Question 6.MF.2.1: A puck with mass m = 1.00 kg traveling at v = 3.00 m/s in th...
A puck with mass m = 1.00 kg traveling at v = 3.00 m/s in the positive x-direction on a frictionless surface strikes a second puck of mass M = 2.00 kg, at rest at the origin. The first puck continues at v_{f}=1.00 m / s at an angle of Φ = -30° with respect to the positive x-axis. Find the speed V and direction θ of the second puck.
Nothing is said of energy conservθation, so only momentum can be assumed conserved:
x-direction: m v=m v_{f} \cos \phi+M V \cos \theta (1)
y-direction: 0=m v_{f} \sin \phi+M V \sin \theta (2)
All quantities in Equations (1) and (2) are given except V and θ. First, solve Equation (1) for MV cos θ and Equation (2) for MV sin θ:
\begin{gathered}M V \cos \theta=m v-m v_{f} \cos \phi=2.13 kg m / s (3) \\M V \sin \theta=-m v_{f} \sin \phi=0.500 kg m / s (4)\end{gathered}
Divide Equation (4) by Equation (3), obtaining
\begin{aligned}\tan \theta &=\frac{M V \sin \theta}{M V \cos \theta}=\frac{0.500 kg \cdot m / s }{2.13 kg \cdot m / s }=0.235 \\\theta &=\tan ^{-1}(0.235)=13.2^{\circ}\end{aligned}
To obtain the speed, substitute this result back into either Equation (3) or Equation (4), getting V = 1.09 m/s.
In certain cases the identity \cos ^{2} \theta+\sin ^{2} \theta=1 is useful. Then Equation (3) and (4) can be squared and added, and the identity applied to eliminate θ in favor of V. Substituting the answer for V back into either Equation (3) or (4) then yields an equation for θ.