EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
10th Edition
ISBN: 8220106740163
Author: SERWAY
Publisher: CENGAGE L
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Chapter 19, Problem 36AP

(a)

To determine

The appropriate models for the system of two bullets for the time interval before to after the interval.

(a)

Expert Solution
Check Mark

Answer to Problem 36AP

It is an isolated system. The kinetic energy is converted into internal energy. Here, the momentum is conserved and the collision is a perfectly inelastic collision.

Explanation of Solution

Given info: The mass of one bullet is 12.0g , mass of second bullet is 8.00g , speed of one bullet which moves to the right direction is 300m/s , speed of second bullet which moves to the left direction is 400m/s , the temperature at collision is 30°C .

It is an isolated system. The energy of the system is conserved but the kinetic energy is converted into internal energy. No external force exists. Here, the momentum is conserved and the collision is a perfectly inelastic collision.

Conclusion:

Therefore, it is an isolated system. The kinetic energy is converted into internal energy. Here, the momentum is conserved and the collision is a perfectly inelastic collision.

(b)

To determine

The speed of the combined bullets after the collision.

(b)

Expert Solution
Check Mark

Answer to Problem 36AP

The speed of the combined bullets after the collision is 20.0m/s which moves to the right.

Explanation of Solution

Given info: The mass of one bullet is 12.0g , mass of second bullet is 8.00g , speed of one bullet which moves to the right direction is 300m/s , speed of second bullet which moves to the left direction is 400m/s , the temperature at collision is 30°C .

Write the equation for conservation of momentum.

m1v1+m2v2=(m1+m2)vv=m1v1+m2v2m1+m2

Here,

m1 is the mass of one bullet.

m2 is the mass of one bullet.

v1 is the speed of one bullet.

v2 is the speed of second bullet.

v is the speed of combined bullets.

Consider positive sign when bullet moves to the right and negative sign when bullet moves to the left.

Substitute 12.0g for m1 , 300m/s for v1 , 8.00g for m2 and 400m/s for v2 in the above equation to get the speed of the combined bullets after the collision. v=(12.0g)(300m/s)+(8.00g)(400m/s)12.0g+8.00g=20.0m/s

Conclusion:

Therefore, the speed of the combined bullets after the collision is 20.0m/s which moves to the right.

(c)

To determine

The amount of initial kinetic energy which is converted into internal energy of the system after the collision.

(c)

Expert Solution
Check Mark

Answer to Problem 36AP

The amount of initial kinetic energy which is converted into internal energy of the system after the collision is 1176J .

Explanation of Solution

Given info: The mass of one bullet is 12.0g , mass of second bullet is 8.00g , speed of one bullet which moves to the right direction is 300m/s , speed of second bullet which moves to the left direction is 400m/s , the temperature at collision is 30°C .

Write the equation for change in kinetic energy to calculate the amount of kinetic energy which is converted into internal energy of the system after the collision. ΔK.E.=|K.EfinalK.Einitial|=|12(m1+m2)v212(m1v12+m2v22)|

Substitute 12.0g for m1 , 300m/s for v1 , 8.00g for m2 , 400m/s for v2 and 20m/s for v in the above equation to get the amount of kinetic energy which is converted into internal energy.

ΔK.E.=|[{12((12.0g+8.00g)(1kg1000g))(20.0m/s)2}{12[(12.0g(1kg1000g))(300m/s)2+(8.00g(1kg1000g))(400m/s)2]}]|=|4J1180J|=|1176J|=1176J

Conclusion:

Therefore, the amount of initial kinetic energy which is converted into internal energy of the system after the collision is 1176J .

(d)

To determine

Whether all the lead is melt due to the collision or not.

(d)

Expert Solution
Check Mark

Answer to Problem 36AP

No, all the lead is not melt due to the collision.

Explanation of Solution

Given info: The mass of one bullet is 12.0g , mass of second bullet is 8.00g , speed of one bullet which moves to the right direction is 300m/s , speed of second bullet which moves to the left direction is 400m/s , the temperature at collision is 30°C .

Write the equation to calculate the energy Q required to change the temperature of mass m.

Q=mcΔT=mc(TfTi)

Here,

c is the specific heat of the aluminum.

Tf is the melting temperature of lead.

Ti is the temperature at collision.

The specific heat of aluminum is 128J/kg°C .

The melting point of lead is 327.5°C .

Substitute 20.0g for m, 128J/kg°C for c , 327.5°C for Tf and 30°C for Ti in the above equation to get the energy required to change the temperature of mass m .

Q=(20.0g(1kg1000g))(128J/kg°C)((327.5°C)(30°C))=761.6J

Initial energy generated by the collision is 1176J , so the available heat for the melting process is calculated by,

Qavailable=EQ=1176J761.6J=414.5J

Write the equation to calculate the amount of heat required to melt all of the lead.

Q=mL

Here,

L is the latent heat of the fusion for lead.

The latent heat of fusion for lead is equal to 2.32×104J/kg .

Substitute 20.0g for m and 2.32×104J/kg for L in the above equation. Q=(20.0g(1kg1000g))(2.32×104J/kg)=464J

Here, only 414.5J heat is available, so the all the lead does not melt.

Conclusion:

Therefore, all the lead is not melt due to the collision.

(e)

To determine

The temperature of the combined bullets after the collision.

(e)

Expert Solution
Check Mark

Answer to Problem 36AP

The temperature of the combined bullets after the collision is 327.5°C .

Explanation of Solution

Given info: The mass of one bullet is 12.0g , mass of second bullet is 8.00g , speed of one bullet which moves to the right direction is 300m/s , speed of second bullet which moves to the left direction is 400m/s , the temperature at collision is 30°C .

The available heat to melt the bullets is 414.5J and heat required to melt the combined bullets is 464J .

The temperature of the combined bullets is the melting temperature of the bullets which is equal to 327.5°C .

Conclusion:

Therefore, the temperature of the combined bullets after the collision is 327.5°C .

(f)

To determine

The phase of the combined bullets after the collision.

(f)

Expert Solution
Check Mark

Answer to Problem 36AP

The phase of the combined bullets after the collision is such that 2.2g of solid lead and 17.8g of liquid lead.

Explanation of Solution

Given info: The mass of one bullet is 12.0g , mass of second bullet is 8.00g , speed of one bullet which moves to the right direction is 300m/s , speed of second bullet which moves to the left direction is 400m/s , the temperature at collision is 30°C .

Write the equation to calculate the total melted mass of lead.

Q=mLm=QL

Substitute 414.5J for Q and 2.32×104J/kg for L in the above equation to get the mass of liquid lead.

m=414.5J2.32×104J/kg=0.01786kg×103g1kg=17.8kg

Calculate the mass of solid lead.

msolid=m1m

Substitute 20.0g for m1 and 17.8g for m in the above equation to get the mass of solid lead.

msolid=20.0g17.8g=2.2g

Conclusion:

Therefore, the phase of the combined bullets after the collision is such that 2.2g of solid lead and 17.8g of liquid lead.

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Chapter 19 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

Ch. 19 - If water with a mass mk at temperature Tk is...Ch. 19 - An aluminum calorimeter with a mass of 100 g...Ch. 19 - An electric drill with a steel drill bit of mass m...Ch. 19 - A 3.00-g copper coin at 25.0C drops 50.0 m to the...Ch. 19 - How much energy is required to change a 40.0-g ice...Ch. 19 - Prob. 11PCh. 19 - A 3.00-g lead bullet at 30.0C is fired at a speed...Ch. 19 - In an insulated vessel, 250 g of ice at 0C is...Ch. 19 - Prob. 14PCh. 19 - One mole of an ideal gas is warmed slowly so that...Ch. 19 - (a) Determine the work done on a gas that expands...Ch. 19 - A thermodynamic system undergoes a process in...Ch. 19 - Why is the following situation impossible? An...Ch. 19 - A 2.00-mol sample of helium gas initially at 300...Ch. 19 - (a) How much work is done on the steam when 1.00...Ch. 19 - A 1.00-kg block of aluminum is warmed at...Ch. 19 - In Figure P19.22, the change in internal energy of...Ch. 19 - A student is trying to decide what to wear. His...Ch. 19 - A concrete slab is 12.0 cm thick and has an area...Ch. 19 - Two lightbulbs have cylindrical filaments much...Ch. 19 - Prob. 26PCh. 19 - (a) Calculate the R-value of a thermal window made...Ch. 19 - Prob. 28PCh. 19 - Gas in a container is at a pressure of 1.50 atm...Ch. 19 - Prob. 30APCh. 19 - You have a particular interest in automobile...Ch. 19 - You are working in a condensed-matter laboratory...Ch. 19 - Prob. 33APCh. 19 - Prob. 34APCh. 19 - Prob. 35APCh. 19 - Prob. 36APCh. 19 - An ice-cube tray is filled with 75.0 g of water....Ch. 19 - Prob. 38APCh. 19 - An iron plate is held against an iron wheel so...Ch. 19 - One mole of an ideal gas is contained in a...Ch. 19 - Prob. 41APCh. 19 - Prob. 42APCh. 19 - Prob. 43APCh. 19 - A student measures the following data in a...Ch. 19 - (a) The inside of a hollow cylinder is maintained...Ch. 19 - A spherical shell has inner radius 3.00 cm and...Ch. 19 - Prob. 47CP
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