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The morning after a winter ice storm in Dallas, a 1400kg automobile going west on Chestnut Street at 35.0 km/h collides with a 2800kg truck going south across Chestnut Street at 50.0 km/h .
What is the magnitude and direction of their velocity after colliding?

2007-11-02 11:21:54 · 2 answers · asked by brandon patrick 1 in Science & Mathematics Physics

2 answers

I will assume they stick together

Look at each component of velocity separately

West
1400*35=4200*vx
vx=35/3 km/hr West

South
2800*50=4200*vy
vy=100/3 South

Direction is
atan(100/35) South of West
=70.71 degrees South of West

Magnitude
=sqrt(100^2+35^2)/3
=35.3 Km/hr

j

2007-11-05 06:18:13 · answer #1 · answered by odu83 7 · 0 0

The law of conservation of momentum states that the total momentum of a closed system of objects is constant. (Here, the effect of the pavement is negligible because of the ice.)

Intuition tells us that the direction will be between West and South, but not perfectly SouthWest because one car might have more momentum than the other.

First draw the initial momentum vectors. Notice they form a right angle with respect to each other. Applying the law of conservation of momentum, we know the final momentum vector is equal to the initial momentum vectors.

1) Solve for velocity magnitude

From Pythagoras' Theorem we know that the magnitude of the final momentum vector is sqrt [(vector1)^2 + (vector2)^2]. Plugging in the values this is sqrt [(1400*35)^2 + (2800*50)^2] = 140698. The magnitude of their velocity, then, is their combined momentem (140698) divided by their combined mass (4200), i.e., 35.3 km/h.

2) Solve for velocity direction.

Expressing the direction as an angle, theta, west with respect to a directly South vector, we can use trigonometry to find theta.

tan(theta) = (1400*35)/(2800*50)

theta = 19.3 degrees west of directly South.

2007-11-05 14:19:10 · answer #2 · answered by RJ Hunt 2 · 0 0

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