Could you please answer these questions? Thanks 1) 5.00 X 10\' Kg: mm/5 = (4.00

Question

Could you please answer these questions? Thanks

1) 5.00 X 10' Kg: mm/5 = (4.00 X 10 Kg), SINU the final y-momentum. Divide Equation (2) by Equation (1) and solve for 0. 5.00 x 10* kg · m/s tan 0 = = 1.33 3.75 x 10“ kg • m/s e = 53.1° Substitute this angle back into Equation (2) to find v. 1} = 5.00 x 104 kg • m/s (4.00 x 10° kg) sin 53.10 = 15.6 m/s LEARN MORE REMARKS It's also possible to first find the x- and y-components va and voy of the resultant velocity. The magnitude and direction of the resultant velocity can then be found with the Pythagorean theorem, vy = Vva + v67, and the inverse tangent function e = tan (v/vs). Setting up this alternate approach is a simple matter of substituting v = vy cos and vs. = vs sind 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 1.52 x 10 kg traveling east at a speed of 25.4 m/s collides at an intersection with a 2.53 x 10 kg van traveling north at a speed of 17.2 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 m/s direction º counterclockwise from the +x-axis

Explanation / Answer

if the car and van had identical mass and speed resultant angle is 45 deg

PRACTICE IT) initial momentum [x-direction] = mtot* vx

1.52 * 103 * 25.4 = (1.52 * 103 + 2.53 * 103) vx

vx = 9.53 m/s

initial momentum [ y-direction] = mtot * vy

2.53 * 103 * 17.2 = (1.52 * 103 + 2.53 * 103) vy

vy = 10.74 m/s

magnitude = sqrt [9.532 + 10.742]

magnitude = 14.3 m/s

tan theta = 10.74 / 9.53

theta = 48.40

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