Question 8.9: An automobile collision A 1000 kg car traveling north at 15 ......
An automobile collision
A 1000 \mathrm{~kg} car traveling north at 15 \mathrm{~m} / \mathrm{s} collides with a 2000 \mathrm{~kg} truck traveling east at 10 \mathrm{~m} / \mathrm{s}. The occupants, wearing seat belts, are uninjured, but the two vehicles move away from the impact point as one. The insurance adjustor asks you to find the velocity of the wreckage just after impact. What is your answer?
IDENTIFY and SET UP Any horizontal external forces (such as friction) on the vehicles during the collision are very small compared with the forces that the colliding vehicles exert on each other. (We’ll verify this below.) So we can treat the cars as an isolated system, and the momentum of the system is conserved. Figure 8.20 shows our sketch and the x – and y-axes. We can use Eqs. (8.15)
P_x=p_{A x}+p_{B x}+\cdots, \quad P_y=p_{A y}+p_{B y}+\cdots, \quad P_z=p_{A z}+p_{B z}+\cdots (8.15)
to find the total momentum \overrightarrow{{P}} before the collision. The momentum has the same value just after the collision; hence we can find the velocity \vec{V} just after the collision (our target variable) by using \overrightarrow{{P}}=M \overrightarrow{{V}}, where M=m_{\mathrm{C}}+m_{\mathrm{T}}=3000 \mathrm{~kg} is the mass of the wreckage.
EXECUTE From Eqs. (8.15), the components of \overrightarrow{{P}} are
\begin{aligned} P_{x} & =p_{\mathrm{C} x}+p_{\mathrm{T} x}=m_{\mathrm{C}} v_{\mathrm{C} x}+m_{\mathrm{T}} v_{\mathrm{T} x} \\ & =(1000 \mathrm{~kg})(0)+(2000 \mathrm{~kg})(10 \mathrm{~m} / \mathrm{s})=2.0 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} \\ P_{y} & =p_{\mathrm{C} y}+p_{\mathrm{T} y}=m_{\mathrm{C}} v_{\mathrm{C} y}+m_{\mathrm{T}} v_{\mathrm{T} y} \\ & =(1000 \mathrm{~kg})(15 \mathrm{~m} / \mathrm{s})+(2000 \mathrm{~kg})(0)=1.5 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} \end{aligned}
The magnitude of \overrightarrow{{P}} is
\begin{aligned} P & =\sqrt{\left(2.0 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\right)^{2}+\left(1.5 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\right)^{2}} \\ & =2.5 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} \end{aligned}
and its direction is given by the angle \theta shown in Fig. 8.20:
\tan \theta=\frac{P_{y}}{P_{x}}=\frac{1.5 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}{2.0 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}=0.75 \quad \theta=37^{\circ}
From \overrightarrow{{P}}=M \overrightarrow{{V}}, the direction of the velocity \overrightarrow{{V}} just after the collision is also \theta=37^{\circ}. The velocity magnitude is
V=\frac{P}{M}=\frac{2.5 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}{3000 \mathrm{~kg}}=8.3 \mathrm{~m} / \mathrm{s}
EVALUATE As you can show, the initial kinetic energy is 2.1 \times 10^{5} \mathrm{~J} and the final value is 1.0 \times 10^{5} \mathrm{~J}. In this inelastic collision, the total kinetic energy is less after the collision than before.
We can now justify our neglect of the external forces on the vehicles during the collision. The car’s weight is about 10,000 \mathrm{~N}; if the coefficient of kinetic friction is 0.5 , the friction force on the car during the impact is about 5000 \mathrm{~N}. The car’s initial kinetic energy is \frac{1}{2}(1000 \mathrm{~kg})(15 \mathrm{~m} / \mathrm{s})^{2}=1.1 \times 10^{5} \mathrm{~J}, so -1.1 \times 10^{5} \mathrm{~J} of work must be done to stop it. If the car crumples by 0.20 \mathrm{~m} in stopping, a force of magnitude \left(1.1 \times 10^{5} \mathrm{~J}\right) /(0.20 \mathrm{~m})=5.5 \times 10^{5} \mathrm{~N} would be needed; that’s 110 times the friction force. So it’s reasonable to treat the external force of friction as negligible compared with the internal forces the vehicles exert on each other.
KEYCONCEPT You can use conservation of momentum in collision problems even though external forces act on the system. That’s because the external forces are typically small compared to the internal forces that the colliding objects exert on each other.
