Question

part 8 of 13
evaluate known quantities in the conservation equations. (cont.)
before the collision, puck 1 travels in the +y - direction with a speed of 1.05 m/s so that ( v_{y1}=+1.05m/s ).
what is the value of ( v_{x1} )?
1.45
incorrect after the collision, the two pucks travel together with a speed of 1.85 m/s in a direction ( 28.0^{circ} ) above the negative x - axis. m/s
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Explanation:

Step1: Analyze the x - direction momentum conservation

Before the collision, Puck 1 has no x - component of velocity (\(v_{x1}=0\) since it travels in the \(+y\) - direction). Let the mass of each puck be \(m\).
The momentum before the collision in the \(x\) - direction \(p_{x,i}=m\times v_{x1}+m\times v_{x2}\) (where \(v_{x1}\) is the \(x\) - component of velocity of Puck 1 before collision and \(v_{x2}\) is the \(x\) - component of velocity of Puck 2 before collision). After the collision, the two - puck system has a velocity \(v = 1.85\ m/s\) at an angle \(\theta=28.0^{\circ}\) above the negative \(x\) - axis. The \(x\) - component of the velocity of the combined system after the collision \(v_{x,f}=-v\cos\theta\). Using the law of conservation of momentum \(p_{x,i}=p_{x,f}\), \(m\times v_{x1}+m\times v_{x2}=(m + m)\times v_{x,f}\). Since \(v_{x1} = 0\), \(v_{x2}=2v_{x,f}\).

Step2: Calculate \(v_{x,f}\)

We know that \(v = 1.85\ m/s\) and \(\theta = 28.0^{\circ}\). Then \(v_{x,f}=-v\cos\theta=- 1.85\times\cos(28.0^{\circ})\). Using \(\cos(28.0^{\circ})\approx0.883\), \(v_{x,f}=-1.85\times0.883=-1.63355\ m/s\).

Step3: Find \(v_{x2}\)

Since \(v_{x2}=2v_{x,f}\), \(v_{x2}=2\times(- 1.63355)=-3.27\ m/s\)

Answer:

\(-3.27\ m/s\)