New Tab x) WileyPLUS s:/ledugen.wileyplus.com/edugen/student/mainfr.uni Gradeboo

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New Tab x) WileyPLUS s:/ledugen.wileyplus.com/edugen/student/mainfr.uni Gradebook ment MESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VERSION BACK Chapter 07, Problem 45 Starting with an initial speed of 4.58 m/s at a height of 0.426 m, a 1.62-kg ball swings downward and strikes a 4.55-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 1.62-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 1.62-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.55-kg ball just after the collision. (d) How high does the 1.62-kg ball swing after the collision, ignoring air resistance? (e) How high does the 4.55-kg ball swing after the collision, ignoring air resistance? (a) Number (b) Number (c) Number (d) Number Units Units Units Units Units (e) Number blem K TO TEXT

Explanation / Answer

a)

h = height from which the ball was released = 0.426 m

vo = initial velocity at the time of release = 4.58 m/s

vf = final velocity just before impact

m = mass = 1.62 kg

using conservation of energy

(0.5) m vo2 + mgh = (0.5) m vf2

(0.5) vo2 + gh = (0.5) vf2

(0.5) (4.58)2 + (9.8 x 0.426) = (0.5) vf2

vf = 5.4 m/s

b)

m = mass of lighter ball = 1.62 kg

vo = initial velocity of ligher ball just before impact = 5.4 m/s

vf = final velocity of ligher ball just after impact = ?

M = mass of the bigger ball = 4.55 kg

Vo = initial velocity of bigger ball before impact = 0 m/s

Vf = final velocity of bigger ball after impact = ?

Using conservation of momentum '

m vo + M Vo = m vf + M Vf

(1.62) (5.4) + (4.55) (0) = 1.62 vf + 4.55 Vf

Vf = (8.75 - 1.62 vf )/4.55 eq-1

Using conservation of kinetic energy

m v2o + M V2o = m v2f + M V2f

(1.62) (5.4)2 + (4.55) (0)2 = 1.62 v2f + 4.55 V2f

using eq-1

(1.62) (5.4)2 = 1.62 v2f + (4.55) ((8.75 - 1.62 vf )/4.55)2

vf = - 2.6 m/s

negative sign indicates opposite direction

c)

using eq-1

Vf = (8.75 - 1.62 vf )/4.55 =  (8.75 - 1.62 (- 2.6))/4.55 = 2.85 m/s

d)

h' = height gained by the lighter ball

h' = vf2 /(2g)

h' = (- 2.6)2/(2 x 9.8) = 0.34 m

e)

H = height gained by the lighter ball

H = Vf2 /(2g)

H = (2.85)2/(2 x 9.8) = 0.41 m

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