Question

1.A mass of 15.50 kg comes into contact with a force of 2655 N, directed North, for 0.05 seconds. After this contact, the object obtains a velocity of 45.00 m/s [N45W]. Determine the initial velocity of the object.

Answer 1

In order to calculate the initial velocity, first let's calculate the acceleration generated by this force, using the second law of Newton:

$\begin{gathered} F=m\cdot a\\ \\ 2655=15.5\cdot a\\ \\ a=\frac{2655}{15.5}=171.29\text{ m/s} \end{gathered}$

Let's calculate the change in velocity caused by this acceleration in 0.05 seconds:

$\Delta V=a\cdot t=171.29\cdot0.05=8.56\text{ m/s}$

Now, let's draw a diagram to understand the velocity vectors:

Let's calculate the vertical and horizontal components of the final velocity:

$\begin{gathered} V_y=V\cdot\sin135°\\ \\ V_y=45\cdot\frac{\sqrt{2}}{2}=31.82\text{ m/s}\\ \\ \\ \\ V_x=V\cdot\cos135° \\ V_x=45\cdot(-\frac{\sqrt{2}}{2})=-31.82\text{ m/s}\operatorname{\\} \end{gathered}$

The horizontal component of the final velocity is equal to the initial horizontal velocity:

$V_{ix}=V_x=-31.82\text{ m/s}$

The vertical component of the final velocity is equal to the sum of the initial vertical velocity and the velocity added by the acceleration:

$\begin{gathered} V_y=V_{iy}+\Delta V\\ \\ V_{iy}=V_y-\Delta V=31.82-8.56=23.26\text{ m/s} \end{gathered}$

Now, we can calculate the initial velocity using the Pythagorean theorem:

$\begin{gathered} V_i^2=V_{ix}^2+V_{iy}^2\\ \\ V_i^2=(-31.82)^2+23.26^2\\ \\ V_i^2=1012.51+541.03\\ \\ V_i^2=1553.54\\ \\ V_i=39.41\text{ m/s} \end{gathered}$

To calculate the direction, we can do the following:

$\begin{gathered} \theta=\tan^{-1}(\frac{V_{iy}}{V_{ix}})\\ \\ \theta=\tan^{-1}(\frac{23.26}{-31.82})\\ \\ \theta=\tan^{-1}(-0.73099)\\ \\ \theta=143.83° \end{gathered}$

So the direction is N 53.83 W.