Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Textbook Question
Chapter 9, Problem 37AP

Review. A 60.0-kg person running at an initial speed of 4.00 m/s jumps onto a 120-kg cart initially at rest (Fig. P9.37). The person slides on the cart’s top surface and finally comes to rest relative to the cart. The coefficient of kinetic friction between the person and the cart is 0.400. Friction between the cart and ground can be ignored. (a) Find the final velocity of the person and cart relative to the ground. (b) Find the friction force acting on the person while he is sliding across the top surface of the cart. (c) How long does the friction force act on the person? (d) Find the change in momentum of the person and the change in momentum of the cart. (c) Determine the displacement of the person relative to the ground while he is sliding on the cart. (f) Determine the displacement of the cart relative to the ground while the person is sliding. (g) Find the change in kinetic energy of the person. (h) Find the change in kinetic energy of the cart. (i) Explain why the answers to (g) and (h) differ. (What kind of collision is this one, and what accounts for the loss of mechanical energy)

Figure P9.37

Chapter 9, Problem 37AP, Review. A 60.0-kg person running at an initial speed of 4.00 m/s jumps onto a 120-kg cart initially

(a)

Expert Solution
Check Mark
To determine

The final velocity of the person and cart relative to the ground.

Answer to Problem 37AP

The final velocity of the person and cart relative to the ground is 1.33i^m/s.

Explanation of Solution

The mass of the person is 60.0kg, and the initial speed is 4.00m/s. The mass of the cart is 120kg and the cart is at rest. The coefficient of kinetic friction between the cart and the person is 0.400.

Write the expression of conservation of momentum.

    mpvi=(mp+mc)vf

Here, mp is the mass of the person, mc is the mass of the cart, vi is the initial velocity of the person and vf is the final velocity of the person and cart relative to the ground.

Substitute 60.0kg for mp, 4.00m/s for vi and 120kg for mc in above equation to find vf.

  60.0kg×4.00m/s=(60.0kg+120kg)vfvf=1.33m/s=1.33i^m/s

Conclusion:

Therefore, the final velocity of the person and cart relative to the ground is 1.33i^m/s.

(b)

Expert Solution
Check Mark
To determine

The frictional force acting on the person while he is sliding.

Answer to Problem 37AP

The frictional force acting on the person while he is sliding is 235i^N.

Explanation of Solution

Write the expression to calculate the frictional force.

    fk=μmpg

Here, μ is the coefficient of kinetic friction between the cart and the person and g is the acceleration due to gravity.

Substitute 0.400 for μ, 60.0kg for mp and 9.81m/s2 for g in above equation to find fk.

    fk=0.400×60.0kg×9.81m/s2=235N=235i^N

The negative sign indicates that the frictional force is acting toward negative x axis.

Conclusion:

Therefore, the frictional force acting on the person while he is sliding is 235i^N.

(c)

Expert Solution
Check Mark
To determine

The time duration in which the frictional force is acting on the person.

Answer to Problem 37AP

The time duration in which the frictional force is acting on the person is 0.680s.

Explanation of Solution

Write the expression of Impulse-momentum equation.

    fkΔt=mp(vfvi)

Here, Δt is the time duration in which the frictional force is acting on the person.

Substitute 235N for fk, 1.33m/s for vf, 4.00m/s for vi and 60.0kg for mp in above equation to find Δt.

    235N×Δt=60.0kg×(1.33m/s4.00m/s)Δt=0.680s

Conclusion:

Therefore, the time duration in which the frictional force is acting on the person is 0.680s.

(d)

Expert Solution
Check Mark
To determine

The change in momentum of the person and cart.

Answer to Problem 37AP

The change in momentum of the person and cart is 160i^Ns and 160i^Ns respectively.

Explanation of Solution

Write the expression to calculate the change in momentum of the person.

    Δpp=mp(vfvi)

Substitute 1.33i^m/s for vf, 4.00i^m/s for vi and 60.0kg for mp in above equation to find Δpp.

    Δpp=60.0kg×(1.334.00)i^m/s160i^Ns

Write the expression to calculate the change in momentum of the cart.

    Δpc=mc(vfvi)

Substitute 1.33i^m/s for vf, 0 for vi and 120kg for mc in above equation to find Δpc.

    Δpp=120kg×(1.330)i^m/s160i^Ns

Conclusion:

Therefore, the change in momentum of the person and cart is 160i^Ns.and 160i^Ns respectively.

(e)

Expert Solution
Check Mark
To determine

The displacement of the person relative to the ground during sliding on the cart.

Answer to Problem 37AP

The displacement of the person relative to the ground during sliding on the cart is 1.81m.

Explanation of Solution

Write the expression to calculate the displacement of the person.

    Δdp=12(vi+vf)Δt

Substitute 1.33i^m/s for vf, 4.00i^m/s for vi and 0.680s for Δt in above equation to find Δdp.

    Δdp=12(1.33i^m/s+4.00i^m/s)×0.680s=1.81m

Conclusion:

Therefore, the displacement of the person relative to the ground during sliding on the cart is 1.81m.

(f)

Expert Solution
Check Mark
To determine

The displacement of the cart relative to the ground during the person sliding on the cart.

Answer to Problem 37AP

The displacement of the cart relative to the ground during the person sliding on the cart is 0.454m.

Explanation of Solution

Write the expression to calculate the displacement of the cart.

    Δdc=12(vi+vf)Δt

Substitute 1.33i^m/s for vf, 0 for vi and 0.680s for Δt in above equation to find Δdc.

    Δdc=12(1.33i^m/s+0)×0.680s=0.454m

Conclusion:

Therefore, the displacement of the cart relative to the ground during the person sliding on the cart is 0.454m.

(g)

Expert Solution
Check Mark
To determine

The change in kinetic energy of the person.

Answer to Problem 37AP

The change in kinetic energy of the person is 427J.

Explanation of Solution

Write the expression to calculate the change in kinetic energy of the person.

    ΔKEP=12mp(vf2vi2)

Substitute 1.33m/s for vf, 4.00m/s for vi and 60.0kg for mp in above equation to find ΔKEP.

`    ΔKEP=12×60.0kg×[(1.33m/s)2(4.00m/s)2]=427J

Conclusion:

Therefore, the change in kinetic energy of the person is 427J.

(h)

Expert Solution
Check Mark
To determine

The change in kinetic energy of the cart.

Answer to Problem 37AP

The change in kinetic energy of the cart is 107J.

Explanation of Solution

Write the expression to calculate the change in kinetic energy of the cart.

    ΔKEc=12mc(vf2vi2)

Substitute 1.33m/s for vf, 0 for vi and 120kg for mc in above equation to find ΔKEc.

`    ΔKEc=12×120kg×[(1.33m/s)20]107J

Conclusion:

Therefore, the change in kinetic energy of the cart is 107J.

(i)

Expert Solution
Check Mark
To determine

The reason due to which the answer in part (g) and (h) are different.

Answer to Problem 37AP

The collision between the person and the cart is perfectly inelastic collision and the loss of energy is due to frictional force.

Explanation of Solution

The force acting on the person must be equal in magnitude and opposite in direction to the force exerted by the cart on the person.

According to the conservation of linear momentum, the change in momentum of the person and the cart must be equal in magnitude and must add to zero. The change in kinetic energy of the person and cart must be equal for elastic collision but in this case change in kinetic energy of the person and cart is not equal, which shows that this is inelastic collision.

The reason of change in kinetic energy of both object of not being same is, the displacement of person and the cart is not same due to frictional force acting on the person.

Due to the frictional force the loss of energy in form of heat is the internal energy.

Write the expression to calculate the internal energy.

    E=KEpKEc

Substitute 427J for KEp and 107J for KEc in above equation.

    E=427J107J=320J

Thus, the collision between the person and the cart is perfectly inelastic collision due to the loss of energy.

Conclusion:

Therefore, the collision between the person and the cart is perfectly inelastic collision and the loss of energy is due to frictional force.

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Chapter 9 Solutions

Physics for Scientists and Engineers

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