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KonstantinChe [14]
3 years ago
6

A 35 g steel ball is held by a ceiling-mounted electromagnet 4.0 m above the floor. A compressed-air cannon sits on the floor, 4

.4 m to one side of the point directly under the ball. When a button is pressed, the ball drops and, simultaneously, the cannon fires a 25 g plastic ball. The two balls collide 1.2 m above the floor. What was the launch speed of the plastic ball?
Physics
1 answer:
HACTEHA [7]3 years ago
4 0

Answer:

7.9 m/s

Explanation:

When both balls collide, they have spent the same time for their motions.

Motion of steel ball

This is purely under gravity. It is vertical.

Initial velocity, <em>u </em>= 0 m/s

Distance, <em>s</em> = 4.0 m - 1.2 m = 2.8 m

Acceleration, <em>a</em> = g

Using the equation of motion

s = ut+\frac{1}{2}at^2

2.8 \text{ m} = 0+\dfrac{gt^2}{2}

t = \sqrt{\dfrac{5.6}{g}}

Motion of plastic ball

This has two components: a vertical and a horizontal.

The vertical motion is under gravity.

Considering the vertical motion,

Initial velocity, <em>u </em>= ?

Distance, <em>s</em> = 1.2 m

Acceleration, <em>a</em> = -<em>g                   </em> (It is going up)

Using the equation of motion

s = ut+\frac{1}{2}at^2

1.2\text{ m} = ut-\frac{1}{2}gt^2

Substituting the value of <em>t</em> from the previous equation,

1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}

u\sqrt{\dfrac{5.6}{g}} = 4.0

Taking <em>g</em> = 9.8 m/s²,

u = \dfrac{4.0}{0.756} = 5.29 \text{ m/s}

This is the vertical component of the initial velocity

Considering the horizontal motion which is not accelerated,

horizontal component of the initial velocity is horizontal distance ÷ time.

u_h = \dfrac{4.4\text{ m}}{0.756\text{ s}} = 5.82\text{ m/s}

The initial velocity is

v_i = \sqrt{u^2+u_h^2} = \sqrt{(5.29\text{ m/s})^2+(5.82\text{ m/s})^2} = 7.9 \text{ m/s}

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for interfering pattern from above , for constructive interference of reflected wave from both sides of the film , the condition is

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