Chapter 9, Problem 9.77AP

Two blocks of masses m₁ = 2.00 kg and m₂ = 4.00 kg are released from rest at a height of h = 5.00 m on a friction less track as shown in Figure P9.77. When they meet on the level portion of the track, they undergo a head-on, elastic collision. Determine the maximum heights to which m, and rise on the curved portion of the track after the collision.

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To determine

The maximum height attained by block $m_1$ and $m_2$ on the curved portion of the track after collision.

Answer to Problem 9.77AP

The maximum height attained by block $m_1$ is $13.9\text{m}$ and block $m_2$ is $0.556\text{m}$

Explanation of Solution

Given info: The mass of block $m_1$ is $2.00\text{kg}$, mass of block $m_2$ is $4.00\text{kg}$ and height between the block and surface level is $5.00\text{m}$.

Write the expression for potential energy of block is,

$U=mgh$

Here,

$h$ is the height between the block and surface level.

$m$ is the mass of block.

$g$ is the acceleration due to gravity.

Write the expression for kinetic energy of block is,

$K=\frac{1}{2}m{v}^2$

Here,

$m$ mass of block.

$v$ is the speed of block.

Write the expression for velocity of blocks using conservation of energy,

$U=K$

Here,

$U$ is the potential energy of block.

$K$ is the kinetic energy of block.

Substitute $mgh$ for $U$ and $\frac{1}{2}m{v}^2$ for $K$ in above expression.

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

Thus, the expression for velocity of block is $\sqrt{2gh}$.

Write the expression for initial velocity of block $1$ and block $2$(as it is clear from the expression for velocity of block that the velocity does not depend upon mass of block. Hence, the initial velocity of block $1$ will be equal to block $2$),

$v=\sqrt{2gh}$

Here,

$v$ is the initial speed of block $m_1$ and block $m_2$.

$g$ is the acceleration due to gravity.

$h$ is the height between the block and surface level.

The value of acceleration due to gravity is $9.8\text{m}/{\text{s}}^{\text{2}}$

Substitute $5.00\text{m}$ for $h$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above expression.

$\begin{array}{c}v=\sqrt{2\times 9.8\text{m}/{\text{s}}^2\times 5.00\text{m}}\\ =9.90\text{m}/\text{s}\end{array}$

Thus, the initial speed of block $m_1$ and block $m_2$ is $9.90\text{m}/\text{s}$.

Write the expression for conservation of linear momentum of block.

$m_{1}v+m_2\left(-v\right)=m_1{v}_1+m_2{v}_2$

Here,

$m_1$ is the mass of block $m_1$.

$m_2$ is the mass of block $m_2$.

$v$ is the initial speed of block $m_1$ and block $m_2$.

$v_1$ is the final speed of block $m_1$ .

$v_2$ is the final speed of block $m_2$.

Substitute $2.00\text{kg}$ for $m_1$, $4.00\text{kg}$ for $m_2$ and $9.90\text{m}/\text{s}$ for $v$ in above expression.

$\begin{array}{c}\left(2.00\text{kg}\times 9.90\text{m}/\text{s}\right)+\left(4.00\text{kg}\times \left(-9.90\text{m}/\text{s}\right)\right)=\left(2.00\text{kg}\right)v_1+\left(4.00\text{kg}\right)v_2\\ 2v_1+4v_2=-19.8\\ v_1+2v_2=-9.9\\ v_1=-9.9-2v_2\end{array}$ (1)

Thus, the expression for final velocity of block $m_1$ is $29.7-2v_2$.

Write the expression for conservation of kinetic energy of the block.

$\begin{array}{c}\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{\left(-v\right)}^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2\\ \left(m_1+m_2\right)v^2=m_1{v}_1^2+m_2{v}_2^2\end{array}$

Here,

$m_1$ is the mass of block $m_1$.

$m_2$ is the mass of block $m_2$.

$v$ is the initial speed of block $m_1$ and block $m_2$.

$v_1$ is the final speed of block $m_1$ .

$v_2$ is the final speed of block $m_2$.

Substitute $2.00\text{kg}$ for $m_1$, $4.00\text{kg}$ for $m_2$, $9.90\text{m}/\text{s}$ for $v$ and $29.7-2v_2$ for $v_1$ in above expression.

$\begin{array}{c}\left(2.00\text{kg}+4.00\text{kg}\right)\times {\left(9.90\text{m}/{\text{s}}^{\text{2}}\right)}^2=\left(2.00\text{kg}\times {\left(-9.9-2v_2\right)}^2\right)+\left(4.00\text{kg}\times v_2^2\right)\\ 3\times {9.90}^2={\left(-9.9-v_2\right)}^2+2v_2^2\\ v_2^2+6.6v_2-196=0\\ v_2=3.30\text{m}/\text{s}\end{array}$

Thus, the final velocity of block $m_2$ is $3.30\text{m}/\text{s}$.

Write the expression for final velocity of block $m_1$.

$v_1=-9.9-2v_2$

Here,

$v_1$ is the final speed of block $m_1$ .

$v_2$ is the final speed of block $m_2$.

Substitute $3.30\text{m}/\text{s}$ for $v_2$ in above expression.

$\begin{array}{c}{v}_1=-9.9-\left(2\times 3.30\right)\\ =-16.5\text{m}/\text{s}\end{array}$

Thus, the final velocity of block $m_1$ is $-16.5\text{m}/\text{s}$.

Write the expression to calculate the maximum height attained by block $m_1$ after collision.

$h_1=\frac{v_1^2}{2g}$

Here,

$h_1$ is the maximum height attained by block $m_1$ after collision.

$v_1$ is the final speed of block $m_1$ .

$g$ is the acceleration due to gravity.

The value of acceleration due to gravity is $9.8\text{m}/{\text{s}}^{\text{2}}$.

Substitute $-16.5\text{m}/\text{s}$ for $v_1$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in above expression.

$\begin{array}{c}{h}_1=\frac{{\left(-16.5\text{m}/\text{s}\right)}^2}{2\times 9.8\text{m}/{\text{s}}^{\text{2}}}\\ =13.9\text{m}\end{array}$

Thus, the maximum height attained by block $m_1$ after collision is $13.9\text{m}$.

Write the expression to calculate the maximum height attained by block $m_2$ after collision.

$h_2=\frac{v_2^2}{2g}$

Here,

$h_2$ is the maximum height attained by block $m_2$ after collision.

$v_2$ is the final speed of block $m_2$ .

$g$ is the acceleration due to gravity.

The value of acceleration due to gravity is $9.8\text{m}/{\text{s}}^{\text{2}}$.

Substitute $3.30\text{m}/\text{s}$ for $v_2$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in above expression.

$\begin{array}{c}{h}_2=\frac{{\left(3.30\text{m}/\text{s}\right)}^2}{2\times 9.8\text{m}/{\text{s}}^{\text{2}}}\\ =0.556\text{m}\end{array}$

Thus, the maximum height attained by block $m_2$ after collision is $0.556\text{m}$.

Conclusion:

Therefore, the maximum height attained by block $m_1$ is $13.9\text{m}$ and block $m_2$ is $0.556\text{m}$.