The velocity of the system just after the collision using Newton’s second law and the comparison of the answer with the result of Example 11.4.

Question

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Answer 1

Question

Chapter 11, Problem 32PQ

To determine

The velocity of the system just after the collision using Newton’s second law and the comparison of the answer with the result of Example 11.4.

Expert Solution & Answer

Answer to Problem 32PQ

The velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

Explanation of Solution

The free-body diagram of the system is shown in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 11, Problem 32PQ

The two-train system only moves in $x$ direction.

Write the expression for the Newton’s second law in $x$ direction.

$ΣFx=Max$ (I)

Here, $ΣFx$ is the magnitude of the net force in $x$ direction, $M$ is the mass of the two-train system and $ax$ is the acceleration in $x$ direction.

Refer to figure 1. The only force acting in $x$ direction is the force of kinetic friction and it acts in $−x$ direction.

Write the equation for $ΣFx$ .

$ΣFx=−Fk$

Here, $Fk$ is the kinetic friction.

Put the above equation in equation (I).

$−Fk=Max$ (II)

The kinetic friction is proportional to the normal force and the normal force is in turn equal to the weight of the train.

Write the expression for $Fk$ .

$Fk=μkFg$ (III)

Here, $μk$ is the coefficient of kinetic friction and $Fg$ is the weight of the train.

Write the equation for $Fg$ .

$Fg=Mg$

Here, $g$ is the acceleration due to gravity.

Put the above equation in equation (III).

$Fk=μkMg$

Put the above equation in equation (II) and rewrite it for $ax$ .

$−μkMg=Maxax=−μkg$ (IV)

Initially only the freight train has momentum. Assume that the two trains move together with velocity $vc$ after the collision. Apply the conservation of momentum for the two-train system before and after the collision.

$m1v1ii^=Mvci^$

Here, $m1$ is the mass of the freight train, $v1i$ is the initial speed of the freight train, $m2$ is the mass of the passenger train and $vc$ is the speed with the two trains move after collision.

Rewrite the above equation for $vc$ .

$vc=m1Mv1i$

Now consider the motion of the system just after the collision to the moment it comes to rest.

Replace $vc$ in the above equation by $vx,0$ where $vx,0$ denotes the speed of the system just after the collision.

$vx,0=m1Mv1i$ (V)

The speed of the two-train system as it comes to rest is zero.

Write the expression for the final speed of the system as it comes to rest.

$vx=0$ (VI)

Here, $vx$ is the final speed of the system as it comes to rest.

Write the constant-acceleration equation of motion.

$vx2=vx,02+2axΔx$

Here, $Δx$ is the distance travelled by the two-train system before it comes to rest.

Put equations (IV) to (VI) in the above equation and rewrite it for $v1i$ .

$0=(m1Mv1i)2+2(−μkg)Δxm1Mv1i=2μkgΔxv1i=Mm12μkgΔx$ (VII)

To find the range of possible initial freight train velocities, the extreme values of the coefficient of kinetic friction must be used.

Write the equation for $M$ .

$M=m1+m2$ (VIII)

Here, $m2$ is the mass of the passenger train.

Conclusion:

Given that the value of $m1$ is $5.221×106 kg$ , the value of $m2$ is $4.1×105 kg$ , the minimum value of the coefficient of kinetic friction is $0.3$ and its maximum value is $0.4$ and the distance travelled by the system before getting stopped is $370 ft$ . The value of $g$ is $9.81 m/s2$ .

Substitute $5.221×106 kg$ for $m1$ and $4.1×105 kg$ for $m2$ in equation (VIII) to find the value of $M$ .

$M=5.221×106 kg+4.1×105 kg=5.63×106 kg$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.3$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the minimum value of the speed of the two-train system.

$v1i(min)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(1.1×102 m)=9 m/s$

Here, $v1i(min)$ is the minimum speed of the system.

The system moves in $x$ direction. Write the expression for the minimum velocity of the system.

$v→1i(min)=v1i(min)i^$

Here, $v→1i(min)$ is the minimum velocity of the system.

Substitute $9 m/s$ for $v1i(min)$ in the above equation to find $v→1i(min)$ .

$v→1i(min)=9i^ m/s$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.4$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the maximum value of the speed of the two-train system.

$v1i(max)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(1.1×102 m)=30 m/s$

Here, $v1i(max)$ is the maximum speed of the system.

It is given that the train crosses the red signal at a speed of $48 mph$ . Convert this speed into units of m/s.

$48 mph=48 mph1609.344 m1 mile1 h(60×60) s=21 m/s$

Since the train passed the red signal with the speed of $21 m/s$ , the maximum speed cannot be greater than this and the upper limit will be replaced by this.

Write the expression for the maximum velocity of the system.

$v→1i(max)=v1i(max)i^$

Here, $v→1i(max)$ is the maximum velocity of the system.

Substitute $21 m/s$ for $v1i(max)$ in the above equation to find $v→1i(max)$ .

$v→1i(max)=21i^ m/s$

The range of the speed of the two-train system is $(9≤v→1i≤21)i^ m/s$ and the result obtained in Example 11.4 is the same.

Therefore, the velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

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Answer 2

Question

Chapter 11, Problem 32PQ

To determine

The velocity of the system just after the collision using Newton’s second law and the comparison of the answer with the result of Example 11.4.

Expert Solution & Answer

Answer to Problem 32PQ

The velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

Explanation of Solution

The free-body diagram of the system is shown in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 11, Problem 32PQ

The two-train system only moves in $x$ direction.

Write the expression for the Newton’s second law in $x$ direction.

$ΣFx=Max$ (I)

Here, $ΣFx$ is the magnitude of the net force in $x$ direction, $M$ is the mass of the two-train system and $ax$ is the acceleration in $x$ direction.

Refer to figure 1. The only force acting in $x$ direction is the force of kinetic friction and it acts in $−x$ direction.

Write the equation for $ΣFx$ .

$ΣFx=−Fk$

Here, $Fk$ is the kinetic friction.

Put the above equation in equation (I).

$−Fk=Max$ (II)

The kinetic friction is proportional to the normal force and the normal force is in turn equal to the weight of the train.

Write the expression for $Fk$ .

$Fk=μkFg$ (III)

Here, $μk$ is the coefficient of kinetic friction and $Fg$ is the weight of the train.

Write the equation for $Fg$ .

$Fg=Mg$

Here, $g$ is the acceleration due to gravity.

Put the above equation in equation (III).

$Fk=μkMg$

Put the above equation in equation (II) and rewrite it for $ax$ .

$−μkMg=Maxax=−μkg$ (IV)

Initially only the freight train has momentum. Assume that the two trains move together with velocity $vc$ after the collision. Apply the conservation of momentum for the two-train system before and after the collision.

$m1v1ii^=Mvci^$

Here, $m1$ is the mass of the freight train, $v1i$ is the initial speed of the freight train, $m2$ is the mass of the passenger train and $vc$ is the speed with the two trains move after collision.

Rewrite the above equation for $vc$ .

$vc=m1Mv1i$

Now consider the motion of the system just after the collision to the moment it comes to rest.

Replace $vc$ in the above equation by $vx,0$ where $vx,0$ denotes the speed of the system just after the collision.

$vx,0=m1Mv1i$ (V)

The speed of the two-train system as it comes to rest is zero.

Write the expression for the final speed of the system as it comes to rest.

$vx=0$ (VI)

Here, $vx$ is the final speed of the system as it comes to rest.

Write the constant-acceleration equation of motion.

$vx2=vx,02+2axΔx$

Here, $Δx$ is the distance travelled by the two-train system before it comes to rest.

Put equations (IV) to (VI) in the above equation and rewrite it for $v1i$ .

$0=(m1Mv1i)2+2(−μkg)Δxm1Mv1i=2μkgΔxv1i=Mm12μkgΔx$ (VII)

To find the range of possible initial freight train velocities, the extreme values of the coefficient of kinetic friction must be used.

Write the equation for $M$ .

$M=m1+m2$ (VIII)

Here, $m2$ is the mass of the passenger train.

Conclusion:

Given that the value of $m1$ is $5.221×106 kg$ , the value of $m2$ is $4.1×105 kg$ , the minimum value of the coefficient of kinetic friction is $0.3$ and its maximum value is $0.4$ and the distance travelled by the system before getting stopped is $370 ft$ . The value of $g$ is $9.81 m/s2$ .

Substitute $5.221×106 kg$ for $m1$ and $4.1×105 kg$ for $m2$ in equation (VIII) to find the value of $M$ .

$M=5.221×106 kg+4.1×105 kg=5.63×106 kg$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.3$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the minimum value of the speed of the two-train system.

$v1i(min)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(1.1×102 m)=9 m/s$

Here, $v1i(min)$ is the minimum speed of the system.

The system moves in $x$ direction. Write the expression for the minimum velocity of the system.

$v→1i(min)=v1i(min)i^$

Here, $v→1i(min)$ is the minimum velocity of the system.

Substitute $9 m/s$ for $v1i(min)$ in the above equation to find $v→1i(min)$ .

$v→1i(min)=9i^ m/s$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.4$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the maximum value of the speed of the two-train system.

$v1i(max)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(1.1×102 m)=30 m/s$

Here, $v1i(max)$ is the maximum speed of the system.

It is given that the train crosses the red signal at a speed of $48 mph$ . Convert this speed into units of m/s.

$48 mph=48 mph1609.344 m1 mile1 h(60×60) s=21 m/s$

Since the train passed the red signal with the speed of $21 m/s$ , the maximum speed cannot be greater than this and the upper limit will be replaced by this.

Write the expression for the maximum velocity of the system.

$v→1i(max)=v1i(max)i^$

Here, $v→1i(max)$ is the maximum velocity of the system.

Substitute $21 m/s$ for $v1i(max)$ in the above equation to find $v→1i(max)$ .

$v→1i(max)=21i^ m/s$

The range of the speed of the two-train system is $(9≤v→1i≤21)i^ m/s$ and the result obtained in Example 11.4 is the same.

Therefore, the velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

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Answer 3

Question

Chapter 11, Problem 32PQ

To determine

The velocity of the system just after the collision using Newton’s second law and the comparison of the answer with the result of Example 11.4.

Expert Solution & Answer

Answer to Problem 32PQ

The velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

Explanation of Solution

The free-body diagram of the system is shown in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 11, Problem 32PQ

The two-train system only moves in $x$ direction.

Write the expression for the Newton’s second law in $x$ direction.

$ΣFx=Max$ (I)

Here, $ΣFx$ is the magnitude of the net force in $x$ direction, $M$ is the mass of the two-train system and $ax$ is the acceleration in $x$ direction.

Refer to figure 1. The only force acting in $x$ direction is the force of kinetic friction and it acts in $−x$ direction.

Write the equation for $ΣFx$ .

$ΣFx=−Fk$

Here, $Fk$ is the kinetic friction.

Put the above equation in equation (I).

$−Fk=Max$ (II)

The kinetic friction is proportional to the normal force and the normal force is in turn equal to the weight of the train.

Write the expression for $Fk$ .

$Fk=μkFg$ (III)

Here, $μk$ is the coefficient of kinetic friction and $Fg$ is the weight of the train.

Write the equation for $Fg$ .

$Fg=Mg$

Here, $g$ is the acceleration due to gravity.

Put the above equation in equation (III).

$Fk=μkMg$

Put the above equation in equation (II) and rewrite it for $ax$ .

$−μkMg=Maxax=−μkg$ (IV)

Initially only the freight train has momentum. Assume that the two trains move together with velocity $vc$ after the collision. Apply the conservation of momentum for the two-train system before and after the collision.

$m1v1ii^=Mvci^$

Here, $m1$ is the mass of the freight train, $v1i$ is the initial speed of the freight train, $m2$ is the mass of the passenger train and $vc$ is the speed with the two trains move after collision.

Rewrite the above equation for $vc$ .

$vc=m1Mv1i$

Now consider the motion of the system just after the collision to the moment it comes to rest.

Replace $vc$ in the above equation by $vx,0$ where $vx,0$ denotes the speed of the system just after the collision.

$vx,0=m1Mv1i$ (V)

The speed of the two-train system as it comes to rest is zero.

Write the expression for the final speed of the system as it comes to rest.

$vx=0$ (VI)

Here, $vx$ is the final speed of the system as it comes to rest.

Write the constant-acceleration equation of motion.

$vx2=vx,02+2axΔx$

Here, $Δx$ is the distance travelled by the two-train system before it comes to rest.

Put equations (IV) to (VI) in the above equation and rewrite it for $v1i$ .

$0=(m1Mv1i)2+2(−μkg)Δxm1Mv1i=2μkgΔxv1i=Mm12μkgΔx$ (VII)

To find the range of possible initial freight train velocities, the extreme values of the coefficient of kinetic friction must be used.

Write the equation for $M$ .

$M=m1+m2$ (VIII)

Here, $m2$ is the mass of the passenger train.

Conclusion:

Given that the value of $m1$ is $5.221×106 kg$ , the value of $m2$ is $4.1×105 kg$ , the minimum value of the coefficient of kinetic friction is $0.3$ and its maximum value is $0.4$ and the distance travelled by the system before getting stopped is $370 ft$ . The value of $g$ is $9.81 m/s2$ .

Substitute $5.221×106 kg$ for $m1$ and $4.1×105 kg$ for $m2$ in equation (VIII) to find the value of $M$ .

$M=5.221×106 kg+4.1×105 kg=5.63×106 kg$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.3$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the minimum value of the speed of the two-train system.

$v1i(min)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(1.1×102 m)=9 m/s$

Here, $v1i(min)$ is the minimum speed of the system.

The system moves in $x$ direction. Write the expression for the minimum velocity of the system.

$v→1i(min)=v1i(min)i^$

Here, $v→1i(min)$ is the minimum velocity of the system.

Substitute $9 m/s$ for $v1i(min)$ in the above equation to find $v→1i(min)$ .

$v→1i(min)=9i^ m/s$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.4$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the maximum value of the speed of the two-train system.

$v1i(max)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(1.1×102 m)=30 m/s$

Here, $v1i(max)$ is the maximum speed of the system.

It is given that the train crosses the red signal at a speed of $48 mph$ . Convert this speed into units of m/s.

$48 mph=48 mph1609.344 m1 mile1 h(60×60) s=21 m/s$

Since the train passed the red signal with the speed of $21 m/s$ , the maximum speed cannot be greater than this and the upper limit will be replaced by this.

Write the expression for the maximum velocity of the system.

$v→1i(max)=v1i(max)i^$

Here, $v→1i(max)$ is the maximum velocity of the system.

Substitute $21 m/s$ for $v1i(max)$ in the above equation to find $v→1i(max)$ .

$v→1i(max)=21i^ m/s$

The range of the speed of the two-train system is $(9≤v→1i≤21)i^ m/s$ and the result obtained in Example 11.4 is the same.

Therefore, the velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

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Answer 4

Question

Chapter 11, Problem 32PQ

To determine

The velocity of the system just after the collision using Newton’s second law and the comparison of the answer with the result of Example 11.4.

Expert Solution & Answer

Answer to Problem 32PQ

The velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

Explanation of Solution

The free-body diagram of the system is shown in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 11, Problem 32PQ

The two-train system only moves in $x$ direction.

Write the expression for the Newton’s second law in $x$ direction.

$ΣFx=Max$ (I)

Here, $ΣFx$ is the magnitude of the net force in $x$ direction, $M$ is the mass of the two-train system and $ax$ is the acceleration in $x$ direction.

Refer to figure 1. The only force acting in $x$ direction is the force of kinetic friction and it acts in $−x$ direction.

Write the equation for $ΣFx$ .

$ΣFx=−Fk$

Here, $Fk$ is the kinetic friction.

Put the above equation in equation (I).

$−Fk=Max$ (II)

The kinetic friction is proportional to the normal force and the normal force is in turn equal to the weight of the train.

Write the expression for $Fk$ .

$Fk=μkFg$ (III)

Here, $μk$ is the coefficient of kinetic friction and $Fg$ is the weight of the train.

Write the equation for $Fg$ .

$Fg=Mg$

Here, $g$ is the acceleration due to gravity.

Put the above equation in equation (III).

$Fk=μkMg$

Put the above equation in equation (II) and rewrite it for $ax$ .

$−μkMg=Maxax=−μkg$ (IV)

Initially only the freight train has momentum. Assume that the two trains move together with velocity $vc$ after the collision. Apply the conservation of momentum for the two-train system before and after the collision.

$m1v1ii^=Mvci^$

Here, $m1$ is the mass of the freight train, $v1i$ is the initial speed of the freight train, $m2$ is the mass of the passenger train and $vc$ is the speed with the two trains move after collision.

Rewrite the above equation for $vc$ .

$vc=m1Mv1i$

Now consider the motion of the system just after the collision to the moment it comes to rest.

Replace $vc$ in the above equation by $vx,0$ where $vx,0$ denotes the speed of the system just after the collision.

$vx,0=m1Mv1i$ (V)

The speed of the two-train system as it comes to rest is zero.

Write the expression for the final speed of the system as it comes to rest.

$vx=0$ (VI)

Here, $vx$ is the final speed of the system as it comes to rest.

Write the constant-acceleration equation of motion.

$vx2=vx,02+2axΔx$

Here, $Δx$ is the distance travelled by the two-train system before it comes to rest.

Put equations (IV) to (VI) in the above equation and rewrite it for $v1i$ .

$0=(m1Mv1i)2+2(−μkg)Δxm1Mv1i=2μkgΔxv1i=Mm12μkgΔx$ (VII)

To find the range of possible initial freight train velocities, the extreme values of the coefficient of kinetic friction must be used.

Write the equation for $M$ .

$M=m1+m2$ (VIII)

Here, $m2$ is the mass of the passenger train.

Conclusion:

Given that the value of $m1$ is $5.221×106 kg$ , the value of $m2$ is $4.1×105 kg$ , the minimum value of the coefficient of kinetic friction is $0.3$ and its maximum value is $0.4$ and the distance travelled by the system before getting stopped is $370 ft$ . The value of $g$ is $9.81 m/s2$ .

Substitute $5.221×106 kg$ for $m1$ and $4.1×105 kg$ for $m2$ in equation (VIII) to find the value of $M$ .

$M=5.221×106 kg+4.1×105 kg=5.63×106 kg$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.3$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the minimum value of the speed of the two-train system.

$v1i(min)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(1.1×102 m)=9 m/s$

Here, $v1i(min)$ is the minimum speed of the system.

The system moves in $x$ direction. Write the expression for the minimum velocity of the system.

$v→1i(min)=v1i(min)i^$

Here, $v→1i(min)$ is the minimum velocity of the system.

Substitute $9 m/s$ for $v1i(min)$ in the above equation to find $v→1i(min)$ .

$v→1i(min)=9i^ m/s$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.4$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the maximum value of the speed of the two-train system.

$v1i(max)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(1.1×102 m)=30 m/s$

Here, $v1i(max)$ is the maximum speed of the system.

It is given that the train crosses the red signal at a speed of $48 mph$ . Convert this speed into units of m/s.

$48 mph=48 mph1609.344 m1 mile1 h(60×60) s=21 m/s$

Since the train passed the red signal with the speed of $21 m/s$ , the maximum speed cannot be greater than this and the upper limit will be replaced by this.

Write the expression for the maximum velocity of the system.

$v→1i(max)=v1i(max)i^$

Here, $v→1i(max)$ is the maximum velocity of the system.

Substitute $21 m/s$ for $v1i(max)$ in the above equation to find $v→1i(max)$ .

$v→1i(max)=21i^ m/s$

The range of the speed of the two-train system is $(9≤v→1i≤21)i^ m/s$ and the result obtained in Example 11.4 is the same.

Therefore, the velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

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Answer 5

Chapter 11, Problem 32PQ

To determine

The velocity of the system just after the collision using Newton’s second law and the comparison of the answer with the result of Example 11.4.

Expert Solution & Answer

Answer to Problem 32PQ

The velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

Explanation of Solution

The free-body diagram of the system is shown in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 11, Problem 32PQ

The two-train system only moves in $x$ direction.

Write the expression for the Newton’s second law in $x$ direction.

$ΣFx=Max$ (I)

Here, $ΣFx$ is the magnitude of the net force in $x$ direction, $M$ is the mass of the two-train system and $ax$ is the acceleration in $x$ direction.

Refer to figure 1. The only force acting in $x$ direction is the force of kinetic friction and it acts in $−x$ direction.

Write the equation for $ΣFx$ .

$ΣFx=−Fk$

Here, $Fk$ is the kinetic friction.

Put the above equation in equation (I).

$−Fk=Max$ (II)

The kinetic friction is proportional to the normal force and the normal force is in turn equal to the weight of the train.

Write the expression for $Fk$ .

$Fk=μkFg$ (III)

Here, $μk$ is the coefficient of kinetic friction and $Fg$ is the weight of the train.

Write the equation for $Fg$ .

$Fg=Mg$

Here, $g$ is the acceleration due to gravity.

Put the above equation in equation (III).

$Fk=μkMg$

Put the above equation in equation (II) and rewrite it for $ax$ .

$−μkMg=Maxax=−μkg$ (IV)

Initially only the freight train has momentum. Assume that the two trains move together with velocity $vc$ after the collision. Apply the conservation of momentum for the two-train system before and after the collision.

$m1v1ii^=Mvci^$

Here, $m1$ is the mass of the freight train, $v1i$ is the initial speed of the freight train, $m2$ is the mass of the passenger train and $vc$ is the speed with the two trains move after collision.

Rewrite the above equation for $vc$ .

$vc=m1Mv1i$

Now consider the motion of the system just after the collision to the moment it comes to rest.

Replace $vc$ in the above equation by $vx,0$ where $vx,0$ denotes the speed of the system just after the collision.

$vx,0=m1Mv1i$ (V)

The speed of the two-train system as it comes to rest is zero.

Write the expression for the final speed of the system as it comes to rest.

$vx=0$ (VI)

Here, $vx$ is the final speed of the system as it comes to rest.

Write the constant-acceleration equation of motion.

$vx2=vx,02+2axΔx$

Here, $Δx$ is the distance travelled by the two-train system before it comes to rest.

Put equations (IV) to (VI) in the above equation and rewrite it for $v1i$ .

$0=(m1Mv1i)2+2(−μkg)Δxm1Mv1i=2μkgΔxv1i=Mm12μkgΔx$ (VII)

To find the range of possible initial freight train velocities, the extreme values of the coefficient of kinetic friction must be used.

Write the equation for $M$ .

$M=m1+m2$ (VIII)

Here, $m2$ is the mass of the passenger train.

Conclusion:

Given that the value of $m1$ is $5.221×106 kg$ , the value of $m2$ is $4.1×105 kg$ , the minimum value of the coefficient of kinetic friction is $0.3$ and its maximum value is $0.4$ and the distance travelled by the system before getting stopped is $370 ft$ . The value of $g$ is $9.81 m/s2$ .

Substitute $5.221×106 kg$ for $m1$ and $4.1×105 kg$ for $m2$ in equation (VIII) to find the value of $M$ .

$M=5.221×106 kg+4.1×105 kg=5.63×106 kg$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.3$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the minimum value of the speed of the two-train system.

$v1i(min)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(1.1×102 m)=9 m/s$

Here, $v1i(min)$ is the minimum speed of the system.

The system moves in $x$ direction. Write the expression for the minimum velocity of the system.

$v→1i(min)=v1i(min)i^$

Here, $v→1i(min)$ is the minimum velocity of the system.

Substitute $9 m/s$ for $v1i(min)$ in the above equation to find $v→1i(min)$ .

$v→1i(min)=9i^ m/s$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.4$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the maximum value of the speed of the two-train system.

$v1i(max)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(1.1×102 m)=30 m/s$

Here, $v1i(max)$ is the maximum speed of the system.

It is given that the train crosses the red signal at a speed of $48 mph$ . Convert this speed into units of m/s.

$48 mph=48 mph1609.344 m1 mile1 h(60×60) s=21 m/s$

Since the train passed the red signal with the speed of $21 m/s$ , the maximum speed cannot be greater than this and the upper limit will be replaced by this.

Write the expression for the maximum velocity of the system.

$v→1i(max)=v1i(max)i^$

Here, $v→1i(max)$ is the maximum velocity of the system.

Substitute $21 m/s$ for $v1i(max)$ in the above equation to find $v→1i(max)$ .

$v→1i(max)=21i^ m/s$

The range of the speed of the two-train system is $(9≤v→1i≤21)i^ m/s$ and the result obtained in Example 11.4 is the same.

Therefore, the velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

Answer 6

Chapter 11, Problem 32PQ

To determine

The velocity of the system just after the collision using Newton’s second law and the comparison of the answer with the result of Example 11.4.

Expert Solution & Answer

Answer to Problem 32PQ

The velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

Explanation of Solution

The free-body diagram of the system is shown in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 11, Problem 32PQ

The two-train system only moves in $x$ direction.

Write the expression for the Newton’s second law in $x$ direction.

$ΣFx=Max$ (I)

Here, $ΣFx$ is the magnitude of the net force in $x$ direction, $M$ is the mass of the two-train system and $ax$ is the acceleration in $x$ direction.

Refer to figure 1. The only force acting in $x$ direction is the force of kinetic friction and it acts in $−x$ direction.

Write the equation for $ΣFx$ .

$ΣFx=−Fk$

Here, $Fk$ is the kinetic friction.

Put the above equation in equation (I).

$−Fk=Max$ (II)

The kinetic friction is proportional to the normal force and the normal force is in turn equal to the weight of the train.

Write the expression for $Fk$ .

$Fk=μkFg$ (III)

Here, $μk$ is the coefficient of kinetic friction and $Fg$ is the weight of the train.

Write the equation for $Fg$ .

$Fg=Mg$

Here, $g$ is the acceleration due to gravity.

Put the above equation in equation (III).

$Fk=μkMg$

Put the above equation in equation (II) and rewrite it for $ax$ .

$−μkMg=Maxax=−μkg$ (IV)

Initially only the freight train has momentum. Assume that the two trains move together with velocity $vc$ after the collision. Apply the conservation of momentum for the two-train system before and after the collision.

$m1v1ii^=Mvci^$

Here, $m1$ is the mass of the freight train, $v1i$ is the initial speed of the freight train, $m2$ is the mass of the passenger train and $vc$ is the speed with the two trains move after collision.

Rewrite the above equation for $vc$ .

$vc=m1Mv1i$

Now consider the motion of the system just after the collision to the moment it comes to rest.

Replace $vc$ in the above equation by $vx,0$ where $vx,0$ denotes the speed of the system just after the collision.

$vx,0=m1Mv1i$ (V)

The speed of the two-train system as it comes to rest is zero.

Write the expression for the final speed of the system as it comes to rest.

$vx=0$ (VI)

Here, $vx$ is the final speed of the system as it comes to rest.

Write the constant-acceleration equation of motion.

$vx2=vx,02+2axΔx$

Here, $Δx$ is the distance travelled by the two-train system before it comes to rest.

Put equations (IV) to (VI) in the above equation and rewrite it for $v1i$ .

$0=(m1Mv1i)2+2(−μkg)Δxm1Mv1i=2μkgΔxv1i=Mm12μkgΔx$ (VII)

To find the range of possible initial freight train velocities, the extreme values of the coefficient of kinetic friction must be used.

Write the equation for $M$ .

$M=m1+m2$ (VIII)

Here, $m2$ is the mass of the passenger train.

Conclusion:

Given that the value of $m1$ is $5.221×106 kg$ , the value of $m2$ is $4.1×105 kg$ , the minimum value of the coefficient of kinetic friction is $0.3$ and its maximum value is $0.4$ and the distance travelled by the system before getting stopped is $370 ft$ . The value of $g$ is $9.81 m/s2$ .

Substitute $5.221×106 kg$ for $m1$ and $4.1×105 kg$ for $m2$ in equation (VIII) to find the value of $M$ .

$M=5.221×106 kg+4.1×105 kg=5.63×106 kg$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.3$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the minimum value of the speed of the two-train system.

$v1i(min)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(1.1×102 m)=9 m/s$

Here, $v1i(min)$ is the minimum speed of the system.

The system moves in $x$ direction. Write the expression for the minimum velocity of the system.

$v→1i(min)=v1i(min)i^$

Here, $v→1i(min)$ is the minimum velocity of the system.

Substitute $9 m/s$ for $v1i(min)$ in the above equation to find $v→1i(min)$ .

$v→1i(min)=9i^ m/s$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.4$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the maximum value of the speed of the two-train system.

$v1i(max)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(1.1×102 m)=30 m/s$

Here, $v1i(max)$ is the maximum speed of the system.

It is given that the train crosses the red signal at a speed of $48 mph$ . Convert this speed into units of m/s.

$48 mph=48 mph1609.344 m1 mile1 h(60×60) s=21 m/s$

Since the train passed the red signal with the speed of $21 m/s$ , the maximum speed cannot be greater than this and the upper limit will be replaced by this.

Write the expression for the maximum velocity of the system.

$v→1i(max)=v1i(max)i^$

Here, $v→1i(max)$ is the maximum velocity of the system.

Substitute $21 m/s$ for $v1i(max)$ in the above equation to find $v→1i(max)$ .

$v→1i(max)=21i^ m/s$

The range of the speed of the two-train system is $(9≤v→1i≤21)i^ m/s$ and the result obtained in Example 11.4 is the same.

Therefore, the velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

Answer 7

Chapter 11, Problem 32PQ

To determine

The velocity of the system just after the collision using Newton’s second law and the comparison of the answer with the result of Example 11.4.

Answer 8

Expert Solution & Answer

Answer to Problem 32PQ

The velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The free-body diagram of the system is shown in figure 1.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 11, Problem 32PQ

The two-train system only moves in $x$ direction.

Write the expression for the Newton’s second law in $x$ direction.

$ΣFx=Max$ (I)

Here, $ΣFx$ is the magnitude of the net force in $x$ direction, $M$ is the mass of the two-train system and $ax$ is the acceleration in $x$ direction.

Refer to figure 1. The only force acting in $x$ direction is the force of kinetic friction and it acts in $−x$ direction.

Write the equation for $ΣFx$ .

$ΣFx=−Fk$

Here, $Fk$ is the kinetic friction.

Put the above equation in equation (I).

$−Fk=Max$ (II)

The kinetic friction is proportional to the normal force and the normal force is in turn equal to the weight of the train.

Write the expression for $Fk$ .

$Fk=μkFg$ (III)

Here, $μk$ is the coefficient of kinetic friction and $Fg$ is the weight of the train.

Write the equation for $Fg$ .

$Fg=Mg$

Here, $g$ is the acceleration due to gravity.

Put the above equation in equation (III).

$Fk=μkMg$

Put the above equation in equation (II) and rewrite it for $ax$ .

$−μkMg=Maxax=−μkg$ (IV)

Initially only the freight train has momentum. Assume that the two trains move together with velocity $vc$ after the collision. Apply the conservation of momentum for the two-train system before and after the collision.

$m1v1ii^=Mvci^$

Here, $m1$ is the mass of the freight train, $v1i$ is the initial speed of the freight train, $m2$ is the mass of the passenger train and $vc$ is the speed with the two trains move after collision.

Rewrite the above equation for $vc$ .

$vc=m1Mv1i$

Now consider the motion of the system just after the collision to the moment it comes to rest.

Replace $vc$ in the above equation by $vx,0$ where $vx,0$ denotes the speed of the system just after the collision.

$vx,0=m1Mv1i$ (V)

The speed of the two-train system as it comes to rest is zero.

Write the expression for the final speed of the system as it comes to rest.

$vx=0$ (VI)

Here, $vx$ is the final speed of the system as it comes to rest.

Write the constant-acceleration equation of motion.

$vx2=vx,02+2axΔx$

Here, $Δx$ is the distance travelled by the two-train system before it comes to rest.

Put equations (IV) to (VI) in the above equation and rewrite it for $v1i$ .

$0=(m1Mv1i)2+2(−μkg)Δxm1Mv1i=2μkgΔxv1i=Mm12μkgΔx$ (VII)

To find the range of possible initial freight train velocities, the extreme values of the coefficient of kinetic friction must be used.

Write the equation for $M$ .

$M=m1+m2$ (VIII)

Here, $m2$ is the mass of the passenger train.

Conclusion:

Given that the value of $m1$ is $5.221×106 kg$ , the value of $m2$ is $4.1×105 kg$ , the minimum value of the coefficient of kinetic friction is $0.3$ and its maximum value is $0.4$ and the distance travelled by the system before getting stopped is $370 ft$ . The value of $g$ is $9.81 m/s2$ .

Substitute $5.221×106 kg$ for $m1$ and $4.1×105 kg$ for $m2$ in equation (VIII) to find the value of $M$ .

$M=5.221×106 kg+4.1×105 kg=5.63×106 kg$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.3$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the minimum value of the speed of the two-train system.

$v1i(min)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.3)(9.81 m/s2)(1.1×102 m)=9 m/s$

Here, $v1i(min)$ is the minimum speed of the system.

The system moves in $x$ direction. Write the expression for the minimum velocity of the system.

$v→1i(min)=v1i(min)i^$

Here, $v→1i(min)$ is the minimum velocity of the system.

Substitute $9 m/s$ for $v1i(min)$ in the above equation to find $v→1i(min)$ .

$v→1i(min)=9i^ m/s$

Substitute $5.63×106 kg$ for $M$ , $5.221×106 kg$ for $m1$ , $0.4$ for $μk$ , $9.81 m/s2$ for $g$ and $370 ft$ for $Δx$ in equation (I) to find the maximum value of the speed of the two-train system.

$v1i(max)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(370 ft0.3048 m1 ft)=5.63×106 kg5.221×106 kg2(0.4)(9.81 m/s2)(1.1×102 m)=30 m/s$

Here, $v1i(max)$ is the maximum speed of the system.

It is given that the train crosses the red signal at a speed of $48 mph$ . Convert this speed into units of m/s.

$48 mph=48 mph1609.344 m1 mile1 h(60×60) s=21 m/s$

Since the train passed the red signal with the speed of $21 m/s$ , the maximum speed cannot be greater than this and the upper limit will be replaced by this.

Write the expression for the maximum velocity of the system.

$v→1i(max)=v1i(max)i^$

Here, $v→1i(max)$ is the maximum velocity of the system.

Substitute $21 m/s$ for $v1i(max)$ in the above equation to find $v→1i(max)$ .

$v→1i(max)=21i^ m/s$

The range of the speed of the two-train system is $(9≤v→1i≤21)i^ m/s$ and the result obtained in Example 11.4 is the same.

Therefore, the velocity of the system just after the collision using Newton’s second law is $(9≤v→1i≤21)i^ m/s$ and it is equal to the result of Example 11.4.

Answer 12

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The velocity of the system just after the collision using Newton’s second law and the comparison of the answer with the result of Example 11.4.

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The velocity of the system just after the collision using Newton’s second law and the comparison of the answer with the result of Example 11.4.

Answer to Problem 32PQ

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions