Chapter 7, Problem 79GP

(a)

To determine

The speed of the two blocks after collision.

Answer to Problem 79GP

The speed of the small block after collision is $4.38\text{ m/s}$ .

The speed of the bigger block after collision is $4.02\text{ m/s}$ .

Explanation of Solution

Given:

The mass of small block is $m=2.20\text{ kg}$ .

The angle is $\theta =30°$ .

The mass of bigger block is $M=7.00\text{ kg}$ .

The height is $h=3.6\text{ m}$ .

Formula used:

The expression for the velocity is,

  $v=\sqrt{2gh}$

Here, g is the acceleration due to gravity and h is the height.

The expression for relative velocity is,

  $u_1-u_2=-\left(v_1-v_2\right)$

Here, $u_1$ is the initial velocity of small block, $u_2$ is the initial velocity of the bigger block, $v_1$ is the final velocity of small block, and $v_2$ is the final velocity of the bigger block.

The expression for law of conservation of momentum is,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Here, $m_1$ is the mass of the small block, $m_2$ is the mass of bigger block, $u_1$ is the initial velocity of small block, $u_2$ is the initial velocity of the bigger block, $v_1$ is the final velocity of small block, and $v_2$ is the final velocity of the bigger block.

Calculation:

Consider the initial velocity of the bigger block $u_2=0$ .

Consider the acceleration due to gravity as $g=9.8{\text{m/s}}^2$ .

The speed of the small block is,

  $\begin{array}{l}{u}_1=\sqrt{2gh}\\ u_1=\sqrt{2\times \left(9.8{\text{m/s}}^2\right)\left(3.6\text{ m}\right)}\\ u_1=8.4\text{ m/s}\end{array}$

The relative velocity is,

  $\begin{array}{l}{u}_1-u_2=-\left(-v_1-v_2\right)\\ u_1-\left(0\right)=v_2+v_1\\ u_1=v_2+v_1\\ v_1=u_1-v_2\end{array}$ …… (1)

Apply the conservation of momentum.

  $\begin{array}{l}m{u}_1+M{u}_2=-m{v}_1+M{v}_2\\ m{u}_1+M\left(0\right)=-m\left(u_1-v_2\right)+M{v}_2\\ m{u}_1=-m{u}_1+m{v}_2+M{v}_2\\ 2m{u}_1=\left(m+M\right)v_2\end{array}$

  $\begin{array}{l}{v}_2=\frac{2\times \left(2.20\text{kg}\right)\left(8.4\text{m/s}\right)}{\left(2.20\text{kg}\right)+\left(7.00\text{ kg}\right)}\\ v_2=4.02\text{ m/s}\end{array}$

The speed of the small block after collision using Equation (1) is,

  $\begin{array}{l}{v}_1=u_1-v_2\\ v_1=\left(8.4\text{m/s}\right)-\left(4.02\text{m/s}\right)\\ v_1=4.38\text{m/s}\end{array}$

Conclusion:

Thus, the speed of the small block after collision is $v_1=4.38\text{ m/s}$ and the speed of the bigger block after collision is $v_2=4.02\text{ m/s}$ .

(b)

To determine

The distance in the incline that smaller mass will go.

Answer to Problem 79GP

The distance in the incline that smaller mass will go is $s=1.96\text{m}$ .

Explanation of Solution

Given:

The mass of small block is $m=2.20\text{ kg}$ .

The angle is $\theta =30°$ .

The mass of bigger block is $M=7.00\text{ kg}$ .

The height is $h=3.6\text{ m}$ .

Formula used:

The expression for the acceleration is,

  $a=g\sin \theta$

Here, g is the acceleration due to gravity and $\theta$ is the angle.

The expression for velocity is,

  $u^2=v^2-2as$

Here, v is the final velocity, a is the acceleration due to gravity, and s is the distance.

Calculation:

From part (a), the speed of the small block after collision is $v_1=4.38\text{ m/s}$ .

The acceleration is,

  $\begin{array}{l}a=g\sin \theta \\ a=\left(9.8{\text{ m/s}}^2\right)\sin 30°\\ a=4.9{\text{ m/s}}^2\end{array}$

The distance in the incline the smaller mass will go is,

  $\begin{array}{l}{v}_1^2-2as=0\\ {\left(4.38\text{ m/s}\right)}^2-2\times \left(4.9{\text{m/s}}^2\right)s=0\\ s=1.96\text{m}\end{array}$

Conclusion:

Thus, the distance in the incline that smaller mass will go is $s=1.96\text{m}$ .