How Do You Calculate the Velocity of Wreckage After a Head-On Collision?

AI Thread Summary
To calculate the velocity of wreckage after a head-on collision, the equation vf = (m1*v1 + m2*v2) / (m1 + m2) is used, where m represents mass and v represents velocity. In this scenario, Vehicle 1 (1245 kg) is traveling south at 20.0 m/s, and Vehicle 2 (910 kg) is traveling north at 26.0 m/s. The key error in the attempted solution was not accounting for the direction of the velocities, which affects the final result. Correctly applying the equation with directional values yields a final velocity of 0.58 m/s for the wreckage, aligning with the lecturer's answer. Accurate direction consideration is crucial in such calculations.
Chica1975
Messages
63
Reaction score
0

Homework Statement



2 vehicles made a totally inelastic head-on collision on the highway. Vehicle 1 of mass 1245kg was traveling southward at 20.0m/s. Vehicle 2 of mass 910kg was traveling northward at 26.0m/s. what was the velocity of the tangled wreckage?


Homework Equations


vf= ma1va1+mb2vb2/ma+mb


The Attempt at a Solution


I plugged in the information as per the equation above, however, I keep getting a different answer to that of my lecturer - I get 22.5336 m/s
The correct answer is 0.58 m/s

PLEASE HELP WHERE AM I GOING WRONG?
 
Physics news on Phys.org
you have to take into consideration the direction of their velocity. You did everything right but did not take into account their direction of motion. I used what you gave and i got your lecture answer
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top