The total kinetic energy of the two blocks.

Question

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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.

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Phys 25

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.

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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.

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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.

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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.