Chapter 8, Problem 48P

Two gliders are set in motion on a horizontal air track. A spring of force constant k is attached to the back end of the second glider. As shown in Figure P8.48, the first glider, of mass m₁, moves to the right with speed v₁, and the second glider, of mass m₂, moves more slowly to the right with speed v₂. When m₁ collides with the spring attached to m₂, the spring compresses by a distance x_max, and the gliders then move apart again. In terms of v₁, v₂, m₁, m₂, and k, find (a) the speed rat maximum compression, (b) the maximum compression x_max, and (c) the velocity of each glider after m₁ has lost contact with the spring.

To determine

The speed at the maximum compression for the spring.

Answer to Problem 48P

The speed at the maximum compression for the spring is $v=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$.

Explanation of Solution

Write the expression for conservation of momentum.

  $p_i=p_f$        (I)

Here, $p_f$ is the final momentum of the system and $p_i$ is the initial momentum of the system.

Rewrite the above expression for mass and velocity of gliders.

  $m_1{v}_1+m_2{v}_{\text{2}}=\left(m_1+m_2\right)v$        (II)

Here, $m_1$ is the mass of the first glider, $v_1$ is the velocity of the first glider, $v$ is the velocity of the gliders collides, $m_2$ is the mass of the second glider, and $v_{\text{2}}$ is the velocity of the second glider.

Conclusion:

Rewrite the equation (II) for v.

  $v=\frac{m_1{v}_1+m_2{v}_{\text{2}}}{m_1+m_2}$

Therefore, the speed at the maximum compression for the spring is $v=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$.

To determine

The maximum compression of the spring.

Answer to Problem 48P

The maximum compression of the spring is $x_{\mathrm{m}}=\left(v_1-v_2\right)\sqrt{\frac{m_1{m}_2}{k\left(m_1+m_2\right)}}$.

Explanation of Solution

Write the expression for law of conservation of energy for the two gliders.

  $\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=\frac{1}{2}\left(m_1+m_2\right)v^2+\frac{1}{2}k{x}_m^2$        (III)

Here, $k$ is the spring constant and $x_{\mathrm{m}}$ is the maximum extension distance of the spring.

Rewrite the above expression.

  $m_1{v}_1^2+m_2{v}_2^2=\left(m_1+m_2\right)v^2+k{x}_m^2$        (IV)

Conclusion:

Substitute $\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$ for $v$ from part (a) in equation (IV).

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

Rewrite the above equation for $x_{\max }$.

$\begin{array}{c}{m}_1{v}_1^2+m_2{v}_2^2-\frac{{\left(m_1{v}_1+m_2{v}_2\right)}^2}{\left(m_1+m_2\right)}=k{x}_m^2\\ x_{\mathrm{m}}^2=\left(\frac{1}{k\left(m_1+m_2\right)}\right)\left[\left(m_1+m_2\right)m_1{v}_1^2+\left(m_1+m_2\right)m_2{v}_2^2-{\left(m_1{v}_1\right)}^2-{\left(m_2{v}_2\right)}^2-2m_1{m}_2{v}_1{v}_2\right]\\ x_m=\sqrt{\frac{m_1{m}_2\left(v_1^2+v_2^2-2v_1{v}_2\right)}{k\left(m_1+m_2\right)}}\\ =\left(v_1-v_2\right)\sqrt{\frac{m_1{m}_2}{k\left(m_1+m_2\right)}}\end{array}$

Therefore, the maximum compression of the spring is $x_{\mathrm{m}}=\left(v_1-v_2\right)\sqrt{\frac{m_1{m}_2}{k\left(m_1+m_2\right)}}$.

To determine

The velocity of the each glider after first gliders has lost the contact with spring.

Answer to Problem 48P

The velocity of the second glider is $v_{2f}=\frac{2m_1{v}_1+\left(m_2-m_1\right)v_2}{m_1+m_2}$ and first glider is

$v_{1f}=\frac{2m_2{v}_2+\left(m_1-m_2\right)v_1}{m_1+m_2}$.

Explanation of Solution

Write the expression for conservation of momentum.

  $m_1\left(v_1-v_{1f}\right)=m_2\left(v_{2f}-v_2\right)$        (V)

Here, $v_{1f}$ is the velocity of the first glider after lost contact with spring and $v_{\text{2}f}$ is the velocity of the second glider after lost contact with spring.

Write the expression for law of conservation of energy for the each gliders.

  $\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{v}_{2f}^2$        (VI)

Rewrite the equation (VI).

  $\begin{array}{c}{m}_1{v}_1^2+m_2{v}_2^2=m_1{v}_{1f}^2+m_2{v}_{2f}^2\\ m_1{v}_1^2-m_1{v}_{1f}^2=m_2{v}_{2f}^2-m_2{v}_2^2\\ m_1\left(v_1^2-v_{1f}^2\right)=m_2\left(v_{2f}^2-v_2^2\right)\end{array}$

Conclusion:

Simplify the above relation and substitute equation (V).

  $\begin{array}{c}{m}_2\left(v_{2f}-v_2\right)\left(v_1+v_{1f}\right)=m_2\left(v_{2f}-v_2\right)\left(v_{2f}+v_2\right)\\ \left(v_1+v_{1f}\right)=\left(v_{2f}+v_2\right)\\ v_{1f}=v_{2f}+v_2-v_1\end{array}$

Substitute the above relation in equation (V) and rearrange for $v_{2f}$ and $v_{1f}$.

  $\begin{array}{c}{m}_1\left(v_1-v_{2f}-v_2+v_1\right)=m_2\left(v_{2f}-v_2\right)\\ m_1\left(2v_1-v_{2f}-v_2\right)=m_2\left(v_{2f}-v_2\right)\\ v_{2f}=\frac{2m_1{v}_1+\left(m_2-m_1\right)v_2}{m_1+m_2}\end{array}$

  $v_{1f}=\frac{2m_2{v}_2+\left(m_1-m_2\right)v_1}{m_1+m_2}$

Therefore, the velocity of the second glider is $v_{2f}=\frac{2m_1{v}_1+\left(m_2-m_1\right)v_2}{m_1+m_2}$ and first glider is

$v_{1f}=\frac{2m_2{v}_2+\left(m_1-m_2\right)v_1}{m_1+m_2}$.