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Mechanics: Motion of a ball rolled on incline and projectile

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A small steel ball is rolled down an inclined plane from a table. The inclined plane is a slotted ruler. One end of the ruler is 10cm off of the table. The ball leaves the table horizontally and lands 46cm from the table on the floor. The table is 72 cm from the ground.

1. Using projectile motion equations, determine the speed of the ball the instant it leaves the table.
2. Compare the potential energy of the ball prior to its release to the kinetic energy of the ball just as it leaves the table.

A second ball is placed in the slot of the ruler at the edge of the table (the end of the inclined plane). The first ball rolls down the incline and collides with the second ball. The ball lands 44cm from the table on the floor.

3. Determine the speed at which the ball is projected off the table.
4. Using the conservation of energy, determine the speed of the first ball at the bottom of the ruler just before it hits the second ball.
5. Since the balls are exactly the same, the conservation of momentum dictates that the momentum of the first ball is transferred to the second ball. Using the above data, show this relationship.

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Solution Summary

The ball is rolling over an incline and then become a projectile. The velocity and energies are calculated at different stages. Details provided in the solution.

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A small steel ball is rolled down an inclined plane from a table. The inclined plane is a slotted ruler. One end of the ruler is 10cm off of the table. The ball leaves the table horizontally and lands 46cm from the table on the floor. The table is 72 cm from the ground.

1. Using projectile motion equations, determine the speed of the ball the instant it leaves the table.

The initial vertical component of the ball as it leaves the table is zero and the acceleration is g = 9.8 m/s2 in downward direction, hence the time taken by the ball to reach the ground t is given by using the equation of motion {s = ut + (1/2)*at2} for vertical direction as

0.72 = 0*t + (1/2)*9.8*t2

Gives s.

Now as there is no force is acting on the ball in horizontal direction the horizontal velocity v0 remains constant and hence the horizontal distance covered in this time t during the fall is given by the equation

x = v0*t

Gives v0 = x/t = ...

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