MODEL Define the system to be the two gliders. This is an isolated system, so its total momentum is conserved in the collision. The gliders stick together, so this is a perfectly inelastic collision.

VISUALIZE FIGURE 11.17 shows a pictorial representation. We’ve chosen to let the 200 g glider (glider 1) start out moving to the right, so $(v_{\text{i}x}){}_1$ is a positive $3.0\text{m}\text{/}\text{s}$. The gliders move to the left after the collision, so their common final velocity is $v_{\text{f}x}=-0.40\text{m}\text{/}\text{s}$.

FIGURE 11.17 The before-and-after pictorial representation of an inelastic collision.

SOLVE The law of conservation of momentum, $P_{\text{f}x}=P_{\text{i}x}$, is

where we made use of the fact that the combined mass $m_1+m_2$ moves together after the collision. We can easily solve for the initial velocity of the 400 g glider:

The negative sign indicates that the 400 g glider started out moving to the left. The initial speed of the glider, which we were asked to find, is $2.1\text{m}\text{/}\text{s}$.

MODEL This is a two-part problem. Part one, the impact of the bullet on the block, is an inelastic collision. For this part, we define the system to be bullet $+$ block. Momentum is conserved, but mechanical energy is not because some of the energy is transformed into thermal energy. For part two, the subsequent swing, mechanical energy is conserved for the system bullet $+$ block $+$ earth (there’s no friction). The total momentum is conserved, including the momentum of the earth, but that’s not helpful. The momentum of the block with the bullet—which is all that we can calculate—is not conserved because the block is acted on by the external forces of tension and gravity.

VISUALIZE FIGURE 11.18 is a pictorial representation in which we’ve identified before-and-after quantities for both the collision and the swing.

FIGURE 11.18 A ballistic pendulum is used to measure the speed of a bullet.

SOLVE The momentum conservation equation $P_{\text{f}}=P_{\text{i}}$ applied to the inelastic collision gives

The wood block is initially at rest, with ${\left(v_{0x}\right)}_{\text{W}}=0$, so the bullet’s velocity is

where $v_{1x}$ is the velocity of the $\text{block}+\text{bullet}$ immediately after the collision, as the pendulum begins to swing. If we can determine $v_{1x}$ from an analysis of the swing, then we will be able to calculate the speed of the bullet. Turning our attention to the swing, the energy conservation equation $K_{\text{f}}+U_{\text{Gf}}=K_{\text{i}}+U_{\text{Gi}}$ is

We used the total mass $(m_{\text{W}}+m_{\text{B}})$ of the block and embedded bullet, but notice that it cancels out. We also dropped the x-subscript on $v_1$ because for energy calculations we need only speed, not velocity. The speed is zero at the top of the swing $(v_2=0)$, and we’ve defined the y-axis such that $y_1=0\text{m}$. Thus

The initial speed is found simply from the maximum height of the swing. You can see from the geometry of Figure 11.18 that

With this, the initial velocity of the pendulum, immediately after the collision, is

Having found $v_{1x}$ from an energy analysis of the swing, we can now calculate that the speed of the bullet was

REVIEW It would have been very difficult to solve this problem using Newton’s laws, but it yielded to a straightforward analysis based on the concepts of momentum and energy.