Exp CENI ETAL RCE OBJECT: TODETERMINE THE CENTRIPETAL FORCEREQUIRED TO KEEP AN O

Question

Exp CENI ETAL RCE OBJECT: TODETERMINE THE CENTRIPETAL FORCEREQUIRED TO KEEP AN OBJECT MovING IN A HORIZONTAL CIRCLE WITH UNIFORM SPEED. STUDY: Chapter 7 in your text, College Physics SewayNuille INTRODUCTION: Centripetal force is the name given to the force directed toward the center of the circular path around which an object is moving. Figure 1. A panicle of mass m oving in a circle with coastant speed v. m The magnitude of the centripetal force on a particle of mass m moving at uniform speed vin a circle R is given by: CI) ama," R (For a derivation of the above equation see chapter 5, sections 3 in your textbook.) The speed of an object traveling in a circle of radius R can be d by multiplying the distance the object goes in one revolution. 2mR by the number of revolutions the object makes per second, this value for vinto equation (1) we have substituting (3) measuring mm f and R. one may calculate the centripetal o:ce Fe by means Eq. 3 By

Explanation / Answer

6 a)

m = mass of bullet = 40 g = 0.040 kg

v = velocity of bullet before collision = ?

M = mass of block = 400 g = 0.4 kg

Vi = velocity of block before collision = 0 m/s               since block was at rest

V = velocity of bullet-block combination after collision = ?

h = height gained = 15 cm = 0.15 m

using conservation of energy

kinetic energy of bullet-block combination after collision = Potential energy gained

(0.5) (m + M) V2 = (m + M) gh

V = sqrt(2gh)

V = sqrt(2 x 9.8 x 0.15)

V = 1.71 m/s

using conservation of momentum for collision betweeb bullet and block

m v + M Vi = (m + M) V

(0.04) v + (0.4) (0) = (0.04 + 0.4) (1.71)

v = 18.81 m/s

7.

for automobile :

m = mass = 2500 lb

v = velocity = 70 mph = 102.7 ft/s

momentum of automobile = mv = 2500 x 102.7 = 256750 = 2.6 x 105kgm/s

kinetic energy = (0.5) m v2 = (0.5) (2500) (102.7)2 = 1.32 x 107 J

for projectile :

M = 100 lb

v = 2000 ft/s

momentum of projectile = mv = 100 x 2000 = 200000 kgm/s = 2 x 105 kgm/s

kinetic energy = (0.5) m v2 = (0.5) (100) (2000)2 = 2 x 108 J

Ratio of momentum = 2.6 x 105 /(2 x 105) = 1.3

Ratio of KE = 1.32 x 107 /( 2 x 108 ) = 0.066

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