Question 8.7: Suppose the following experiment is performed. A 0.250-kg ob...

Solving m_1  v_1 =  m_1v^′_1  cos θ_1  +  m_2v^′_2  cos θ_2  and  0  =  m_1v^′_1  sin θ_1  + m_2v^′_2  sin θ_2  for  v^′_2  sin θ_2 and taking the ratio yields an equation (because \left(tan θ = \frac{sin θ}{cos θ}\right) in which all but one quantity is known:

tan θ_2 = \frac{v^′_1 sin θ_1}{ v^′_1 cos θ_1 − v_1} .                     (8.68)

Entering known values into the previous equation gives

tan θ_2 = \frac{(1.50 m/s)(0.7071)}{(1.50 m/s)(0.7071) − 2.00 m/s} = −1.129.                    (8.69)

Thus,

θ_2 = tan^{−1}(−1.129) = 311.5º ≈ 312º.                   (8.70)

Angles are defined as positive in the counter clockwise direction, so this angle indicates that m_2 is scattered to the right in Figure 8.12, as expected (this angle is in the fourth quadrant). Either equation for the x – or y -axis can now be used to solve for v^′_2 , but the latter equation is easiest because it has fewer terms.

v^′_2 = −\frac{ m_1}{m_2} v^′_1 \frac{sin θ_1}{sin θ_2}                  (8.71)

Entering known values into this equation gives

v^′_2 = −\left(\frac{0.250 kg}{ 0.400 kg}\right) (1.50 m/s)\left(\frac{0.7071}{−0.7485}\right) .                       (8.72)

Thus,

v^′_2 = 0.886 m/s.               (8.73)

Discussion
It is instructive to calculate the internal kinetic energy of this two-object system before and after the collision. (This calculation is left as an end-of-chapter problem.) If you do this calculation, you will find that the internal kinetic energy is less after the collision, and so the collision is inelastic. This type of result makes a physicist want to explore the system further