EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 8, Problem 41P

(a)

To determine

The total Kinetic Energy of the two blocks.

(a)

Expert Solution
Check Mark

Answer to Problem 41P

The total kinetic energy of the two blocks is 43.5J .

Explanation of Solution

Given:

The mass of block 1 is m1=3kg .

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

The mass of block 2 is m2=3kg .

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

Formula used:

The expression for kinetic energy is given as,

  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=12×3kg×(5m/s)2+12×3kg×(2m/s)2=43.5kgms2(J kg m s 2 )

Further solving the above equation,

  KE=43.5J

Conclusion:

Therefore, the total kinetic energy of the two blocks is 43.5J .

(b)

To determine

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

(b)

Expert Solution
Check Mark

Answer to Problem 41P

The velocity of centre of mass of the two block system is (1.5m/s)i^ .

Explanation of Solution

Given:

The mass of block 1 is m1=3kg .

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

The mass of block 2 is m2=3kg .

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

Formula used:

The expression for momentum is given as,

  p=mv

Calculation:

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

  Mvcm=m1v1+m2v2vcm=m1v1+m2v2m1+m2vcm=3kg×( 5m/s )i^+3kg×( 2m/s )i^3kg+3kgvcm=(1.5m/s)i^

Conclusion:

Therefore, the velocity of centre of mass of the two block system is (1.5m/s)i^ .

(c)

To determine

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

(c)

Expert Solution
Check Mark

Answer to Problem 41P

The velocity of the block 1 relative to the centre of mass is (3.5m/s)i^ and the velocity of the block 2 with respect to centre of mass is (3.5m/s)i^ .

Explanation of Solution

Given:

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

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

Formula used:

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

  vrel=vvcm

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=vvcmv1,rel=(5m/s)i^(1.5m/s)i^=(3.5m/s)i^

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

  vrel=vvcmv2,rel=(2m/s)i^(1.5m/s)i^=(3.5m/s)i^

Conclusion:

Therefore, the velocity of the block 1 relative to the centre of mass is (3.5m/s)i^ and the velocity of the block 2 with respect to centre of mass is (3.5m/s)i^ .

(d)

To determine

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

(d)

Expert Solution
Check Mark

Answer to Problem 41P

The kinetic energy of the blocks relative to the centre of mass is 36.75J .

Explanation of Solution

Given:

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

The velocity of block 2 in negative x direction v2=2m/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=12×3kg×(3.5m/s)2+12×3kg×(3.5m/s)2=36.75kgms2(J kg m s 2 )

Further solving the above equation,

  KE=36.75J

Conclusion:

Therefore, the kinetic energy of the blocks relative to the centre of mass is 36.75J .

(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
Check Mark

Answer to Problem 41P

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=3kg .

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

The mass of block 2 is m2=3kg .

The velocity of block 2 in negative x direction v2=2m/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(6kg)×(1.5m/s)2=6.75kgms2( 1J kg m s 2 )=6.75J

The above result is equal to KEKErel .

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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Chapter 8 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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