Chapter 6, Problem 71AP

A block with mass m₁ = 0.500 kg is released from rest on a frictionless track at a distance h₁, = 2.50 m above the top of a table. It then collides elastically with an object having mass m₂ = 1.00 kg that is initially at rest on the table, as shown in Figure P6.71. (a) Determine the velocities of the two objects just after the collision. (b) How high up the track does the 0.500-kg object travel back after the collision? (c) How far away from the bottom of the table does the1.00-kg object land, given that the height of the table h₂ = 2.00 m? (d) How far away from the bottom of the table does the 0.500-kg object eventually band?

Figure P6.71

To determine

The velocity of each object just after the collision.

Answer to Problem 71AP

The velocity of first object just after collision is $-2.33\text{m/s}$ and the velocity of second object just after collision is $4.67\text{m/s}$.

Explanation of Solution

Given info:

The acceleration due to gravity is $9.80{\text{m/s}}^2$.

The height of the track is $2.50\text{m}$.

The first mass is $0.500\text{kg}$.

The mas of the second mass is $1.00\text{kg}$.

Explanation:

The Formula to calculate the conservation of mechanical energy,

$\frac{1}{2}{m}_1{v}_{1i}^2+mg\left(0\right)=\frac{1}{2}{m}_1\left(0\right)+m_{1}g{h}_1$ (I)

• $h_1$ is the height of the track
• $m_1$ is the first mass
• $g$ is the acceleration due to gravity
• $v_{1i}$ is the speed of first object

Rearranging the formula gives the speed as,

$v_{1i}=\sqrt{2g{h}_1}$

Substitute $9.80{\text{m/s}}^2$ for $g$ and $2.50\text{m}$ for $h_1$ in equation (I) to find $v_{1i}$ .

$\begin{array}{c}{v}_{1i}=\sqrt{2\left(9.80{\text{m/s}}^2\right)\left(2.50\text{m}\right)}\\ =7.00\text{m/s}\end{array}$

The Formula to calculate the conservation of momentum just before to the collision to just after the collision,

$m_1{v}_{1f}+m_2{v}_{2f}=m_1{v}_{1i}+0$ (II)

• $v_{1f}$ is the speed of first bock just after collision
• $v_{2f}$ is the speed of second bock just after collision

Substitute $0.500\text{kg}$ for $m_1$, $1.00\text{kg}$ for $m_2$ and $7.00\text{m/s}$ for $v_{1i}$ in equation (II) .

$\left(0.500\text{kg}\right)v_{1f}+\left(1.00\text{kg}\right)v_{2f}=\left(0.500\text{kg}\right)\left(7.00\text{m/s}\right)+0$

Therefore,

$v_{1f}+2v_{2f}=7.00\text{m/s}$

For perfectly elastic head on collision, with second object at rest initially,

$v_{2f}=v_{1f}+v_{1i}$ (III)

Substitute $7.00\text{m/s}$ for $v_{1i}$.

$v_{2f}=v_{1f}+7.00\text{m/s}$

Substituting equation (III) into (II) gives,

$v_{1f}=2\left(v_{1f}+7.00\text{m/s}\right)=7.00\text{m/s}$

Therefore, the speed of first block just after collision is.

$\begin{array}{c}{v}_{1f}=\frac{-7.00\text{m/s}}{3}\\ =-2.33\text{m/s}\end{array}$

Substitute $7.00\text{m/s}$ for $v_{1i}$ and $-2.33\text{m/s}$ for $v_{1f}$ in equation (III).

$\begin{array}{c}{v}_{2f}=-2.33\text{m/s}+7.00\text{m/s}\\ \text{=4}\text{.67}\text{m/s}\end{array}$

Thus, the velocity of first object just after collision is $-2.33\text{m/s}$ and the velocity of second object just after collision is $4.67\text{m/s}$.

Conclusion:

The velocity of first object just after collision is $-2.33\text{m/s}$ and the velocity of second object just after collision is $4.67\text{m/s}$.

To determine

The height at which the 0.500 kg object travel back after the collision.

Answer to Problem 71AP

The height at which the 0.500 kg object travel back after the collision is $0.277\text{m}$.

Explanation of Solution

The Formula to calculate the conservation of energy,

$\frac{1}{2}{m}_1{\left(0\right)}^2+m_{1}g{h}_1^{\text{'}}=\frac{1}{2}{m}_1{v}_{1f}^2+mg\left(0\right)$

• $h_1^{\text{'}}$ is the height at which the 0.500 kg object travel back after the collision

Rearranging the equation gives the maximum height as,

$h_1^{\text{'}}=\frac{v_{1f}^2}{2g}$ (IV)

Substitute $-2.33\text{m/s}$ for $v_{1f}$ and $9.80{\text{m/s}}^2$ for $g$ in equation (IV) to calculate the maximum height the first object reaches $h_1^{\text{'}}$.

$\begin{array}{c}{h}_1^{\text{'}}=\frac{{\left(-2.33\text{m/s}\right)}^2}{2\left(9.80{\text{m/s}}^2\right)}\\ =0.277\text{m}\end{array}$

Thus, the height at which the 0.500 kg object travel back after the collision is $0.277\text{m}$.

Conclusion:

The height at which the 0.500 kg object travel back after the collision is $0.277\text{m}$.

To determine

The distance at which the 1.00 kg object lands.

Answer to Problem 71AP

The distance at which the 1.00 kg object lands is $2.98\text{m}$.

Explanation of Solution

Given info:

The height of the table is $2.00\text{m}$.

Explanation:

The Formula to calculate the distance at which the 1.00 kg object lands,

$\Delta x=v_{2f}\sqrt{\frac{2\Delta y}{a_y}}$

• $\Delta x$ is the distance at which the 1.00 kg object lands
• $\Delta y$ is the height of the table
• $a_y$ is the acceleration due to gravity acting on the block

Substitute $4.67\text{m/s}$ for $v_{2f}$, $-2.00\text{m}$ for $\Delta y$ and $-9.80{\text{m/s}}^2$ for $a_y$ to find $\Delta x$.

$\begin{array}{c}\Delta x=\left(4.67\text{m/s}\right)\sqrt{\frac{2\left(-2.00\text{m/s}\right)}{-9.80\text{m/s}}}\\ =2.98\text{m}\end{array}$

Thus, the distance at which the 1.00 kg object lands is $2.98\text{m}$.

Conclusion:

The distance at which the 1.00 kg object lands is $2.98\text{m}$.

To determine

The distance at which the 0.500 kg object lands.

Answer to Problem 71AP

The distance at which the 0.500 kg object lands is $1.49\text{m}$.

Explanation of Solution

The Formula to calculate the distance at which the 0.500 kg object lands,

$\Delta x=v_{1f}\sqrt{\frac{2\Delta y}{a_y}}$

• $\Delta x$ is the distance at which the 1.00 kg object lands
• $\Delta y$ is the height of the table
• $a_y$ is the acceleration due to gravity acting on the block

Substitute $2.33\text{m/s}$ for $v_{2f}$, $-2.00\text{m}$ for $\Delta y$ and $-9.80{\text{m/s}}^2$ for $a_y$ to find $\Delta x$.

$\begin{array}{c}\Delta x=\left(2.33\text{m/s}\right)\sqrt{\frac{2\left(-2.00\text{m/s}\right)}{-9.80\text{m/s}}}\\ =1.49\text{m}\end{array}$

Thus, the distance at which the 0.500 kg object lands is $1.49\text{m}$.

Conclusion:

The distance at which the 0.500 kg object lands is $1.49\text{m}$.