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3 years ago

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A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediat

ely after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction relative to the +x-axis does the third piece move?
(a) 39.8º from the +x-axis
(b) 36.9° from the +x-axis
(c) 39.9° from the +x-axis
(d) 216.9° from the +x-axis
(e) 219.8° from the +X-axis

1 answer:

djyliett [7]3 years ago

8 0

Answer:

M1 Vx1 + M2 Vx2 + M3 Vx3 = 0 conservation of momentum in x direction

Vx3 = -(M1 Vx1 + M2 Vx2 ) / M3

Vx3 = - 320 * 2 / 100 = -6.4 m/s M2 has no x-component of momentum

Likewise:

Vy3 = -(M1 Vy1 + M2 Vy2 ) / M3

Vy3 = - 355 * 1.5 / 100 = -5.33 m/s

tan theta = -5.33 / -6.4 = .833 where theta is in the third quadrant and measured from the negative x-axis

theta = 39.8 deg

180 + 39.8 = 219.8 from the positive x-axis

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(a) 3.35 s

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a = g = -9.8 m/s^2

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The initial height of the projectile is

h = 55.0 m

The vertical position of the projectile at time t is

$y = h + \frac{1}{2}at^2$

By requiring y = 0, we find the time t at which the projectile reaches the position y=0, which corresponds to the ground:

$0 = h + \frac{1}{2}at^2\\t=\sqrt{-\frac{2h}{a}}=\sqrt{-\frac{2(55.0 m)}{(-9.8 m/s^2)}}=3.35 s$

(b) 78.4 m

The distance travelled by the projectile from the base of the building to the point it lands depends only on the horizontal motion.

The horizontal motion is a uniform motion with constant velocity -

The horizontal velocity of the projectile is

$v_x = 23.4 m/s$

the time it takes the projectile to reach the ground is

t = 3.35 s

So, the horizontal distance covered by the projectile is

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(c) 23.4 m/s, -32.8 m/s

The motion of the projectile consists of two independent motions:

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$v_x = 23.4 m/s$

so this value is also the value of the horizontal velocity just before the projectile reaches the ground.

- Along the vertical direction, the motion is acceleration, so the vertical velocity is given by

$v_y = u_y +at$

where

$u_y = 0$ is the initial vertical velocity

Using

a = g = -9.8 m/s^2

and

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We find the vertical velocity of the projectile just before reaching the ground

$v_y = 0 + (-9.8 m/s^2)(3.35 s)=-32.8 m/s$

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Kinetic energy (K) is the energy associated with bodies that are in motion, depends on the mass and speed of the body and is calculated using the formula:

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for this problem We replace in the equation (1)

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We replace in the equation (1) :

$Ke=\frac{1}{2}*me*ve^{2}$

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