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Stolb23 [73]
3 years ago
9

Two balls of equal size are dropped from the same height from the roof of a building. One ball has twice the mass of the other.

When the balls reach the ground, how do the kinetic energies of the two balls compare
Physics
1 answer:
frutty [35]3 years ago
3 0

Answer:

The kinetic energy of the more massive ball is greater by a factor of 2.

Explanation:

By conservation of energy, we know that the initial energy = final energy. At first, the balls are dropped from a height with no initial velocity so their initial energy is all potential energy. When they reach the bottom, all their energy is kinetic energy. So all of their energy is changed from potential to kinetic energy. This means that the ball with greater potential energy will have a greater kinetic energy.

Potential energy = mgh. Since g = gravity is a constant and h = height is the same, the only difference is mass. Since mass is directly proportional to potential energy, the greater the mass, the greater the potential energy, so the more massive ball has a greater initial potential energy and will have a greater kinetic energy at the bottom.

Additionally, let B1 = lighter ball with mass m and let B2 = heavier ball with mass m2. Since we know that intial potential energy = final kinetic energy. We can rewrite it as potential energy = kinetic energy = mass * gravity constant * height. For B1, it is mgh and for B2 it is 2mgh, so B2's kinetic energy is twice that of B1.

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A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block
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Answer:

a) v'=-0.227\ m.s^{-1}

b) v=1.36\ m.s^{-1}

Explanation:

Given:

mass of the lighter block, m'=0.02\kg

velocity of the lighter block, u'=0.68\ m.s^{-1}

mass of the heavier block, m=0.04\ kg

velocity of the heavier block, u=0\ m.s^{-1}

a)

Using conservation of linear momentum:

m'.u'+m.u=m'.v'+m.v

where:

v'= final velocity of the lighter block

v= final velocity of the heavier block

m'.u'=m'.v'+m.v

m'(u'-v')=m.v ........................(1)

Since kinetic energy is conserved in elastic collision:

\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2

m'(u'^2-v'^2)=m.v^2

m'(u'-v')(u'+v')= m.v^2

divide the above equation by eq. (1)

v=u'+v' .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

m'(u'-v')=m(u'+v')

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v'=-0.227\ m.s^{-1} (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

m'(u'-v+u')=m.v

2m'.u'=(m-m')v

2\times 0.02\times 0.68=(0.04-0.02)\times v

v=1.36\ m.s^{-1}

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Answer:

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Explanation:

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Hello!

Answer: 
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Explanation:

We are assuming that the floor (field) is completely horizontal since there's no information about that in the statement. 

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Where:

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8 0
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