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xeze [42]
3 years ago
14

Object A has of mass 7.20 kilograms, and object B has a mass of 5.75 kilograms. The two objects move along a straight line towar

d each other with velocities +2.00 meters/second and -1.30 meters/second respectively. What is the total kinetic energy of the objects after the collision, if the collision is perfectly elastic?
Physics
2 answers:
cestrela7 [59]3 years ago
8 0
Two things:
1) All collisions conserve momentum
2) Elastic collisions also conserve kinetic energy.

So basically the kinetic energy before the collision is the same as the kinetic energy after the collision.

KE = ½mv²

For object A:
KE = ½(7.20 kg)(2.00 m/s)² = 14.4 J

For object B:
KE = ½(5.75 kg)(-1.30 m/s)² = 4.86 J

total kinetic energy = 14.4 J + 4.86 J = 19.3 J
Genrish500 [490]3 years ago
7 0
The equation for Kinetic Energy in an Elastic Collision is this:
1/2(m₁v₁) + 1/2(m₂v₂) = 1/2(7.20kg)(2.00m/s) + 1/2(5.75kg)(-1.30m/s) =
3.4625m/s

I think....technically the equation above equates to the same thing except measuring the final velocity of the given masses, however you were not given that in the equation, making the first half of the full equation the only viable solution.

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Answer:

229,098.96 J

Explanation:

mass of water (m) = 456 g = 0.456 kg

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latent heat of fusion =33.5 x 10^(4) J/kg

specific heat of water = 4186 J/kg

for the water to be converted to ice it must undergo three stages:

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        Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J

  • the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp

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  • the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp

        Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J

The quantity of heat removed from all three stages would be added to get the total heat removed.

Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J

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