Total Kinetic Energy B/F & A/F Collision

AI Thread Summary
The discussion focuses on calculating the total kinetic energy before and after a collision between a truck and a car. The truck has a mass of 3000 kg and initial velocity of 5.0 m/s, while the car is moving at 2.0 m/s. After the collision, the truck's velocity decreases to 3 m/s, and the car's velocity increases to 6 m/s. The kinetic energy formula, 1/2 mv^2, is highlighted as essential for the calculations, but the mass of the car is not provided, suggesting the need to apply conservation laws to find it. The conversation emphasizes understanding both kinetic energy calculations and the principles of conservation in collisions.
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Homework Statement



The truck of mass 3000Kg, moving at 5.0m/s on a level on a level, icy road, bumps into the rear of a car moving at 2.0m/s in the same direction. After the impact the truck has a velocity of 3m/s and the car a velocity of 6m/s both forward. Calculate the total kinetic energy before and after the collisioin

Homework Equations



Dont know the equation

The Attempt at a Solution

 
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Well, the equation for kinetic energy is 1/2 mv^2.
This is a relatively simple equation to use; for total kinetic energy you must calculate the KE for each body involved.
However, since in this problem the mass of the car isn't given, you'll probably have to find that first (hint: what conservation law involves mass and velocity?).
Hope this helps!
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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