Chapter 6, Problem 54AP

Consider a frictionless track as shown in Figure P6.62. A block of mass m₁ = 5.00 kg is released from . It makes a head-on elastic collision at  with a block of mass m₂= 10.0 kg that is initially at rest. Calculate the maximum height to which m₁ rises after the collision.

Figure P6.62

To determine

The height to which $m_1$ rises after collision.

Answer to Problem 54AP

The height to which $m_1$ rises is $0.556\text{m}$.

Explanation of Solution

Given Info:

Mass of the block at position A is $5.00\text{kg}$ , the mass of the block at B is $10.0\text{kg}$ , initial speed of mass at position B is zero.

According to conservation of energy , kinetic energy will be equal to the potential energy of the mass.

Formula to calculate the kinetic energy of the mass 1 is at position A is ,

$KE=\frac{1}{2}{m}_1{v}_1{}^2$ (I)

• $m_1$ is the block released from top.
• $v_1$ is the speed of the mass 1 at position A before collision

Formula to calculate the potential energy of the mass 1 at position A is,

$PE=m_{1}gh$ (II)

• g is the acceleration due to gravity
• h is the height of the mass $m_1$

According to the principle of conservation of energy, Potential energy of mass 1 at position A is converted into its equivalent the kinetic energy at position B.

Equate (I) and (II) to calculate $v_1$.

$\begin{array}{c}\frac{1}{2}{m}_1{v}_1{}^2=m_{1}gh\\ v_1{}^2=2gh\\ v_1=\sqrt{2gh}\end{array}$

Substitute $9.80\text{m}\cdot {\text{s}}^{-2}$ for g and $5.00\text{m}$ for h in the above equation to calculate $v_1$.

$\begin{array}{c}{v}_1=\sqrt{2\left(9.80\text{m}\cdot {\text{s}}^{-2}\right)\left(5.00\text{m}\right)}\\ =\sqrt{98.0{\text{m}}^2\cdot {\text{s}}^{-2}}\\ =9.899\text{m}\cdot {\text{s}}^{-1}\\ \approx 9.90\text{m}\cdot {\text{s}}^{-1}\end{array}$

Collisions conserve momentum.

Apply conservation of momentum before and after collision for the masses $m_1$ and $m_2$

$m_1{v}_1+m_2{v}_2=m_1{v}_1{}^{\prime }+m_2{v}_2{}^{\prime }$ (III)

• $v_2$ is the speed of the mass $m_2$ at position B
• $v_1{}^{\prime }$ is the speed of mass $m_1$ after collision
• $v_2{}^{\prime }$ is the speed of mass $m_2$ after collision
• $m_{\text{2}}$ is the mass of the mass of the block at rest at bottom

Initial speed of mass2 is zero.

Substitute $0\text{m}/\text{s}$ for $v_2$ in the above equation and rewrite in terms of ${v^{\prime }}_2$.

$\begin{array}{c}{m}_1{v}_1+m_2\left(0\text{m}/\text{s}\right)=m_1{v}_1{}^{\prime }+m_2{v}_2{}^{\prime }\\ m_2{v}_2{}^{\prime }=m_1{v}_1-m_1{v}_1{}^{\prime }\\ =m_1\left(v_1-{v^{\prime }}_1\right)\\ {v^{\prime }}_2=\frac{m_1}{m_2}\left(v_1-{v^{\prime }}_1\right)\end{array}$

But for elastic collision

$v_1-v_2=-\left(v_1{}^{\prime }-v_2{}^{\prime }\right)$

Since initial speed of $m_2$ is zero,

Substitute $0\text{m}/\text{s}$ for $v_2$ in the above equation and rewrite in terms of ${v^{\prime }}_2$.

$v_2{}^{\prime }=v_1+v_1{}^{\prime }$ (IV)

Substitute the above equation (IV) in (III) to calculate $v_1{}^{\prime }$.

$\begin{array}{c}{m}_1{v}_1+m_2{v}_2=m_1{v}_1{}^{\prime }+m_2\left(v_1+v_1{}^{\prime }\right)\\ m_1{v}_1=m_1{v}_1{}^{\prime }+m_2{v}_1+m_2{v}_1{}^{\prime }\\ v_1\left(m_1-m_2\right)=v_1{}^{\prime }\left(m_1+m_2\right)\\ v_1{}^{\prime }=v_1\left(\frac{m_1-m_2}{m_1+m_2}\right)\end{array}$

Substitute $9.90\text{m}\cdot {\text{s}}^{-1}$ for $v_1$ , $5.00\text{kg}$ for $m_1$ and $10.0\text{kg}$ for $m_2$ to calculate $v_1{}^{\prime }$,

$\begin{array}{c}{v}_1{}^{\prime }=\left(9.90\text{m}\cdot {\text{s}}^{-1}\right)\left(\frac{5.00\text{kg}-10.0\text{kg}}{5.00\text{kg+}10.0\text{kg}}\right)\\ =-\left(9.90\text{m}\cdot {\text{s}}^{-1}\right)\left(\frac{1}{3}\right)\\ =-3.30\text{m}\cdot {\text{s}}^{-1}\end{array}$

Conclusion:

Formula to calculate kinetic energy of the mass $m_1$ after collision is,

$K{E}^{\prime }=\frac{1}{2}{m}_1{\left(v_1{}^{\prime }\right)}^2$ (V)

Formula to calculate potential energy of the mass $m_1$ after collision is ,

$P{E}^{\prime }=m_{1}g{h}^{\prime }$ (VI)

• $h^{\prime }$ is the height of the mass $m_1$ rises after collision

Apply conservation of energy for mass $m_1$ after collision to calculate the maximum height to which $m_1$ rises.

Equate (VI) and (V) to calculate the height of the mass $m_1$ rises.

$\begin{array}{c}{m}_{1}g{h}^{\prime }=\frac{1}{2}{m}_1{\left(v_1{}^{\prime }\right)}^2\\ h^{\prime }=\frac{{\left(v_1{}^{\prime }\right)}^2}{2g}\end{array}$

Substitute $9.80\text{m}\cdot {\text{s}}^{-2}$ for g and $-3.3\text{m}\cdot {\text{s}}^{-1}$ for $v_1{}^{\prime }$ to calculate $h^{\prime }$.

$\begin{array}{c}{h}^{\prime }=\frac{{\left(-3.3\text{m}\cdot {\text{s}}^{-1}\right)}^2}{2\left(9.8\text{m}\cdot {\text{s}}^{-2}\right)}\\ =\frac{10.89}{19.6}\\ =0.5556\text{m}\\ \approx 0.556\text{m}\end{array}$

Therefore, the height to which $m_1$ rises after collision is $0.556\text{m}$