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Two cars collide at an intersection. Car A, with a mass of 2000 kg, is going from west to east, while car B, of mass 1500 kg, is going from north to south at 17.0 m/s. As a result of this collision, the two cars become enmeshed and move as one afterwards. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 60.0 degrees south of east from the point of impact.How fast were the enmeshed cars moving just after the collision?

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Final answer:

To find how fast the enmeshed cars were moving just after the collision, conservation of momentum is used. The velocities are broken into components, which are then used to calculate the final velocity magnitude and direction using trigonometric relationships.

Step-by-step explanation:

The question is concerned with a two-dimensional inelastic collision between two cars at an intersection. When two objects collide and move together after the collision, the situation is best analyzed using conservation of momentum.

To determine how fast the enmeshed cars were moving just after the collision, we can use the principle of conservation of linear momentum, which states that the total momentum of a closed system remains constant if no external forces act upon it. The vector sum of the momenta of Car A and Car B before collision will equal the momentum of the enmeshed cars just after the collision. We can calculate this by breaking down the velocities into x and y components and applying the conservation of momentum in each direction.

The final step is to find the magnitude and direction of the velocity of the enmeshed cars using the resultant components of momentum in each direction. If the cars are moving at an angle of 60 degrees south of east, it can be deduced that the x-component of the final velocity is to the east, and the y-component is to the south. The magnitude of the final velocity (v) of the enmeshed cars can be found using the equation v = \sqrt{(v_x)^2 + (v_y)^2}, where v_x and v_y are the x and y components of the final velocity, respectively. The direction (\theta) with respect to east can be determined by \theta = \arctan(v_y / v_x).

User Bartosz Bierkowski
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8.3k points
4 votes

Answer:


v = 4.2 \hat i - 7.28 \hat j


v = 8.41 m/s

Step-by-step explanation:

let say car A is moving with speed v in direction West to East

so its momentum is given as


P_1 = 2000 v\hat i

Car B is moving with velocity 17 m/s from North to South

So here we will have


P_2 = 1500(17)(-\hat j)

so we have initial total momentum given as


P = 2000 v\hat i - 25500\hat j

now the direction of the momentum is given as 60 degree South of East

so it is


tan60 = (P_y)/(P_x)


\sqrt3 = (25500)/(2000v)


v = 7.36 m/s

now velocity of two cars just after collision is given as


P_1 + P_2 = (m_1 + m_2) v


2000(7.36)\hat i - 25500\hat j = (2000 + 1500) v


v = 4.2 \hat i - 7.28 \hat j

magnitude of velocity is


v = √(4.2^2 + 7.28^2)


v = 8.41 m/s

User EldarGranulo
by
7.7k points

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