Question 6.7: Collision at an Intersection Goal Analyze a two-dimensional ...
Collision at an Intersection Goal Analyze a two-dimensional inelastic collision.
Problem A car with mass 1.50 × 10³ kg traveling east at a speed of 25.0 m/s collides at an intersection with a 2.50 × 10³-kg van traveling north at a speed of 20.0 m/s, as shown in Figure 6.14. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that friction between the vehicles and the road can be neglected.
Strategy Use conservation of momentum in two dimensions. (Kinetic energy is not conserved.) Choose coordinates as in Figure 6.14. Before the collision, the only object having momentum in the x-direction is the car, while the van carries all the momentum in the y-direction. After the totally inelastic collision, both vehicles move together at some common speed v_{f} \text { and angle } \theta . Solve for these two unknowns, using the two components of the conservation of momentum equation.
Find the x-components of the initial and final total momenta:
\begin{aligned}\Sigma p_{x i} &=m_{ car } v_{ car }=\left(1.50 \times 10^{3} kg \right)(25.0 m / s ) \\&=3.75 \times 10^{4} kg \cdot m / s\end{aligned}
Set the initial x-momentum equal to the final x-momentum:
\begin{aligned}&\Sigma p_{x f}=\left(m_{c a r}+m_{ van }\right) v_{f} \cos \theta=\left(4.00 \times 10^{3} kg \right) v_{f} \cos \theta \\&3.75 \times 10^{4} kg \cdot m / s =\left(4.00 \times 10^{3} kg \right) v_{f} \cos \theta (1)\end{aligned}
Find the y-components of the initial and final total momenta:
\begin{aligned}\Sigma p_{i y} &=m_{\operatorname{van}} v_{\operatorname{van}}=\left(2.50 \times 10^{3} kg \right)(20.0 m / s ) \\&=5.00 \times 10^{4} kg \cdot m / s\end{aligned}
Set the initial y-momentum equal to the final y-momentum:
\begin{aligned}&\Sigma p_{f y}=\left(m_{ car }+m_{ van }\right) v_{f} \sin \theta=\left(4.00 \times 10^{3} kg \right) v_{f} \sin \theta \\&5.00 \times 10^{4} kg \cdot m / s =\left(4.00 \times 10^{3} kg \right) v_{f} \sin \theta (2)\end{aligned}
Divide Equation (2) by Equation (1) and solve for θ:
\begin{aligned}\tan \theta &=\frac{5.00 \times 10^{4} kg \cdot m / s }{3.75 \times 10^{4} kg \cdot m }=1.33 \\\theta &=53.1^{\circ}\end{aligned}
Substitute this angle back into Equation (2) to find v_{f} :
v_{f}=\frac{5.00 \times 10^{4} kg \cdot m / s }{\left(4.00 \times 10^{3} kg \right) \sin 53.1^{\circ}}=15.6 m / s
Remark It’s also possible to first find the x- andy-components v_{f x} \text { and } v_{f y} of the resultant velocity. The magnitude and direction of the resultant velocity can then be found with the Pythagorean theorem, v_{f}=\sqrt{v_{f x}^{2}+v_{f y}^{2}} , and the inverse tangent function \theta=\tan ^{-1}\left(v_{f y} / v_{f x}\right) . Setting up this alternate approach is a simple matter of substituting v_{f x}=v_{f} \cos \theta \text { and } v_{f y}=v_{f} \sin \theta in Equations (1) and (2).