Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 11, Problem 38AP

(a)

To determine

The velocity of Jacob and Ethan immediate thereafter.

(a)

Expert Solution
Check Mark

Answer to Problem 38AP

The velocity of Jacob and Ethan immediate thereafter is (0.25 m/s)i^ .

Explanation of Solution

Given information:

The mass of Jacob is 45.0 kg , the mass of Ethan is 31.0 kg , the speed of Jacob is 8.0 m/s and the speed of Ethan is 11.0 m/s .

The momentum is conserved in the isolated system of two boys then,

pi=pfm1v1i^m2v2i^=(m1+m2)vf

  • m1 is the mass of Jacob.
  • m2 is the mass of Ethan.
  • v1 is the speed of Jacob.
  • v2 is the speed of Ethan.
  • vf is the final velocity vector.

Substitute 45.0 kg for m1 , 31.0 kg for m2 , 8.0 m/s for v1 and 11.0 m/s for v2 in above equation to find vf .

(45.0 kg)(8.0 m/s)i^(31.0 kg)(11.0 m/s)i^=(45.0 kg+31.0 kg)vf(19.0 kgm/s)i^=(76.0 kg)vfvf=(0.25 m/s)i^

Conclusion:

Therefore, the velocity of Jacob and Ethan immediate thereafter is (0.25 m/s)i^ .

(b)

To determine

The fraction of their original kinetic energy is still mechanical energy after their collision.

(b)

Expert Solution
Check Mark

Answer to Problem 38AP

The fraction of their original kinetic energy is still mechanical energy after their collision is 0.00071 .

Explanation of Solution

Section 1:

To determine: The initial kinetic energy of the system.

Answer: The initial kinetic energy of the system is 3315.5 J .

Given information:

The mass of Jacob is 45.0 kg , the mass of Ethan is 31.0 kg , the speed of Jacob is 8.0 m/s and the speed of Ethan is 11.0 m/s .

The formula to calculate initial kinetic energy of the system is,

KEi=12m1v12+12m2v22

Substitute 45.0 kg for m1 , 31.0 kg for m2 , 8.0 m/s for v1 and 11.0 m/s for v2 in above equation to find KEi .

KEi=12(45.0 kg)(8.0 m/s)2+12(31.0 kg)(11.0 m/s)2=3315.5 J

Section 2:

To determine: The final kinetic energy of the system.

Answer: The final kinetic energy of the system is 2.375 J .

Given information:

The mass of Jacob is 45.0 kg , the mass of Ethan is 31.0 kg , the speed of Jacob is 8.0 m/s and the speed of Ethan is 11.0 m/s .

The formula to calculate final kinetic energy of the system is,

KEf=12(m1+m2)vf2

Substitute 45.0 kg for m1 , 31.0 kg for m2 and 0.25 m/s for vf in above equation to find KEf .

KEf=((45.0 kg)+(31.0 kg))(0.25 m/s)2=2.375 J

Section 3:

To determine: The fraction of kinetic energy of the system.

Answer: The fraction of energy of the system is 0.00071 .

Given information:

The mass of Jacob is 45.0 kg , the mass of Ethan is 31.0 kg , the speed of Jacob is 8.0 m/s and the speed of Ethan is 11.0 m/s .

The formula to calculate fraction of kinetic energy is,

KEfKEi=2.375 J3315.5 J=0.00071

Conclusion:

Therefore, the fraction of their original kinetic energy is still mechanical energy after their collision is 0.00071 .

(c)

To determine

The velocity of the centre of mass of Jacob and Ethan.

(c)

Expert Solution
Check Mark

Answer to Problem 38AP

The velocity of the centre of mass of Jacob and Ethan is (0.25 m/s)i^ .

Explanation of Solution

Given information:

The mass of Jacob is 45.0 kg , the mass of Ethan is 31.0 kg , the speed of Jacob is 8.0 m/s and the speed of Ethan is 11.0 m/s .

The velocity of the centre of mass of Jacob and Ethan is still remains same as calculated in part (a) because the conservation of momentum calculations will be same as part (a).

Then, the velocity of the centre of mass of Jacob and Ethan is,

vf=(0.25 m/s)i^

Conclusion:

Therefore, the velocity of the centre of mass of Jacob and Ethan is (0.25 m/s)i^ .

(d)

To determine

The angular speed of Jacob and Ethan.

(d)

Expert Solution
Check Mark

Answer to Problem 38AP

The angular speed of Jacob and Ethan is 15.8 rad/s .

Explanation of Solution

Section 1:

To determine: The position of the centre of mass of the boys.

Answer: The position of the centre of mass of the boys is 0.489 m .

Given information:

The mass of Jacob is 45.0 kg , the mass of Ethan is 31.0 kg , the speed of Jacob is 8.0 m/s and the speed of Ethan is 11.0 m/s .

The position of the centre of mass of the boys is,

yCM=m1y1+m2L(m1+m2)

Substitute 45.0 kg for m1 , 31.0 kg for m2 , 0 for y1 and 1.20 m for L in above equation.

yCM=(45.0 kg)(0)+(31.0 kg)(1.20 m)(45.0 kg+31.0 kg)=0.489 m

Section 2:

To determine: The angular speed of Jacob and Ethan.

Answer: The angular speed of Jacob and Ethan is 15.8 rad/s .

Given information:

The mass of Jacob is 45.0 kg , the mass of Ethan is 31.0 kg , the speed of Jacob is 8.0 m/s and the speed of Ethan is 11.0 m/s .

The Jacob is yCM=0.489 m from the centre of mass and Ethan is (LyCM)=0.711 m from the centre of mass.

The angular momentum is,

m1v1yCM+m2v2(LyCM)=(m1yCM2+m2(LyCM)2)ωω=m1v1yCM+m2v2(LyCM)(m1yCM2+m2(LyCM)2)

Substitute 45.0 kg for m1 , 31.0 kg for m2 , 8.0 m/s for v1 , 11.0 m/s for v2 , 0.711 m for (LyCM) and 0.489 m for yCM in above equation.

ω=(45.0 kg)(8.0 m/s)(0.489 m)+(31.0 kg)(11.0 m/s)(0.711 m)((45.0 kg)(0.489 m)2+(31.0 kg)(0.711 m)2)=418 kgm2/s26.4 kgm2=15.8 rad/s

Conclusion:

Therefore, the angular speed of Jacob and Ethan is 15.8 rad/s .

(e)

To determine

The fraction of their original kinetic energy that is still mechanical energy after they link arms.

(e)

Expert Solution
Check Mark

Answer to Problem 38AP

The fraction of their original kinetic energy that is still mechanical energy after they link arms is 1 .

Explanation of Solution

Section 1:

To determine: The initial and final kinetic energy of the system when they link arms.

Answer: The initial and final kinetic energy of the system when they link arms is 3315.5 J .

Given information:

The mass of Jacob is 45.0 kg , the mass of Ethan is 31.0 kg , the speed of Jacob is 8.0 m/s and the speed of Ethan is 11.0 m/s .

Refer to the section 1 of part (b), the initial kinetic energy is,

KEi=3315.5 J

The formula to calculate final kinetic energy of the system is,

KEf=12(m1+m2)vCM2+12(m1yCM2+m2(LyCM)2)ω2

Substitute 45.0 kg for m1 , 31.0 kg for m2 , 0.25 m/s for vCM , 0.711 m for (LyCM) , 0.489 m for yCM and 15.8 rad/s for ω in above equation to find KEf .

KEf=[12((45.0 kg)+(31.0 kg))(0.25 m/s)2+12((45.0 kg)(0.489 m)2+(31.0 kg)(0.711 m)2)(15.8 rad/s)2]=12(76.0 kg)(0.25 m/s)2+12(26.4 kgm2)(15.8 rad/s)2=3315.5 J

Section 3:

To determine: The fraction of kinetic energy of the system.

Answer: The fraction of energy of the system is 1 .

Given information:

The mass of Jacob is 45.0 kg , the mass of Ethan is 31.0 kg , the speed of Jacob is 8.0 m/s and the speed of Ethan is 11.0 m/s .

The fraction of kinetic energy is,

KEfKEi=3315.5 J3315.5 J=1

Conclusion:

Therefore, the fraction of their original kinetic energy is still mechanical energy after they link arms is 1 .

(f)

To determine

To explain: The reason for the answer of part (b) and part (e) is so different.

(f)

Expert Solution
Check Mark

Answer to Problem 38AP

the answer of part (b) and part (e) is so different because the head on collision between similarly sized objects are grossly inefficient.

Explanation of Solution

The deformation is a process in which one form of energy changed into the other form. The head on collision between similarly sized objects are grossly inefficient. So the answers are so different. If the two kids were the same size and had the same velocity; conservation of the momentum states that they lose all their kinetic energy. On the other hand, glancing blows like this one allow a lot of energy to be transferred into the rotational kinetic energy, causing much less energy to be lost on impact.

Conclusion:

Therefore, the answer of part (b) and part (e) is so different because the head on collision between similarly sized objects are grossly inefficient.

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Chapter 11 Solutions

Physics for Scientists and Engineers with Modern Physics

Ch. 11 - A particle is located at a point described by the...Ch. 11 - A 1.50-kg particle moves in the xy plane with a...Ch. 11 - Prob. 9PCh. 11 - Heading straight toward the summit of Pikes Peak,...Ch. 11 - Review. A projectile of mass m is launched with an...Ch. 11 - Prob. 12PCh. 11 - A particle of mass m moves in a circle of radius R...Ch. 11 - A 5.00-kg particle starts from the origin at time...Ch. 11 - A ball having mass m is fastened at the end of a...Ch. 11 - Prob. 16PCh. 11 - A uniform solid disk of mass m = 3.00 kg and...Ch. 11 - Show that the kinetic energy of an object rotating...Ch. 11 - Prob. 19PCh. 11 - Prob. 20PCh. 11 - Prob. 21PCh. 11 - Prob. 22PCh. 11 - A 60.0-kg woman stands at the western rim of a...Ch. 11 - Prob. 24PCh. 11 - A uniform cylindrical turntable of radius 1.90 m...Ch. 11 - Prob. 26PCh. 11 - A wooden block of mass M resting on a...Ch. 11 - Prob. 28PCh. 11 - A wad of sticky clay with mass m and velocity vi...Ch. 11 - A 0.005 00-kg bullet traveling horizontally with a...Ch. 11 - The angular momentum vector of a precessing...Ch. 11 - A light rope passes over a light, frictionless...Ch. 11 - Prob. 33APCh. 11 - Prob. 34APCh. 11 - We have all complained that there arent enough...Ch. 11 - Prob. 36APCh. 11 - A rigid, massless rod has three particles with...Ch. 11 - Prob. 38APCh. 11 - Two astronauts (Fig. P11.39), each having a mass...Ch. 11 - Two astronauts (Fig. P11.39), each having a mass...Ch. 11 - Native people throughout North and South America...Ch. 11 - Two children are playing on stools at a restaurant...Ch. 11 - You are attending a county fair with your friend...Ch. 11 - Prob. 44APCh. 11 - Global warming is a cause for concern because even...Ch. 11 - The puck in Figure P11.46 has a mass of 0.120 kg....Ch. 11 - Prob. 47APCh. 11 - A solid cube of wood of side 2a and mass M is...Ch. 11 - Prob. 49CPCh. 11 - Prob. 50CP
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