Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 8, Problem 49P

(a)

To determine

The final velocity of the person and the cart.

(a)

Expert Solution
Check Mark

Answer to Problem 49P

The final velocity of the person and the cart is vf=1.33m/si^.

Explanation of Solution

Write the expression for conservation of momentum for inelastic collision.

  mpvi=(mp+mc)vf        (I)

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

Conclusion:

Substitute 60kg for mp, 4.00m/s for vi, and 120kg for mc in the equation (I) to find vf.

  (60kg)(4.00m/s)=(60kg+120kg)vf240kgm/s=(180kg)vfvf=1.33m/s

In vector notation the velocity of the cart and person is,

  vf=1.33m/si^

Therefore, the final velocity of the person and the cart is vf=1.33m/si^.

(b)

To determine

The friction force acting on the person.

(b)

Expert Solution
Check Mark

Answer to Problem 49P

The friction force acting on the person is 235Ni^.

Explanation of Solution

Write the expression for normal force by using Newton’s second law in y direction

  Fy=0Nmg=0        (II)

Here, N is the normal force exerted on the person, m is the mass of the person, and g is the acceleration due to gravity.

Write the expression for frictional force exerted on the person.

  fk=μkN        (III)

Here, fk is the frictional force and μk is the coefficient of kinetic friction.

Conclusion:

Substitute 60kg for m and 9.80m/s2 for g in equation (II) to find N.

  N(60kg)(9.80m/s2)=0N588N=0N=588N

Substitute 588N for N and 0.400 for μk in equation (III) to find fk.

  fk=(588N)(0.400)=235N

In vector notation the frictional force is,

  fk=235Ni^

Therefore, the friction force acting on the person is 235Ni^.

(c)

To determine

The time taken for the frictional force acting on the person.

(c)

Expert Solution
Check Mark

Answer to Problem 49P

The time taken for the frictional force acting on the person is 0.680s.

Explanation of Solution

Write the expression for person’s momentum equal to the impulse.

  pi+I=pf        (IV)

Here, pi is the initial momentum, I is the impulse, and pf is the final momentum.

Write the expression for initial momentum.

  pi=mvi        (V)

Here, m is the mass of the person and vi is the initial speed of the person.

Write the expression for final momentum.

  pf=mvf        (VI)

Here, vf is the final speed of the person and cart.

Write the expression for impulse.

  I=Ft        (VII)

Here, F is the force exerted on the person and t is the time taken for the frictional force acting on the person.

Conclusion:

Substitute the equations (V), (VI) and (VII) in equation (IV).

  mvi+Ft=mvf

Substitute 60kg for m, 4.00m/s for vi, 235N for F, and 1.33m/s for vf in above relation to find t.

  (60kg)(4.00m/s)(235N)t=(60kg)(1.33m/s)240kgm/s(235N)t=79.8kgm/s(235N)t=79.8kgm/s240kgm/st=0.680s

Therefore, the time taken for the frictional force acting on the person is 0.680s.

(d)

To determine

The change in momentum of the person and the cart.

(d)

Expert Solution
Check Mark

Answer to Problem 49P

The change in momentum of the person is 160Nsi^ and change in momentum of the cart is +160Nsi^.

Explanation of Solution

Write the expression for change in momentum of the person.

  pp=m(vfvi)        (VIII)

Write the expression for change in momentum of the cart.

  pc=m(vfvi)        (IX)

Conclusion:

Substitute 60kg for m, 4.00m/s for vi, and 1.33m/s for vf in equation (VIII) to find pp.

  pp=60kg(1.33m/s4.00m/s)=160Nsi^

Substitute 120kg for m, 0m/s for vi, and 1.33m/s for vf in equation (IX) to find pc.

  pc=120kg(1.33m/s0m/s)=+160Nsi^

Therefore, the change in momentum of the person is 160Nsi^ and change in momentum of the cart is +160Nsi^.

(e)

To determine

The displacement of the person relative to the ground.

(e)

Expert Solution
Check Mark

Answer to Problem 49P

The displacement of the person relative to the ground is 1.81m.

Explanation of Solution

Write the expression for displacement of the person relative to the ground.

  xfxi=12(vi+vf)t        (X)

Here, xi is the initial distance and xf is the final distance.

Conclusion:

Substitute 4.00m/s for vi, 0.680s for t, and 1.33m/s for vf in equation (X) to find xfxi.

  xfxi=12(4.00m/s+1.33m/s)(0.680s)=1.81m

Therefore, the displacement of the person relative to the ground is 1.81m.

(f)

To determine

The displacement of the cart relative to the ground.

(f)

Expert Solution
Check Mark

Answer to Problem 49P

The displacement of the cart relative to the ground is 0.454m.

Explanation of Solution

Write the expression for displacement of the cart relative to the ground.

  xfxi=12(vi+vf)t        (XI)

Conclusion:

Substitute 0m/s for vi, 0.680s for t, and 1.33m/s for vf in equation (XI) to find xfxi.

  xfxi=12(0m/s+1.33m/s)(0.680s)=0.454m

Therefore, the displacement of the cart relative to the ground is 0.454m.

(g)

To determine

The change in kinetic energy of the person.

(g)

Expert Solution
Check Mark

Answer to Problem 49P

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

Explanation of Solution

Write the expression for change in kinetic energy of the person.

  ΔKp=12m(vf2vi2)        (XII)

Conclusion:

Substitute 4.00m/s for vi, 60kg for m, and 1.33m/s for vf in equation (XII) to find ΔKp.

  ΔKp=1260kg((1.33m/s)2(4.00m/s)2)=427J

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

(h)

To determine

The change in kinetic energy of the cart.

(h)

Expert Solution
Check Mark

Answer to Problem 49P

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

Explanation of Solution

Write the expression for change in kinetic energy of the cart.

  ΔKc=12m(vf2vi2)        (XIII)

Conclusion:

Substitute 0m/s for vi, 120kg for m, and 1.33m/s for vf in equation (XIII) to find ΔKc.

  ΔKc=12120kg((1.33m/s)2(0m/s)2)=107J

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

(i)

To determine

Why the answers part (g) and (h) are differ.

(i)

Expert Solution
Check Mark

Answer to Problem 49P

Because. the distance moved by the cart is different from the distance moved by the point of application of friction force to the cart.

Explanation of Solution

The force exerted by the person on the cart must be equal in magnitude and opposite in direction to the force exerted by the cart on the person. The changes in momentum of the two objects must be equal in magnitude and must be added to zero.

The change in kinetic energy is different in magnitude and does not add to zero.

Conclusion:

The following situation is represents in two ways,

The distance moved by the cart is different from the distance moved by the point of application of friction force to the cart.

The total change in mechanical energy for both objects add together becomes zero, it is perfectly in elastic collision.

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Chapter 8 Solutions

Principles of Physics: A Calculus-Based Text

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