9.8)(0.12) = 0.294 J

Kinetic Energy = (1/2)(mass)(velocity^2)

velocity = sqrt (2(0.294 J) / (0.250 kg)) = 1.534 m/s

Since momentum is mass x velocity, so

=> (0.250 kg)(1.534 m/s) = (0.750 kg)(v2)

Solving for v2 we get 0.511 m/s as the velocity of the 750 gram mass.

(1/2)(55)(A^2) = (1/2)(0.750)(0.511^2)

A = 5.97 cm

Option (d)

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