Collisions in 2-dimensions

Suppose that an object of mass $m_1$, moving with initial speed $v_{i1}$, strikes a second object, of mass $m_2$, which is initially at rest. Suppose, further, that the collision is not head-on, so that after the collision the first object moves off at an angle $\theta_1$ to its initial direction of motion, whereas the second object moves off at an angle $\theta_2$ to this direction. Let the final speeds of the two objects be $v_{f1}$ and $v_{f2}$, respectively. See Fig. 55.

We are again considering a system in which there is zero net external force (the forces associated with the collision are internal in nature). It follows that the total momentum of the system is a conserved quantity. However, unlike before, we must now treat the total momentum as a vector quantity, since we are no longer dealing with 1-dimensional motion. Note that if the collision takes place wholly within the $x$-$y$ plane, as indicated in Fig. 55, then it is sufficient to equate the $x$- and $y$- components of the total momentum before and after the collision.

Consider the $x$-component of the system's total momentum. Before the collision, the total $x$-momentum is simply $m_1 v_{i1}$, since the second object is initially stationary, and the first object is initially moving along the $x$-axis with speed $v_{i1}$. After the collision, the $x$-momentum of the first object is $m_1 v_{f1} \cos\theta_1$: i.e., $m_1$ times the $x$-component of the first object's final velocity. Likewise, the final $x$-momentum of the second object is $m_2 v_{f2} \cos\theta_2$. Hence, momentum conservation in the $x$-direction yields

$$m_1 v_{i1} = m_1 v_{f1} \cos\theta_1+ m_2 v_{f2} \cos\theta_2.$$ (240)

Consider the $y$-component of the system's total momentum. Before the collision, the total $y$-momentum is zero, since there is initially no motion along the $y$-axis. After the collision, the $y$-momentum of the first object is $-m_1 v_{f1} \sin\theta_1$: i.e., $m_1$ times the $y$-component of the first object's final velocity. Likewise, the final $y$-momentum of the second object is $m_2 v_{f2} \sin\theta_2$. Hence, momentum conservation in the $y$-direction yields

$$m_1 v_{f1} \sin\theta_1= m_2 v_{f2} \sin\theta_2.$$ (241)

For the special case of an elastic collision, we can equate the total kinetic energies of the two objects before and after the collision. Hence, we obtain

$$\frac{1}{2} m_1 v_{i1}^{ 2} =\frac{1}{2} m_1 v_{f1}^{ 2}+\frac{1}{2} m_2 v_{f2}^{ 2}.$$ (242)

Given the initial conditions (i.e., $m_1$, $m_2$, and $v_{i1}$), we have a system of three equations [i.e., Eqs. (240), (241), and (242)] and four unknowns (i.e., $\theta_1$, $\theta_2$, $v_{f1}$, and $v_{f2}$). Clearly, we cannot uniquely solve such a system without being given additional information: e.g., the direction of motion or speed of one of the objects after the collision.

Figure 56: A totally inelastic collision in 2-dimensions.

Figure 56 shows a 2-dimensional totally inelastic collision. In this case, the first object, mass $m_1$, initially moves along the $x$-axis with speed $v_{i1}$. On the other hand, the second object, mass $m_2$, initially moves at an angle $\theta_i$ to the $x$-axis with speed $v_{i2}$. After the collision, the two objects stick together and move off at an angle $\theta_f$ to the $x$-axis with speed $v_f$. Momentum conservation along the $x$-axis yields

$$m_1 v_{i1} + m_2 v_{i2} \cos\theta_i = (m_1+m_2) v_f \cos\theta_f.$$ (243)

Likewise, momentum conservation along the $y$-axis gives

$$m_2 v_{i2} \sin\theta_i = (m_1+m_2) v_f \sin\theta_f.$$ (244)

Given the initial conditions (i.e., $m_1$, $m_2$, $v_{i1}$, $v_{i2}$, and $\theta_i$), we have a system of two equations [i.e., Eqs. (243) and (244)] and two unknowns (i.e., $v_f$ and $\theta_f$). Clearly, we should be able to find a unique solution for such a system.