Chapter 7, Problem 38P

To determine

The ratio of distances each block travels.

Answer to Problem 38P

The ratio of distance traveled by the heavier to the lighter massive block is $1/9$ .

Explanation of Solution

Given:

A wooden block is cut into two pieces with the mass ratio as $1:3$ . Initially, both masses are at rest.

Formula used:

According to the conservation of momentum principle, the total momentum before and after the collision is always constant, i.e.

  $\begin{array}{l}{p}_i=p_f\\ {\displaystyle \sum m_i{u}_i}={\displaystyle \sum m_f{u}_f}\end{array}$

And, kinetic energy between both the cars is lost due to the negative work done by the friction. Thus, work done by the friction is

  $W=F\cdot s=Fs \cos \theta$

Calculation:

Let, the more massive block has mass $m_a$ and less massive has mass $m_b$ that is $m_a=3m_b$ . Momentum is conserved in the explosion. Thus, final velocity using the momentum conservation principle is

  $\begin{array}{l}0+0=m_a{v}_a+m_b{v}_b\\ 0=3m_b{v}_a+m_b{v}_b\\ v_a=-\frac{1}{3}{v}_b\end{array}$

Now, formulating final velocity using work done formula

  $W={(K{E}_f-K{E}_i)}_{after collision}=-\frac{1}{2}m{v}^2$

Also, work done due to friction is

  $\begin{array}{l}{W}_{fr}=Fs \cos \theta =-\mu mgs \cos 180°\\ W_{fr}=-\mu mgs\end{array}$

Now, equating both the equation of work done to evaluate distance traveled after the explosion,

  $\begin{array}{l}-\frac{1}{2}m{v}^2=-\mu mgs\\ s=\frac{1}{2}\frac{v^2}{\mu g}\end{array}$

Now, inserting the value of the final velocity to calculate the ratio of distance traveled by both the masses

  $\begin{array}{l}\frac{s_a}{s_b}=\frac{\frac{v_a{}^2}{\mu g}}{\frac{v_b{}^2}{\mu g}}\\ \frac{s_a}{s_b}=\frac{v_a{}^2}{v_b{}^2}=\frac{{\left(-\frac{1}{3}{v}_b\right)}^2}{v_b{}^2}=\frac{1}{9}\end{array}$

Conclusion:

The ratio of distance traveled by the heavier to the lighter massive block is $1/9$ .