Question

The two spheres are rigidly connected to the rod of negligible mass and are initially at rest on the smooth horizontal surface. A force F is suddenly applied to one sphere in the y-direction and imparts an impulse of 9.8 N·s during a negligibly short period of time. As the spheres pass the dashed position, calculate the magnitude of the velocity of each one.

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity of the both spheres is $v=4.62 m/s$

Explanation:

The free body diagram of this question is shown on the second uploaded image

     Looking at this diagram we can deduce that there is no impulse force along the horizontal direction so

      The mathematical equation for the impulse force along the horizontal axis is  

           $\int\limits {\sum F_s} \, dt = 0$

 The mathematical equation for the impulse force along the horizontal axis is

           $\int\limits {\sum F_y} \, dt = \Delta I_y$

  Where $I_y$ is the impulse  momentum along the y-axis and this is mathematically given as

         $\Delta I_y = 2mv_y$

substituting 9.8 N.s for $\Delta I_y$ and 1.5kg for m (mass of sphere) and the making $v_y$ the subject

        $v_y = \frac{9.8}{2 *1.5}$

            $=3.267 m/s$

The sum of the moment about the point I is mathematically represented as

         $\int\limits {\sum M_I } \, dt =\Delta H_I$

From the free body diagram  $\int\limits {\sum M_I } \, dt = I * 0.3 = 9.8 *0.3=2.94N.m$

$\Delta H_I$ is the angular momentum along the horizontal axis  given as

                  $\Delta H_I$ $= 2mv_x(0.3)$

substituting parameters into above equation

                  $2.94 = 2 * 1.5 * v_x * 0.3$

Making $v_x$ the subject

                 $v_x = \frac{2.94}{2*1.5*0.3}$

                      $=3.267 m/s$

the velocity of the sphere is mathematically represented as

                 $v = \sqrt{(v_x)^2 + (v_y)^2}$

Now substituting values

                $v = \sqrt{3.267^2 + 3.267^2}$

                   $v=4.62 m/s$