Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)
Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)
9th Edition
ISBN: 9781305116429
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 11, Problem 11.46AP

Review. Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45.0 kg, is gliding to the right at 8.00 m/s, and Ethan, mass 31.0 kg, is gliding to the left at 11.0 m/s along the same line. When they meet, they grab each other and hang on. (a) What is their velocity immediately thereafter? (b) What fraction of their original kinetic energy is still mechanical energy after their collision? That was so much fun that the boys repeat the collision with the same original velocities, this time moving along parallel lines 1.20 m apart. At closest approach, they lock arms and start rotating about their common center of mass. Model the boys as particles and their arms as a cord that does not stretch. (c) Find the velocity of their center of mass. (d) Find their angular speed. (e) What fraction of their original kinetic energy is still mechanical energy after they link arms? (f) Why are the answers to parts (b) and (e) so different?

(a)

Expert Solution
Check Mark
To determine

The velocity of Jacob and Ethan immediate thereafter.

Answer to Problem 11.46AP

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

Explanation of Solution

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

Here, m1 is the mass of Jacob, m2 is the mass of Ethan, v1 is the speed of Jacob, v2 is the speed of Ethan and 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)

Expert Solution
Check Mark
To determine

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

Answer to Problem 11.46AP

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

Explanation of Solution

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

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

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)

Expert Solution
Check Mark
To determine

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

Answer to Problem 11.46AP

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

Explanation of Solution

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)

Expert Solution
Check Mark
To determine

The angular speed of Jacob and Ethan.

Answer to Problem 11.46AP

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

Explanation of Solution

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

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)

Expert Solution
Check Mark
To determine

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

Answer to Problem 11.46AP

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

Explanation of Solution

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

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)

Expert Solution
Check Mark
To determine

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

Answer to Problem 11.46AP

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, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)

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