Question

A mass of 15.50 kg comes into contact with a force of 2655 N, directed North, for 0.05 seconds. Afterthis contact, the object obtains a velocity of 45.00 m/s [N45W]. Determine the initial velocity of theobject. Please calculate only the direction for this question.

Answer 1

In order to find the direction of the initial velocity, first let's calculate the change in velocity caused by the acceleration:

Now, let's calculate the horizontal and vertical components of the final velocity.

The direction is N45W, therefore the angle is 90° + 45° = 135°:

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

The change in the velocity occurred in north direction, so let's calculate the initial vertical velocity:

`V_(iy)=V_y-\Delta V=31.82-8.5645=23.2555\text{ m/s}`

Now, to calculate the direction, we can use the arc tangent below:

`\theta=\tan^(-1)((V_(iy))/(V_(ix)))=\tan^(-1)((23.2555)/(-31.82))=-36.16°`

Since the direction is between north and west, we need to add 180° to the result, so the angle will be 143.84°.

This angle is equal to the direction N 53.84° W.