Chapter 7, Problem 80GP

To determine

The upper limit on mass $m$ .

Answer to Problem 80GP

The upper limit on mass $m$ is 2.33 kg.

Explanation of Solution

Given:

  $m=2.20 \text{kg}$ and $M=7 \text{kg}$ .

Let A be mass $m$ and B represent mass $M$ . Mass $M$ is at rest initially.

Formula used:

According to the conservation of momentum principle, the total momentum before and after the collision remains constant that is,

  $p_i=p_f$

Calculation:

For elastic collision, $e=1$ . Using this relation, a relationship is set up between the velocities $v_a-v_b=v_b^{\text{'}}-v_a^{\text{'}}$ . Since mass $M$ is at rest.

  $v_a^{\text{'}}=v_b^{\text{'}}-v_a$

After hitting mass $M$ , mass $m$ will move in backward direction which makes $v_a^{\text{'}}$ negative. For second collision, it is must that mass $m$ goes up the incline in negative direction and then after gaining enough potential energy to gain velocity greater than mass $M$ , mass $m$ will go down the incline with a positive velocity of $-v_a^{\text{'}}$ . Thus, $-v_a^{\text{'}}>v_b^{\text{'}}$ or $v_a^{\text{'}}<-v_b^{\text{'}}$ . Now, using the velocity relationship, we get

  $\begin{array}{l}{v}_b^{\text{'}}-v_a<-v_b^{\text{'}}\\ v_b^{\text{'}}<\frac{1}{2}{v}_a\end{array}$

Now, using momentum conservation for the original collision and using the above inequality based result.

  $\begin{array}{l}m{v}_a=m{v}_a^{\text{'}}+M{v}_b^{\text{'}}\\ m{v}_a=m(v_b^{\text{'}}-v_a)+M{v}_b^{\text{'}}\\ v_b^{\text{'}}=\frac{2m}{m+M}{v}_a\\ \frac{2m}{m+M}{v}_a<\frac{1}{2}{v}_a\\ 4m<m+M\\ m<\frac{1}{3}M\end{array}$

After inserting the mass values, the final result would be

  $m<\frac{7}{3}=2.33kg$

Conclusion:

Thus, upper limit of mass $m$ is 2.33 kg.