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ExtremeBDS [4]
4 years ago
6

A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of

2.10 m/s and rebounds with a speed of 1.90 m/s, determine the following.
a. magnitude of the change in the ball's momentum (Let up be in the positive direction.)
________ kg - m/s
b. change in the magnitude of the ball's momentum (Let negative values indicate a decrease in magnitude.)
_______ kg - m/s
c. Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor?
A. Neither are related to the net force acting on the ball.
B. They both are equally related to the net force acting on the ball.
C. The change in the magnitude of the ball's momentum
D. The magnitude of the change in the ball's momentum
Physics
1 answer:
VARVARA [1.3K]4 years ago
8 0

Answer:

a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.

Explanation:

a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:

\vec p_{o} + Imp = \vec p_{f}

Where:

\vec p_{o}, \vec p_{f} - Initial and final momentums, measured in kilogram-meters per second.

Imp - Impact due to collision, measured in kilogram-meters per second.

The impact experimented by the ball due to collision is:

Imp = \vec p_{f} - \vec p_{o}

By using the definition of momentum, the expression is therefore expanded:

Imp = m \cdot (\vec v_{f}-\vec v_{o})

Where:

m - Mass of the ball, measured in kilograms.

\vec v_{o}, \vec v_{f} - Initial and final velocities, measured in meters per second.

If m = 0.275\,kg, \vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right] and \vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right], the vectorial change of the linear momentum is:

Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right]

Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right]

The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.

b) The magnitudes of initial and final momentums of the ball are, respectively:

p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)

p_{o} = 0.578\,\frac{kg\cdot m}{s}

p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)

p_{o} = 0.523\,\frac{kg\cdot m}{s}

The change in the magnitude of the ball's momentum is:

\Delta p = p_{f}-p_{o}

\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}

\Delta p = -0.055\,\frac{kg\cdot m}{s}

The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.

c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.

In a nutshell, the right choice is option D.

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Alenkasestr [34]

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Explanation:

The impulse exerted by a force is defined as the product between the force itself and the time interval during which the force is applied. Mathematically, it is equal to the change in momentum experienced by the object on which the force is acting:

I=F\Delta t = \Delta p

Where

I is the impulse

F is the force

Delta t is the time interval during which the force is applied

\Delta p is the change in momentum

In this problem,

\Delta t = 3.0 s is the time interval

I=6.0 N\cdot s (east) is the impulse

Therefore, the magnitude of the force is

F=\frac{I}{\Delta t}=\frac{6.0}{3.0}=2.0 N

And the direction is the same as the impulse (east).

Learn more about impulse and change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

7 0
3 years ago
a truck drives up a hill with a 15 incline. if the truck has a constant speed of 22m/s, what are the horizontal and vertical com
Kamila [148]

Answer:

speed of truck (v) =  22 m/s ,

angle of hill (Θ) =15°

Find

Vertical component (Fv) = ?

Harizontal component (Fh) =?

               Vertical component (Fh) = V cosΘ

                                                        = 22. cos 15

                                                        = 21.25 m/s.

               Harizontal component (Fv) = V sinΘ

                                                            = 22. sin 15

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3 0
4 years ago
How fast must an object move before its length appears to be contracted to one-fourth its proper length? (Give your answer in te
Tresset [83]

Answer:

<em>0.97c</em>

<em></em>

Explanation:

From the relativistic equation for length contraction, we have

l = l_{0}\sqrt{1 - \beta }

where

l is the final length of the object

l_{0} is the original length of the object before contraction

β = v^{2} /c^2

where v is the speed of the object

c is the speed of light in free space = 3 x 10^8 m/s

The equation can be re-written as

l/l_{0} = \sqrt{1 - \beta }

For the length to contract to one-fourth of the proper length, then

l/l_{0} = 1/4

substituting into the equation, we'll have

1/4 = \sqrt{1 - \beta }

substituting for β, we'll have

1/4 = \sqrt{1 - v^2/c^2 }

squaring both side of the equation, we'll have

1/16 = 1 - v^2/c^2

v^2/c^2 = 1 - 1/16

v^2/c^2 = 15/16

square root both sides of the equation, we have

v/c = 0.968

v = <em>0.97c</em>

3 0
3 years ago
For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e
Sati [7]

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg  is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

5 0
4 years ago
Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents th
agasfer [191]

Answer:

T_1=T_3=\dfrac{2\pi}{21}

T_2=T_4=\dfrac{2\pi}{42}

Explanation:

Wave 1, y_1=0.12\ cos(3x-21t)

Wave 2, y_2=0.15\ sin(6x+42t)

Wave 3, y_3=0.13\ cos(6x+21t)

Wave 4, y_4=-0.27\ sin(3x-42t)

The general equation of travelling wave is given by :

y=A\ cos(kx\pm \omega t)

The value of \omega will remain the same if we take phase difference into account.

For first wave,

\omega_1=21

\dfrac{2\pi }{T_1}=21

T_1=\dfrac{2\pi}{21}

For second wave,

\omega_2=42

\dfrac{2\pi }{T_2}=42

T_2=\dfrac{2\pi}{42}

For the third wave,

\omega_3=21

\dfrac{2\pi }{T_3}=21

T_3=\dfrac{2\pi}{21}

For the fourth wave,

\omega_4=42

\dfrac{2\pi }{T_4}=42

T_4=\dfrac{2\pi}{42}

It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.

3 0
3 years ago
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