A 2100-kg truck is traveling east through an intersection at 2.0m/s when it is h

Question

A 2100-kg truck is traveling east through an intersection at 2.0m/s when it is hit simultaneously from the side and the rear. One car is a 1200-kg compact traveling north at 5.0m/s. The other is 1500-kg midsize traveling east at 10m/s. The three vehicles become entangled and slide as one body. What are their speed and direction just after the collision?

a.) Draw a representational cartoon of the situation, including before and after if some kind of transformation takes place. Draw a physics representation of situation based on the cartoon representation. Choose and show coordinate axes. Record all quantities in appropriate places on the diagram and in a table. Identify symbolically all given and unknown quantities with appropriate variable names.

b.) State the general topic of physics needed to solve the problem. State all specific physics principles, concepts, laws, or theories needed. State any assumptions or simplifications you need to make. Construct a mathematical representation which is based on the physics principles and simplifying assumptions. Use algebra or other mathematical techniques to manipulate the equations to solve for your unknowns variables. Substitute numbers and units in for the variables you know. Use arithmetic to solve for your unknown quantities or numbers including units at every stage.

c.) Explain why your answer is a reasonable quantity or expression.

Thanks in advance!

Explanation / Answer

Given,

m1 = 2100 kg ; u1 = 2 m/s ; m2 = 1200 kg ; u2 = 5 m/s ; m3 = 1500 kg ; u3 = 10 m/s

The given situation is an example of an inelastic collision, where momentum of the system is conserved buy energy is not.

From conservation of linear momentum

Pi = Pf

m1 u1 + m2 u2 + m3 u3 = (m1 + m2 + m3) v

momentum of the truck going east

p1 = m1 u1 = 2100 x 2 = 4200 kg-m/s

that of midsize going east ; p3 = m3 u3 = 1500 x 10 = 15000 kg-m/s

that of car going north = p2 = 0

Intial momentum of the system becomes:

Pi = p1 + p2 + p3 = 4200 + 15000 + 0 = 19200 kg-m/s

Vew = Pi / (m1 + m2 + m3) = 19200/4800 = 4 m/s (due east west)

The momentum due north will be just due to the car only

Pi = m2 u2 = 1200 x 5 = 6000 kg-m/s

Vns = Pi/(m1 + m2 + m3) = 6000/4800 = 1.25 m/s (due north south)

V = sqrt [(Vew)2 + (Vns)2 ] = sqrt [(4)2 + (1.25)2 ] = 4.2 m/s

Hence, V = 4.2 m/s North East.

b)conservation of momentum has been used. Vector addition and trigonometry has been used.

c)We had magnitude of the speeds and masses so we got a reasonable quantity.

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