Solving a Simple Collision Problem: Finding Velocity and Kinetic Energy Loss

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To solve the collision problem, the velocity of the 10.0 kg object after the collision can be determined using conservation of momentum in both the x and y directions. The initial momentum of the system is calculated from the 8.00 kg mass, and after the collision, the momentum equations can be set up to find the unknown velocity of the 10.0 kg object. The percentage of kinetic energy lost is found by comparing the initial kinetic energy of the system to the kinetic energy after the collision. The calculations reveal both the magnitude and direction of the 10.0 kg object's velocity, as well as the percentage of kinetic energy lost. This problem illustrates key principles of momentum conservation and energy transformation in collisions.
physicsisphun
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Heres the problem:

An 8.00 kg mass moving east at 14.8 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at rest. After the collision, the 8.00 kg object moves south at 3.70 m/s.

a. what is the velocity of the 10.0 kb object after the collision?
magnitude:____ m/s
direction: ____ (degrees)

b. what percentage of the initial kinetic energy is lost in the collision?

thanks guys!
 
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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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