Problem A car with mass 1.50 x10^3 kg traveling east at a speed of 25.0 m/s coll

Question

Problem A car with mass 1.50 x10^3 kg traveling east at a speed of 25.0 m/s collides at an intersection with a 2.50 x10^3 kg van traveling north at a speed of 20.0 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.

Strategy Use conservation of momentum in two dimensions. (Kinetic energy is not conserved.) Choose coordinates as in the figure. 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 vf and angle ?. Solve for these two unknowns, using the two components of the conservation of momentum equation.
SOLUTION
Find the x-components of the initial and final total momenta.
Spxi = mcarvcar = (1.50 103 kg)(25.0 m/s) = 3.75 104 kg · m/s
Spxf = (mcar + mvan)vf cos ? = (4.00 103 kg)vf cos ?
Set the initial x-momentum equal to the final x-momentum.
(1)
3.75 104 kg · m/s = (4.00 103 kg)vf cos ?
Find the y-components of the initial and final total momenta.
Spiy = mvanvvan = (2.50 103 kg)(20.0 m/s) = 5.00 104 kg · m/s
Spfy = (mcar + mvan)vf sin ? = (4.00 103 kg)vf sin ?
Set the initial y-momentum equal to the final y-momentum.
(2)
5.00 104 kg · m/s = (4.00 103 kg)vf sin ?
Divide Equation (2) by Equation (1) and solve for ?.
tan ? = 5.00 104 kg · m/s = 1.33
3.75 104 kg · m/s
? = 53.1°
Substitute this angle back into Equation (2) to find vf.
vf = 5.00 104 kg · m/s = 15.6 m/s
(4.00 103 kg) sin 53.1°
LEARN MORE
Remarks It's also possible to first find the x- and y-components vfx and vfy of the resultant velocity. The magnitude and direction of the resultant velocity can then be found with the Pythagorean theorem, vf = vvfx2 + vfy2, and the inverse tangent function ? = tan-1 (vfy/vfx). Setting up this alternate approach is a simple matter of substituting vfx = vf cos ? and vfy = vf sin ? in Equations (1) and (2).

1) Question If the car and van had identical mass and speed, what would the resultant angle have been?

Consider from the given information how the component of final momentum along the initial direction of the first car should be related to that along the initial direction of the second car. Then use your knowledge of trigonometry to decide what value ? must have.°
PRACTICE IT
2) Use the worked example above to help you solve this problem. A car with mass 1.45 x10^3 kg traveling east at a speed of 26.9 m/s collides at an intersection with a 2.48 x10^3 van traveling north at a speed of 19.7 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.

PLEASE HELP, THE FIRST PART IS A WORKED EXAMPLE, AND ON THE BOTTOM I NUMBERED 1 AND 2 FOR THE ONES THAT I NEED HELP WITH. I will rate as lifesaver for anyone who helps with correct answer and explanations. thanks

Explanation / Answer

(1) if the car and van had the same mass and velocity, the resultant direction would be symmetrical with respect to initial direction. The car was travelling east and the van was ravelling north. Hence the resultant velocity would be towards north-east, ie, 450 from the x-axis, if the east direction is taken as +ve x-axis.

(2) let the final velocity both the car and the van be vx in the x-direction(east) and vy in the y-direction (north).

Conserving the momentum in the x-direction:

the initial momentum in the x-direction= momentum of the car

the initial momentum in the x-direction= (total mass of car and van)*(common velocity in x-dir)

1.45 x103 * 26.9=(1.45 x103 +2.48 x103)vx

=> vx=9.92 m/s

Conserving the momentum in the y-direction:

the initial momentum in the y-direction= momentum of the van

the initial momentum in the y-direction= (total mass of car and van)*(common velocity in y-dir)

2.48 x103 * 19.7=(1.45 x103 +2.48 x103)vy

=> vy=12.43 m/s

if the magnitude of velocity is v, and it makes an angle with the x-direction,

v cos=vx=9.92 m/s

v sin=vy=12.43 m/s

v=15.90 m/s [squaring and adding the 2 equations gives v^2]

tan=vy/vx=1.25   [dividing the 2 equations]

=> =51.40

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