The total kinetic energy of the two blocks.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 8, Problem 42P

(a)

To determine

The total kinetic energy of the two blocks.

(a)

Expert Solution

Answer to Problem 42P

The total kinetic energy of the two blocks is 60 J .

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Here, m is the mass of the body and v is the velocity of the body.

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KE=KE1+KE2=12m1v12+12m2v22=12×3 kg×(5 m/s)2+12×5 kg×(3 m/s)2=60 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=60 J

Conclusion:

Therefore, the total kinetic energy of the two blocks is 60 J .

(b)

To determine

The velocity of centre of mass of the two block system.

(b)

Expert Solution

Answer to Problem 42P

The velocity of centre of mass of the two block system is (3.75 m/s)i^ .

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for momentum is given by,

p=mv

Calculation:

Since, the momentum of the system remains conserved and can be calculated as,

Mvcm=m1v1+m2v2=m1v1+m2v2m1+m2=3 kg×( 5 m/s )i^+5 kg×( 3 m/s )i^3 kg+5 kg=(3.75 m/s)i^

Conclusion:

Therefore, the velocity of centre of mass of the two block system is (3.75 m/s)i^ .

(c)

To determine

The velocity of each block relative to the centre of mass.

(c)

Expert Solution

Answer to Problem 42P

The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

Explanation of Solution

Given:

The velocity of the block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction v2=3 m/s .

Formula used:

The expression for the velocity with respect to centre of mass is given as,

vrel=v−vcm

Here, vrel is the velocity of the blocks with respect to the centre of mass.

Calculation:

The velocity of block 1 with respect to the centre of mass can be calculated as,

vrel=v−vcmv1,rel=(5 m/s)i^−(3.75 m/s)i^=(1.25 m/s)i^

The velocity of block 2 with respect to the centre of mass can be calculated as,

vrel=v−vcmv2,rel=(3 m/s)i^−(3.75 m/s)i^=(−0.75 m/s)i^

Conclusion:

Therefore, The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

(d)

To determine

The kinetic energy of the blocks relative to the centre of mass.

(d)

Expert Solution

Answer to Problem 42P

The kinetic energy of the blocks relative to the centre of mass is 3.74 J .

Explanation of Solution

Given:

The velocity of block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KErel=KE1,rel+KE2,rel=12m1v1,rel2+12m2v2,rel2=12×3 kg×(1.25 m/s)2+12×5 kg×(−0.75 m/s)2=3.74 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=3.74 J

Conclusion:

Therefore, the kinetic energy of the blocks relative to the centre of mass is 3.74 J .

(e)

To determine

The proof that the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

(e)

Expert Solution

Answer to Problem 42P

The kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=3 kg .

The velocity of block 2 in negative x direction v2=2 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KEcm=12(m1+m2)vcm2=12(8 kg)×(3.75 m/s)2=56.25 kg⋅ms2(kg⋅m s 2 =1 J)=56.25 J

The above result is equal to KE−KErel .

Conclusion:

Therefore, the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Four capacitors are connected as shown in the figure below. (Let C = 12.0 μF.) a C 3.00 με Hh. 6.00 με 20.0 με HE (a) Find the equivalent capacitance between points a and b. 5.92 HF (b) Calculate the charge on each capacitor, taking AV ab = 16.0 V. 20.0 uF capacitor 94.7 6.00 uF capacitor 67.6 32.14 3.00 µF capacitor capacitor C ☑ με με The 3 µF and 12.0 uF capacitors are in series and that combination is in parallel with the 6 μF capacitor. What quantity is the same for capacitors in parallel? μC 32.14 ☑ You are correct that the charge on this capacitor will be the same as the charge on the 3 μF capacitor. μC

In the pivot assignment, we observed waves moving on a string stretched by hanging weights. We noticed that certain frequencies produced standing waves. One such situation is shown below: 0 ст Direct Measurement ©2015 Peter Bohacek I. 20 0 cm 10 20 30 40 50 60 70 80 90 100 Which Harmonic is this? Do NOT include units! What is the wavelength of this wave in cm with only no decimal places? If the speed of this wave is 2500 cm/s, what is the frequency of this harmonic (in Hz, with NO decimal places)?

Four capacitors are connected as shown in the figure below. (Let C = 12.0 µF.) A circuit consists of four capacitors. It begins at point a before the wire splits in two directions. On the upper split, there is a capacitor C followed by a 3.00 µF capacitor. On the lower split, there is a 6.00 µF capacitor. The two splits reconnect and are followed by a 20.0 µF capacitor, which is then followed by point b. (a) Find the equivalent capacitance between points a and b. µF(b) Calculate the charge on each capacitor, taking ΔVab = 16.0 V. 20.0 µF capacitor µC 6.00 µF capacitor µC 3.00 µF capacitor µC capacitor C µC

Answer 2

Question

Chapter 8, Problem 42P

(a)

To determine

The total kinetic energy of the two blocks.

(a)

Expert Solution

Answer to Problem 42P

The total kinetic energy of the two blocks is 60 J .

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Here, m is the mass of the body and v is the velocity of the body.

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KE=KE1+KE2=12m1v12+12m2v22=12×3 kg×(5 m/s)2+12×5 kg×(3 m/s)2=60 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=60 J

Conclusion:

Therefore, the total kinetic energy of the two blocks is 60 J .

(b)

To determine

The velocity of centre of mass of the two block system.

(b)

Expert Solution

Answer to Problem 42P

The velocity of centre of mass of the two block system is (3.75 m/s)i^ .

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for momentum is given by,

p=mv

Calculation:

Since, the momentum of the system remains conserved and can be calculated as,

Mvcm=m1v1+m2v2=m1v1+m2v2m1+m2=3 kg×( 5 m/s )i^+5 kg×( 3 m/s )i^3 kg+5 kg=(3.75 m/s)i^

Conclusion:

Therefore, the velocity of centre of mass of the two block system is (3.75 m/s)i^ .

(c)

To determine

The velocity of each block relative to the centre of mass.

(c)

Expert Solution

Answer to Problem 42P

The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

Explanation of Solution

Given:

The velocity of the block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction v2=3 m/s .

Formula used:

The expression for the velocity with respect to centre of mass is given as,

vrel=v−vcm

Here, vrel is the velocity of the blocks with respect to the centre of mass.

Calculation:

The velocity of block 1 with respect to the centre of mass can be calculated as,

vrel=v−vcmv1,rel=(5 m/s)i^−(3.75 m/s)i^=(1.25 m/s)i^

The velocity of block 2 with respect to the centre of mass can be calculated as,

vrel=v−vcmv2,rel=(3 m/s)i^−(3.75 m/s)i^=(−0.75 m/s)i^

Conclusion:

Therefore, The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

(d)

To determine

The kinetic energy of the blocks relative to the centre of mass.

(d)

Expert Solution

Answer to Problem 42P

The kinetic energy of the blocks relative to the centre of mass is 3.74 J .

Explanation of Solution

Given:

The velocity of block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KErel=KE1,rel+KE2,rel=12m1v1,rel2+12m2v2,rel2=12×3 kg×(1.25 m/s)2+12×5 kg×(−0.75 m/s)2=3.74 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=3.74 J

Conclusion:

Therefore, the kinetic energy of the blocks relative to the centre of mass is 3.74 J .

(e)

To determine

The proof that the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

(e)

Expert Solution

Answer to Problem 42P

The kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=3 kg .

The velocity of block 2 in negative x direction v2=2 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KEcm=12(m1+m2)vcm2=12(8 kg)×(3.75 m/s)2=56.25 kg⋅ms2(kg⋅m s 2 =1 J)=56.25 J

The above result is equal to KE−KErel .

Conclusion:

Therefore, the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 8, Problem 42P

(a)

To determine

The total kinetic energy of the two blocks.

(a)

Expert Solution

Answer to Problem 42P

The total kinetic energy of the two blocks is 60 J .

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Here, m is the mass of the body and v is the velocity of the body.

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KE=KE1+KE2=12m1v12+12m2v22=12×3 kg×(5 m/s)2+12×5 kg×(3 m/s)2=60 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=60 J

Conclusion:

Therefore, the total kinetic energy of the two blocks is 60 J .

(b)

To determine

The velocity of centre of mass of the two block system.

(b)

Expert Solution

Answer to Problem 42P

The velocity of centre of mass of the two block system is (3.75 m/s)i^ .

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for momentum is given by,

p=mv

Calculation:

Since, the momentum of the system remains conserved and can be calculated as,

Mvcm=m1v1+m2v2=m1v1+m2v2m1+m2=3 kg×( 5 m/s )i^+5 kg×( 3 m/s )i^3 kg+5 kg=(3.75 m/s)i^

Conclusion:

Therefore, the velocity of centre of mass of the two block system is (3.75 m/s)i^ .

(c)

To determine

The velocity of each block relative to the centre of mass.

(c)

Expert Solution

Answer to Problem 42P

The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

Explanation of Solution

Given:

The velocity of the block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction v2=3 m/s .

Formula used:

The expression for the velocity with respect to centre of mass is given as,

vrel=v−vcm

Here, vrel is the velocity of the blocks with respect to the centre of mass.

Calculation:

The velocity of block 1 with respect to the centre of mass can be calculated as,

vrel=v−vcmv1,rel=(5 m/s)i^−(3.75 m/s)i^=(1.25 m/s)i^

The velocity of block 2 with respect to the centre of mass can be calculated as,

vrel=v−vcmv2,rel=(3 m/s)i^−(3.75 m/s)i^=(−0.75 m/s)i^

Conclusion:

Therefore, The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

(d)

To determine

The kinetic energy of the blocks relative to the centre of mass.

(d)

Expert Solution

Answer to Problem 42P

The kinetic energy of the blocks relative to the centre of mass is 3.74 J .

Explanation of Solution

Given:

The velocity of block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KErel=KE1,rel+KE2,rel=12m1v1,rel2+12m2v2,rel2=12×3 kg×(1.25 m/s)2+12×5 kg×(−0.75 m/s)2=3.74 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=3.74 J

Conclusion:

Therefore, the kinetic energy of the blocks relative to the centre of mass is 3.74 J .

(e)

To determine

The proof that the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

(e)

Expert Solution

Answer to Problem 42P

The kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=3 kg .

The velocity of block 2 in negative x direction v2=2 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KEcm=12(m1+m2)vcm2=12(8 kg)×(3.75 m/s)2=56.25 kg⋅ms2(kg⋅m s 2 =1 J)=56.25 J

The above result is equal to KE−KErel .

Conclusion:

Therefore, the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 8, Problem 42P

(a)

To determine

The total kinetic energy of the two blocks.

(a)

Expert Solution

Answer to Problem 42P

The total kinetic energy of the two blocks is 60 J .

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Here, m is the mass of the body and v is the velocity of the body.

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KE=KE1+KE2=12m1v12+12m2v22=12×3 kg×(5 m/s)2+12×5 kg×(3 m/s)2=60 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=60 J

Conclusion:

Therefore, the total kinetic energy of the two blocks is 60 J .

(b)

To determine

The velocity of centre of mass of the two block system.

(b)

Expert Solution

Answer to Problem 42P

The velocity of centre of mass of the two block system is (3.75 m/s)i^ .

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for momentum is given by,

p=mv

Calculation:

Since, the momentum of the system remains conserved and can be calculated as,

Mvcm=m1v1+m2v2=m1v1+m2v2m1+m2=3 kg×( 5 m/s )i^+5 kg×( 3 m/s )i^3 kg+5 kg=(3.75 m/s)i^

Conclusion:

Therefore, the velocity of centre of mass of the two block system is (3.75 m/s)i^ .

(c)

To determine

The velocity of each block relative to the centre of mass.

(c)

Expert Solution

Answer to Problem 42P

The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

Explanation of Solution

Given:

The velocity of the block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction v2=3 m/s .

Formula used:

The expression for the velocity with respect to centre of mass is given as,

vrel=v−vcm

Here, vrel is the velocity of the blocks with respect to the centre of mass.

Calculation:

The velocity of block 1 with respect to the centre of mass can be calculated as,

vrel=v−vcmv1,rel=(5 m/s)i^−(3.75 m/s)i^=(1.25 m/s)i^

The velocity of block 2 with respect to the centre of mass can be calculated as,

vrel=v−vcmv2,rel=(3 m/s)i^−(3.75 m/s)i^=(−0.75 m/s)i^

Conclusion:

Therefore, The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

(d)

To determine

The kinetic energy of the blocks relative to the centre of mass.

(d)

Expert Solution

Answer to Problem 42P

The kinetic energy of the blocks relative to the centre of mass is 3.74 J .

Explanation of Solution

Given:

The velocity of block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KErel=KE1,rel+KE2,rel=12m1v1,rel2+12m2v2,rel2=12×3 kg×(1.25 m/s)2+12×5 kg×(−0.75 m/s)2=3.74 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=3.74 J

Conclusion:

Therefore, the kinetic energy of the blocks relative to the centre of mass is 3.74 J .

(e)

To determine

The proof that the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

(e)

Expert Solution

Answer to Problem 42P

The kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=3 kg .

The velocity of block 2 in negative x direction v2=2 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KEcm=12(m1+m2)vcm2=12(8 kg)×(3.75 m/s)2=56.25 kg⋅ms2(kg⋅m s 2 =1 J)=56.25 J

The above result is equal to KE−KErel .

Conclusion:

Therefore, the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 8, Problem 42P

(a)

To determine

The total kinetic energy of the two blocks.

(a)

Expert Solution

Answer to Problem 42P

The total kinetic energy of the two blocks is 60 J .

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Here, m is the mass of the body and v is the velocity of the body.

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KE=KE1+KE2=12m1v12+12m2v22=12×3 kg×(5 m/s)2+12×5 kg×(3 m/s)2=60 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=60 J

Conclusion:

Therefore, the total kinetic energy of the two blocks is 60 J .

(b)

To determine

The velocity of centre of mass of the two block system.

(b)

Expert Solution

Answer to Problem 42P

The velocity of centre of mass of the two block system is (3.75 m/s)i^ .

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for momentum is given by,

p=mv

Calculation:

Since, the momentum of the system remains conserved and can be calculated as,

Mvcm=m1v1+m2v2=m1v1+m2v2m1+m2=3 kg×( 5 m/s )i^+5 kg×( 3 m/s )i^3 kg+5 kg=(3.75 m/s)i^

Conclusion:

Therefore, the velocity of centre of mass of the two block system is (3.75 m/s)i^ .

(c)

To determine

The velocity of each block relative to the centre of mass.

(c)

Expert Solution

Answer to Problem 42P

The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

Explanation of Solution

Given:

The velocity of the block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction v2=3 m/s .

Formula used:

The expression for the velocity with respect to centre of mass is given as,

vrel=v−vcm

Here, vrel is the velocity of the blocks with respect to the centre of mass.

Calculation:

The velocity of block 1 with respect to the centre of mass can be calculated as,

vrel=v−vcmv1,rel=(5 m/s)i^−(3.75 m/s)i^=(1.25 m/s)i^

The velocity of block 2 with respect to the centre of mass can be calculated as,

vrel=v−vcmv2,rel=(3 m/s)i^−(3.75 m/s)i^=(−0.75 m/s)i^

Conclusion:

Therefore, The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

(d)

To determine

The kinetic energy of the blocks relative to the centre of mass.

(d)

Expert Solution

Answer to Problem 42P

The kinetic energy of the blocks relative to the centre of mass is 3.74 J .

Explanation of Solution

Given:

The velocity of block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KErel=KE1,rel+KE2,rel=12m1v1,rel2+12m2v2,rel2=12×3 kg×(1.25 m/s)2+12×5 kg×(−0.75 m/s)2=3.74 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=3.74 J

Conclusion:

Therefore, the kinetic energy of the blocks relative to the centre of mass is 3.74 J .

(e)

To determine

The proof that the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

(e)

Expert Solution

Answer to Problem 42P

The kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=3 kg .

The velocity of block 2 in negative x direction v2=2 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KEcm=12(m1+m2)vcm2=12(8 kg)×(3.75 m/s)2=56.25 kg⋅ms2(kg⋅m s 2 =1 J)=56.25 J

The above result is equal to KE−KErel .

Conclusion:

Therefore, the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Answer 6

Chapter 8, Problem 42P

(a)

To determine

The total kinetic energy of the two blocks.

(a)

Expert Solution

Answer to Problem 42P

The total kinetic energy of the two blocks is 60 J .

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Here, m is the mass of the body and v is the velocity of the body.

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KE=KE1+KE2=12m1v12+12m2v22=12×3 kg×(5 m/s)2+12×5 kg×(3 m/s)2=60 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=60 J

Conclusion:

Therefore, the total kinetic energy of the two blocks is 60 J .

Answer 7

Chapter 8, Problem 42P

(a)

To determine

The total kinetic energy of the two blocks.

Answer 8

(a)

Expert Solution

Answer to Problem 42P

The total kinetic energy of the two blocks is 60 J .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Here, m is the mass of the body and v is the velocity of the body.

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KE=KE1+KE2=12m1v12+12m2v22=12×3 kg×(5 m/s)2+12×5 kg×(3 m/s)2=60 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=60 J

Conclusion:

Therefore, the total kinetic energy of the two blocks is 60 J .

Answer 13

(b)

To determine

The velocity of centre of mass of the two block system.

(b)

Expert Solution

Answer to Problem 42P

The velocity of centre of mass of the two block system is (3.75 m/s)i^ .

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for momentum is given by,

p=mv

Calculation:

Since, the momentum of the system remains conserved and can be calculated as,

Mvcm=m1v1+m2v2=m1v1+m2v2m1+m2=3 kg×( 5 m/s )i^+5 kg×( 3 m/s )i^3 kg+5 kg=(3.75 m/s)i^

Conclusion:

Therefore, the velocity of centre of mass of the two block system is (3.75 m/s)i^ .

Answer 14

(b)

To determine

The velocity of centre of mass of the two block system.

Answer 15

(b)

Expert Solution

Answer to Problem 42P

The velocity of centre of mass of the two block system is (3.75 m/s)i^ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=5 kg .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for momentum is given by,

p=mv

Calculation:

Since, the momentum of the system remains conserved and can be calculated as,

Mvcm=m1v1+m2v2=m1v1+m2v2m1+m2=3 kg×( 5 m/s )i^+5 kg×( 3 m/s )i^3 kg+5 kg=(3.75 m/s)i^

Conclusion:

Therefore, the velocity of centre of mass of the two block system is (3.75 m/s)i^ .

Answer 20

(c)

To determine

The velocity of each block relative to the centre of mass.

(c)

Expert Solution

Answer to Problem 42P

The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

Explanation of Solution

Given:

The velocity of the block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction v2=3 m/s .

Formula used:

The expression for the velocity with respect to centre of mass is given as,

vrel=v−vcm

Here, vrel is the velocity of the blocks with respect to the centre of mass.

Calculation:

The velocity of block 1 with respect to the centre of mass can be calculated as,

vrel=v−vcmv1,rel=(5 m/s)i^−(3.75 m/s)i^=(1.25 m/s)i^

The velocity of block 2 with respect to the centre of mass can be calculated as,

vrel=v−vcmv2,rel=(3 m/s)i^−(3.75 m/s)i^=(−0.75 m/s)i^

Conclusion:

Therefore, The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

Answer 21

(c)

To determine

The velocity of each block relative to the centre of mass.

Answer 22

(c)

Expert Solution

Answer to Problem 42P

The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

The velocity of the block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction v2=3 m/s .

Formula used:

The expression for the velocity with respect to centre of mass is given as,

vrel=v−vcm

Here, vrel is the velocity of the blocks with respect to the centre of mass.

Calculation:

The velocity of block 1 with respect to the centre of mass can be calculated as,

vrel=v−vcmv1,rel=(5 m/s)i^−(3.75 m/s)i^=(1.25 m/s)i^

The velocity of block 2 with respect to the centre of mass can be calculated as,

vrel=v−vcmv2,rel=(3 m/s)i^−(3.75 m/s)i^=(−0.75 m/s)i^

Conclusion:

Therefore, The velocity of the block 1 relative to the centre of mass is (1.25 m/s)i^ and the velocity of the block 2 with respect to centre of mass is (−0.75 m/s)i^ .

Answer 27

(d)

To determine

The kinetic energy of the blocks relative to the centre of mass.

(d)

Expert Solution

Answer to Problem 42P

The kinetic energy of the blocks relative to the centre of mass is 3.74 J .

Explanation of Solution

Given:

The velocity of block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KErel=KE1,rel+KE2,rel=12m1v1,rel2+12m2v2,rel2=12×3 kg×(1.25 m/s)2+12×5 kg×(−0.75 m/s)2=3.74 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=3.74 J

Conclusion:

Therefore, the kinetic energy of the blocks relative to the centre of mass is 3.74 J .

Answer 28

(d)

To determine

The kinetic energy of the blocks relative to the centre of mass.

Answer 29

(d)

Expert Solution

Answer to Problem 42P

The kinetic energy of the blocks relative to the centre of mass is 3.74 J .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given:

The velocity of block 1 in positive x direction is v1=5 m/s .

The velocity of block 2 in positive x direction is v2=3 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KErel=KE1,rel+KE2,rel=12m1v1,rel2+12m2v2,rel2=12×3 kg×(1.25 m/s)2+12×5 kg×(−0.75 m/s)2=3.74 kg⋅ms2(kg⋅m s 2 =1 J)

Further solving the above equation,

KE=3.74 J

Conclusion:

Therefore, the kinetic energy of the blocks relative to the centre of mass is 3.74 J .

Answer 34

(e)

To determine

The proof that the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

(e)

Expert Solution

Answer to Problem 42P

The kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=3 kg .

The velocity of block 2 in negative x direction v2=2 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KEcm=12(m1+m2)vcm2=12(8 kg)×(3.75 m/s)2=56.25 kg⋅ms2(kg⋅m s 2 =1 J)=56.25 J

The above result is equal to KE−KErel .

Conclusion:

Therefore, the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Answer 35

(e)

To determine

The proof that the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Answer 36

(e)

Expert Solution

Answer to Problem 42P

The kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given:

The mass of block 1 is m1=3 kg .

The velocity of the block 1 in positive x direction is v1=5 m/s .

The mass of block 2 is m2=3 kg .

The velocity of block 2 in negative x direction v2=2 m/s .

Formula used:

The expression for kinetic energy is given by,

KE=12mv2

Calculation:

The total Kinetic energy of the two blocks can be calculated as,

KEcm=12(m1+m2)vcm2=12(8 kg)×(3.75 m/s)2=56.25 kg⋅ms2(kg⋅m s 2 =1 J)=56.25 J

The above result is equal to KE−KErel .

Conclusion:

Therefore, the kinetic energy in part (a) is greater than the kinetic energy in part (d) by an amount equal to the kinetic energy with respect to centre of mass.