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AlekseyPX
2 years ago
12

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
Physics
2 answers:
mel-nik [20]2 years ago
8 0

Answer:

8.3 m/s, 2196 degree from + X axis

Explanation:

m = 320 g , u = 2 m/s along X axis

m'  = 355 g, u' = 1.5 m/s along Y axis

m'' = 100 g, u'' = v

Let the speed of the third piece is v makes an angle A from the X axis.

use conservation of momentum along X axis

0 = 320 x 2 + 100 x v cos A

v cos A =  - 6.4 ..... (1)

Use conservation of momentum along Y axis

0 = 355 x 1.5 + 100 x v sin A

v sinA = - 5.3 ... (2)

Squaring and adding

v^2 = (-6.4)^2 +(-5.3)^2\\\\v= 8.3 m/s

The angle is given by

tan A = \frac{-5.3}{-6.4}\\\\A = 219.6 degree from + X axis

Olegator [25]2 years ago
6 0

Answer:

Explanation:

There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.

momentum of first piece = .320 kg x 2 m/s

= 0.64 kg m/s along x -axis.

momentum of second piece = .355 kg x 1.5 m/s

= 0.5325 kg m/s along y- axis .

Let the velocity of third piece be v and it is making angle of θ with x -axis .

Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ

vertical  component of its velocity = .100 kg x v sinθ = .1 v sinθ

For making total momentum in the plane zero

.1 v cosθ = 0.64 kg m/s

.1 v sinθ = 0.5325 kg m/s

Dividing

Tanθ = .5325 / .64 = .83

θ = 40⁰.

The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.

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A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 340 N is appl
Taya2010 [7]

Answer:

The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

Explanation:

Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

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Mass of block = 60 kg

Acceleration = 2.0 m/s²

Mass = 100 kg

Horizontal force = 340 N

Let the frictional force be f.

We need to calculate the frictional force

Using balance equation

F-f=ma

Put the value into the formula

340-f=60\times2.0

f=340-60\times2.0

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We need to calculate the coefficient of friction

Using formula of friction force

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We need to calculate the acceleration of the 100 kg block

Using formula of newton's law

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Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

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