Question

part 8 of 13
evaluate known quantities in the conservation equations. (cont.)
after the collision, both pucks travel together with a speed of 1.85 m/s in a direction 28.0° above the negative x - axis (152° from the positive x - axis). the final x - component of their
common velocity is then

what is the value of ?

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Explanation:

Step1: Analyze the y - component of velocity

The formula for the y - component of a velocity vector \(v\) with magnitude \(|v|\) and angle \(\theta\) (measured from the positive x - axis) is \(v_y = v\sin\theta\). Here, \(v = 1.85\ m/s\) and \(\theta=152^{\circ}\) (or we can also use \(\theta = 28^{\circ}\) above the negative x - axis. Since \(\sin(180^{\circ}-\alpha)=\sin\alpha\), \(\sin(152^{\circ})=\sin(28^{\circ})\)).

Step2: Calculate \(v_{fy}\)

Using the formula \(v_{fy}=v\sin\theta\), substitute \(v = 1.85\ m/s\) and \(\theta = 152^{\circ}\) (or \(\theta = 28^{\circ}\) for the non - standard position angle calculation).

$$v_{fy}=(1.85\ m/s)\sin(152^{\circ})=(1.85\ m/s)\sin(28^{\circ})$$

We know that \(\sin(28^{\circ})\approx0.4695\)

$$v_{fy}=1.85\times0.4695\approx0.869\ m/s$$

Answer:

\(0.87\ m/s\)