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Vedmedyk [2.9K]
3 years ago
15

A 300g ball and a 100g ball are dropped from a tower. Both balls are the same size and their is no air resistance. Which one hit

s the ground first?
Physics
1 answer:
kifflom [539]3 years ago
3 0

Answer: The balls would hit the ground at the same time.

Explanation: Since there is no air resistance, we would put Galileo's experiment into motion. Galileo once performed an experiment of dropping two items, with different masses, from the tower of Pisa. Since there was no air resistance, the balls hit the ground at the same time. In this problem, the balls are "free-falling." Freefall is a term used in Physics to describe the motion of a falling object experiencing only the acceleration due to gravity. g=9.8 m/s^2 (acceleration due to gravity.) 9.8 will always be the acceleration in a free fall event like this with no air resistance, since Gravity remains constant.

Newton's second law of motion states that F=ma, (force is equal to mass times acceleration), but since Newton also states that f=mg, we can conclude that ma=mg. Stating that no matter what the mass of the object is, both of the objects will fall at the same time with the same velocity.

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A 20.00-kg lead sphere is hanging from a hook by a thin wire 2.80 m long and is free to swing in a complete circle. Suddenly it
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Explanation:

It is known that motion of objects on a vertical is an example of non-uniform motion.

When an object is at the highest point of the circle then for crossing highest point the centripetal force balances the weight of the object.

Therefore,    \frac{mv^{2}}{r} = mg

At the highest point of the circle, the minimum speed is as follows.

                 v = \sqrt{rg}

                    = \sqrt{2.80 \times 9.8}

                    = 5.23 m/s

As the sphere falls from highest to lowest point of circle as follows. According to the law of conservation of energy,

       K.E_{lowest} = K.E_{highest} + P.E_{highest}

Expression for potential energy is as follows.

          P.E_{highest} = mgh

where,    h = diameter of the circle (2r)

So,              P.E_{highest} = mgh

                  P.E_{highest} = mg(2r)

      \frac{1}{2}mu^{2} = \frac{1}{2}mv^{2} + mg(2r)

where,    u = velocity at the lowest time

Hence, the above equation will be as follows.

               u^{2} = v^{2} + 4gr

               u = \sqrt{v^{2} + 4gr}

                  = \sqrt{(5.23)^{2} + (4 \times 9.8 \times 2.80)}

                  = 11.71 m/s

The dart is embedded into bullet and the collision is inelastic. From the law of conservation of momentum,

       m_{2}v = (m_{1} + m_{2})u

                v = \frac{(m_{1} + m_{2})u}{m_{2}}

                   = \frac{(20 + 5) \times 11.71}{5}

                   = \frac{292.75}{5}

                   = 58.55 m/s

Thus, we can conclude that the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision is 58.55 m/s.

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