Chapter 7, Problem 31P

To determine

To compare: The speed of second projectile with the first.

Answer to Problem 31P

The second projectile is $\sqrt{2}$ times faster than the first.

Explanation of Solution

Given:

In the experiment, projectile 1 results in a maximum height of the pendulum $h_1=2.6\text{cm}$ and the second pendulum swings twice as high as the first one that is $h_2=5.2\text{cm}$

Formula used:

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

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

For the projectile motion velocity is related as $v=\sqrt{2gh}$

Using this in the main formula,

  $u=\frac{m+M}{m}\sqrt{2gh}$

Calculation:

Now, comparing both the velocities with the same mass,

  $\begin{array}{l}\frac{u_2}{u_1}=\frac{\frac{m+M}{m}\sqrt{2g{h}_2}}{\frac{m+M}{m}\sqrt{2g{h}_1}}\\ \frac{u_2}{u_1}=\sqrt{\frac{h_2}{h_1}}=\sqrt{\frac{5.2}{2.6}}=\sqrt{2}\end{array}$

Conclusion:

Therefore, the second projectile is $\sqrt{2}$ times faster than the first.