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remarks its also possible to first find the x - and y - components ( v_{fx} ) and ( v_{fy} ) of the resultant velocity.
the magnitude and direction of the resultant velocity can then be found with the pythagorean theorem, ( v_f )
( = sqrt{v_{fx}^2 + v_{fy}^2} ), and the inverse tangent function ( \theta=\tan^{-1}(v_{fy}/v_{fx}) ). setting up this alternate approach is a
simple matter of substituting ( v_{fx}=v_fcos\theta ) and ( v_{fy}=v_fsin\theta ) in equations (1) and (2).
question if the car and van had identical mass and speed, what would the resultant angle have
been?
practice it
use the worked example above to help you solve this problem. a car with mass ( 1.50\times10^{3} \text{kg} ) traveling
east at a speed of ( 24.3 \text{m/s} ) collides at an intersection with a ( 2.59\times10^{3} \text{kg} ) van traveling north at a speed
of ( 20.6 \text{m/s} ), as shown in the figure. 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.
magnitude ( \text{m/s}
direction ( {}^{circ} ) counterclockwise from the ( +x ) - axis
exercise hints: getting started | im stuck!
a ( 3.21 \text{kg} ) object initially moving in the positive ( x ) - direction with a velocity of ( +5.26 \text{m/s} ) collides with and
sticks to a ( 1.74 \text{kg} ) object initially moving in the negative ( y ) - direction with a velocity of ( -3.07 \text{m/s} ). find the
final components of velocity of the composite object. (indicate the direction with the sign of your answer.)
( v_{fx}=\text{m/s}
( v_{fy}=\text{m/s}

Explanation:

Step1: Apply conservation of momentum in x - direction

The initial momentum in the x - direction is \(p_{ix}=m_{car}v_{car}\). After the collision, \(p_{fx}=(m_{car} + m_{van})v_{fx}\). By conservation of momentum \(p_{ix}=p_{fx}\), so \(v_{fx}=\frac{m_{car}v_{car}}{m_{car}+m_{van}}\).
Substituting \(m_{car}=1.50\times 10^{3}\space kg\), \(v_{car} = 24.3\space m/s\), \(m_{van}=2.59\times 10^{3}\space kg\)

$$v_{fx}=\frac{1.50\times 10^{3}\times24.3}{1.50\times 10^{3}+ 2.59\times 10^{3}}=\frac{36450}{4090}\approx8.91\space m/s$$

Step2: Apply conservation of momentum in y - direction

The initial momentum in the y - direction is \(p_{iy}=m_{van}v_{van}\). After the collision, \(p_{fy}=(m_{car} + m_{van})v_{fy}\). By conservation of momentum \(p_{iy}=p_{fy}\), so \(v_{fy}=\frac{m_{van}v_{van}}{m_{car}+m_{van}}\).
Substituting \(m_{van}=2.59\times 10^{3}\space kg\), \(v_{van}=20.6\space m/s\), \(m_{car}=1.50\times 10^{3}\space kg\)

$$v_{fy}=\frac{2.59\times 10^{3}\times20.6}{1.50\times 10^{3}+ 2.59\times 10^{3}}=\frac{53354}{4090}\approx13.04\space m/s$$

Step3: Find the magnitude of the resultant velocity

Using the Pythagorean theorem \(v_{f}=\sqrt{v_{fx}^{2}+v_{fy}^{2}}\)

$$v_{f}=\sqrt{(8.91)^{2}+(13.04)^{2}}=\sqrt{79.3881 + 170.0416}=\sqrt{249.4297}\approx15.8\space m/s$$

Step4: Find the direction of the resultant velocity

Using \(\theta=\tan^{- 1}(\frac{v_{fy}}{v_{fx}})\)

$$\theta=\tan^{-1}(\frac{13.04}{8.91})\approx\tan^{-1}(1.464)\approx55.6^{\circ}$$

Step5: For the exercise (conservation of momentum in x - direction)

\(p_{ix}=m_{1}v_{1x}\), \(p_{fx}=(m_{1}+m_{2})v_{fx}\). By conservation of momentum \(v_{fx}=\frac{m_{1}v_{1x}}{m_{1}+m_{2}}\)
Substituting \(m_{1} = 3.21\space kg\), \(v_{1x}=+5.26\space m/s\), \(m_{2}=1.74\space kg\)

$$v_{fx}=\frac{3.21\times5.26}{3.21 + 1.74}=\frac{16.8846}{4.95}\approx3.41\space m/s$$

Step6: For the exercise (conservation of momentum in y - direction)

\(p_{iy}=m_{2}v_{2y}\), \(p_{fy}=(m_{1}+m_{2})v_{fy}\). By conservation of momentum \(v_{fy}=\frac{m_{2}v_{2y}}{m_{1}+m_{2}}\)
Substituting \(m_{2}=1.74\space kg\), \(v_{2y}=- 3.07\space m/s\), \(m_{1}=3.21\space kg\)

$$v_{fy}=\frac{1.74\times(-3.07)}{3.21+1.74}=\frac{-5.3418}{4.95}\approx - 1.08\space m/s$$

Answer:

For the "PRACTICE IT" problem:

• magnitude: \(15.8\space m/s\)
• direction: \(55.6^{\circ}\) counterclockwise from the \(+x -\)axis

For the "EXERCISE" problem:

• \(v_{fx}=3.41\space m/s\)
• \(v_{fy}=-1.08\space m/s\)