Chapter 8, Problem 42P

(a)

To determine

The total kinetic energy of the two blocks.

Answer to Problem 42P

The total kinetic energy of the two blocks is $60\text{J}$ .

Explanation of Solution

Given:

The mass of block 1 is $m_1=3\text{kg}$ .

The velocity of the block 1 in positive x direction is $v_1=5\text{m}/\text{s}$ .

The mass of block 2 is $m_2=5\text{kg}$ .

The velocity of block 2 in positive x direction is $v_2=3\text{m}/\text{s}$ .

Formula used:

The expression for kinetic energy is given by,

  $KE=\frac{1}{2}m{v}^2$

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,

  $\begin{array}{c}KE=K{E}_1+K{E}_2\\ =\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2\\ =\frac{1}{2}\times 3\text{kg}\times {\left(5\text{m}/\text{s}\right)}^2+\frac{1}{2}\times 5\text{kg}\times {\left(3\text{m}/\text{s}\right)}^2\\ =60\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}=1\text{J}\right)\end{array}$

Further solving the above equation,

  $KE=60\text{J}$

Conclusion:

Therefore, the total kinetic energy of the two blocks is $60\text{J}$ .

(b)

To determine

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

Answer to Problem 42P

The velocity of centre of mass of the two block system is $\left(3.75\text{m}/\text{s}\right)\widehat{i}$ .

Explanation of Solution

Given:

The mass of block 1 is $m_1=3\text{kg}$ .

The velocity of the block 1 in positive x direction is $v_1=5\text{m}/\text{s}$ .

The mass of block 2 is $m_2=5\text{kg}$ .

The velocity of block 2 in positive x direction is $v_2=3\text{m}/\text{s}$ .

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,

  $\begin{array}{c}M{v}_{cm}=m_1{v}_1+m_2{v}_2\\ =\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}\\ =\frac{3\text{kg}\times \left(5\text{m}/\text{s}\right)\widehat{i}+5\text{kg}\times \left(3\text{m}/\text{s}\right)\widehat{i}}{3\text{kg}+5\text{kg}}\\ =\left(3.75\text{m}/\text{s}\right)\widehat{i}\end{array}$

Conclusion:

Therefore, the velocity of centre of mass of the two block system is $\left(3.75\text{m}/\text{s}\right)\widehat{i}$ .

(c)

To determine

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

Answer to Problem 42P

The velocity of the block 1 relative to the centre of mass is $\left(1.25\text{m}/\text{s}\right)\widehat{i}$ and the velocity of the block 2 with respect to centre of mass is $\left(-0.75\text{m}/\text{s}\right)\widehat{i}$ .

Explanation of Solution

Given:

The velocity of the block 1 in positive x direction is $v_1=5\text{m}/\text{s}$ .

The velocity of block 2 in positive x direction $v_2=3\text{m}/\text{s}$ .

Formula used:

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

  $v_{rel}=v-v_{cm}$

Here, $v_{rel}$ 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,

  $\begin{array}{c}{v}_{rel}=v-v_{cm}\\ v_{1,rel}=\left(5\text{m}/\text{s}\right)\widehat{i}-\left(3.75\text{m}/\text{s}\right)\widehat{i}\\ =\left(1.25\text{m}/\text{s}\right)\widehat{i}\end{array}$

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

  $\begin{array}{c}{v}_{rel}=v-v_{cm}\\ v_{2,rel}=\left(3\text{m}/\text{s}\right)\widehat{i}-\left(3.75\text{m}/\text{s}\right)\widehat{i}\\ =\left(-0.75\text{m}/\text{s}\right)\widehat{i}\end{array}$

Conclusion:

Therefore, The velocity of the block 1 relative to the centre of mass is $\left(1.25\text{m}/\text{s}\right)\widehat{i}$ and the velocity of the block 2 with respect to centre of mass is $\left(-0.75\text{m}/\text{s}\right)\widehat{i}$ .

(d)

To determine

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

Answer to Problem 42P

The kinetic energy of the blocks relative to the centre of mass is $3.74\text{J}$ .

Explanation of Solution

Given:

The velocity of block 1 in positive x direction is $v_1=5\text{m}/\text{s}$ .

The velocity of block 2 in positive x direction is $v_2=3\text{m}/\text{s}$ .

Formula used:

The expression for kinetic energy is given by,

  $KE=\frac{1}{2}m{v}^2$

Calculation:

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

  $\begin{array}{c}K{E}_{rel}=K{E}_{1,rel}+K{E}_{2,rel}\\ =\frac{1}{2}{m}_1{v}_{1,rel}^2+\frac{1}{2}{m}_2{v}_{2,rel}^2\\ =\frac{1}{2}\times 3\text{kg}\times {\left(1.25\text{m}/\text{s}\right)}^2+\frac{1}{2}\times 5\text{kg}\times {\left(-0.75\text{m}/\text{s}\right)}^2\\ =3.74\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}=1\text{J}\right)\end{array}$

Further solving the above equation,

  $KE=3.74\text{J}$

Conclusion:

Therefore, the kinetic energy of the blocks relative to the centre of mass is $3.74\text{J}$ .

(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.

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 $m_1=3\text{kg}$ .

The velocity of the block 1 in positive x direction is $v_1=5\text{m}/\text{s}$ .

The mass of block 2 is $m_2=3\text{kg}$ .

The velocity of block 2 in negative x direction $v_2=2\text{m}/\text{s}$ .

Formula used:

The expression for kinetic energy is given by,

  $KE=\frac{1}{2}m{v}^2$

Calculation:

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

  $\begin{array}{c}K{E}_{cm}=\frac{1}{2}\left(m_1+m_2\right)v_{cm}^2\\ =\frac{1}{2}\left(8\text{kg}\right)\times {\left(3.75\text{m}/\text{s}\right)}^2\\ =56.25\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}=1\text{J}\right)\\ =56.25\text{J}\end{array}$

The above result is equal to $KE-K{E}_{rel}$ .

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.