Part (a) To determine Fraction of its initial energy is lost during the bounce. Answer Solution: 20% Explanation Given: Initial height of ball =2 .0 m Maximum height attained by ball after bounce =1 .6 m Formula Used: P.E of a body at height from a point ( reference point ) = m g h Conservation of mechanical energy. K E 2 + P E 2 = K E 1 + P E 1 When only conservative force is acting. Calculation: Initial height of ball h 1 = 2.0 m P . E 1 = m g h 1 Maximum height attained by ball is h 2 P E 2 = m g h 2 Loss in P.E ( P . E 2 ) = P . E 1 - P . E 2 = m g h 1 - m g h 2 = m g ( h 1 - h 2 ) . Fraction of its potential energy losses. = m g ( h 1 - h 2 ) m g h 1 × 100 = h 1 - h 2 h 1 × 100 = 2 - 1.6 2 × 100 = 0.4 2 × 100 = 20 % Conclusion: The fraction of energy lost during the bounce =20% Part (b) To determine The ball’s speed just before and just after the bounce. Answer Solution: 6.3 m/s (just before bounce), 5.6 m/s (just after bounce) Explanation Given: Initial height of ball =2 .0 m Maximum height attained by ball after bounce =1 .6 m Formula Used: P.E of a body at height from a point ( reference point ) =mgh Conservation of mechanical energy. K E 2 + P E 2 = K E 1 + P E 1 ( when only conservative force is acting ) Calculation: Before and after bounce only gravitational force is acting on ball (conservative force) so apply conservation of energy before bounce and after bounces. Initial mechanical energy of ball ( E i ) = P . E 1 + K . E 1 = m g h 1 + 1 2 m ( 0 ) 2 = m g h 1 . Mechanical energy of ball just before collision ( E f ) = P . E 2 + K . E 2 = m g ( 0 ) + 1 2 m ( v 2 ) 2 Reference position for potential energy is taken ground surface. E i = E f m g h 1 = 1 2 m ( v 2 ) 2 v 2 = 2 g h 1 = 2 × 9.80 × 2 =6 .26 m/s ≈ 6 .3 m/s Speed of ball just before bounce =6 .26m/s 2 after bounce Initial mechanical energy of ball ( E i ) = P . E 1 + K . E 1 = m g ( 0 ) + 1 2 m v 1 2 = 1 2 m v 1 2 Final mechanical energy of ball ( E f ) = P . E 2 + K . E 2 = m g h 2 + 1 2 m ( 0 ) 2 ( at the heighest point velocity=0 ) = m g h 2 1 2 m v 2 2 = m g h 2 ( conservation of energy ) v 2 = 2 g h 2 = 2 × 9.80 × 1.60 =5 .6 m/s . Conclusion: Speed of ball just after bounce =5 .6 m/s Part (c) To determine Where did the energy go Answer Solution: Into heat and sound energy Explanation Loss in mechanical energy is converted into heat and sound energy during collision with Earth’s surface. Conclusion: Loss in mechanical energy is converted into sound and heat energy.

6th Edition

ISBN: 9780130606204

Author: Douglas C. Giancoli

Publisher: Prentice Hall

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Question

Chapter 6, Problem 52P

Part (a)

To determine

Fraction of its initial energy is lost during the bounce.

Part (a)

Expert Solution

Answer to Problem 52P

Solution: 20%

Explanation of Solution

Given:

Initial height of ball $=2.0 m$

Maximum height attained by ball after bounce $=1.6 m$

Formula Used: P.E of a body at height from a point $(reference point)$ $=mgh$

Conservation of mechanical energy.

$KE2+PE2=KE1+PE1$

When only conservative force is acting.

Calculation:

Physics: Principles with Applications, Chapter 6, Problem 52P , additional homework tip 1

Initial height of ball $h1=2.0 m$

$P.E1=mgh1$

Maximum height attained by ball is $h2$

$PE2=mgh2$

Loss in P.E $(P.E2)=P.E1-P.E2$

$=mg h1-mg h2=mg(h1-h2).$

Fraction of its potential energy losses.

$=mg( h 1 - h 2 )mg h1×100=h1-h2h1×100=2-1.62×100=0.42×100=20%$

Conclusion:The fraction of energy lost during the bounce $=20%$

Part (b)

To determine

The ball’s speed just before and just after the bounce.

Part (b)

Expert Solution

Answer to Problem 52P

Solution: 6.3 m/s (just before bounce), 5.6 m/s (just after bounce)

Explanation of Solution

Given: Initial height of ball $=2.0 m$

Maximum height attained by ball after bounce $=1.6 m$

Formula Used: P.E of a body at height from a point $(reference point)$ $=mgh$

Conservation of mechanical energy.

$KE2+PE2=KE1+PE1$

$(when only conservative force is acting)$

Calculation:

Physics: Principles with Applications, Chapter 6, Problem 52P , additional homework tip 2

Before and after bounce only gravitational force is acting on ball (conservative force) so apply conservation of energy before bounce and after bounces.

Initial mechanical energy of ball $(Ei)$

$=P.E1+K.E1=mg h1+12m(0)2=mg h1.$

Mechanical energy of ball just before collision $(Ef)$

$=P.E2+K.E2=mg(0)+12m( v 2)2$

Reference position for potential energy is taken ground surface.

$Ei=Efmg h1=12m( v 2)2v2=2gh1 =2×9.80×2 =6.26 m/s≈6.3 m/s$

Speed of ball just before bounce $=6.26m/s2$ after bounce

Initial mechanical energy of ball $(Ei)$

$=P.E1+K.E1=mg(0)+12mv12=12mv12$

Final mechanical energy of ball $(Ef)$

$=P.E2+K.E2=mg h2+12m(0)2 (at the heighest point velocity=0)=mg h2$

$12mv22=mgh2 (conservation of energy)v2=2gh2 =2×9.80×1.60 =5.6 m/s.$

Conclusion:

Speed of ball just after bounce $=5.6 m/s$

Part (c)

To determine

Where did the energy go

Part (c)

Expert Solution

Answer to Problem 52P

Solution:Into heat and sound energy

Explanation of Solution

Loss in mechanical energy is converted into heat and sound energy during collision with Earth’s surface.

Conclusion:

Loss in mechanical energy is converted into sound and heat energy.

Chapter 6, Problem 51P

Chapter 6, Problem 53P

Chapter 6 Solutions

Physics: Principles with Applications

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