On a horizontal frictionless surface, a small block with a mass 0.200kg has a collision with a block of mass 0.400kg. Immediately after the collision, the 0.200kg block is moving at 12.0m/s in the direction 30° north of east and the 0.400kg is moving at 12.6m/s in the direction 53.1° south of east. a) what is the total kinetic energy of the two blocks after the collision (in joules) b) what is the the x-component of the total momentum of the two blocks after the collision? c) what is the the y-component of the total momentum of the two blocks after the collision?

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A:

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A: See below

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A:

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A: m1 = 1 kg v1 = 2 m/s m2 = 1 kg v2 = 0 m/s Kinetic energy lost = 0.8 J

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A: Given, mass of golf, m = 50g = 0.05 kgvelocity whent it leaves the tee , v = 70 m/stime taken to…

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Q: with a speed of 0.5 m/s. How much kinetic energy is lost in the collision? O 0.50 Joules O 3.75…

A: Inelastic Collision An inelastic collision is a type of collision where the kinetic energy is lost…

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A:

Q: In the figure here, a stationary block explodes into two pieces L and R that slide across a…

A: Given mL=1.1 Kg     mR=?μL=0.49dL=0.3μR=0.47dR=0.38

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A:

Helmholtz Free Energy

The Helmholtz free energy is a thermodynamic potential used to calculate the measure of useful work from a closed system. This system is kept at constant temperature and volume. German physicist Hermann Helmholtz introduced this concept in 1882.