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damaskus [11]
4 years ago
13

A ball of mass 1 kg bounces vertically off the floor. Just before hitting the floor its velocity is 10 m/s downward. Just after

hitting the floor its velocity is 8 m/s upward. What is the magnitude of the impulse experienced by the particle in this collision?
Physics
1 answer:
inysia [295]4 years ago
6 0

Answer:

2kgm/s

Explanation:

Impulse is defined as the change in momentum of a body. It is expressed as;

Impulse = mass × change in velocity

Change in velocity = final velocity v - initial velocity u

Impulse = m(v-u)

Given mass of the ball = 1kg

Final velocity v = 8m/s (after hitting the floor)

Initial velocity = 10m/s (before hitting the floor)

Impulse = 1(8-10)

Impulse = -2kgm/s

Magnitude of the impulse experienced by the particle in the collision is 2kgm/s

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A concave mirror produces a real image that is three times as large as the object. If the object is 20 cm in front of the mirror
TiliK225 [7]

Answer:

The image is produced 60 cm behind the mirror

The focal length of the mirror is 30 cm

Explanation:

u = Object distance =  20 cm

v = Image distance

f = Focal length

m = Magnification = 3

m=-\frac{v}{u}\\\Rightarrow 3=-\frac{v}{20}\\\Rightarrow v=-3\times 20\\\Rightarrow v=-60\ cm

The image is produced 60 cm behind the mirror

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{20}+\frac{1}{-60}\\\Rightarrow \frac{1}{f}=\frac{1}{30}\\\Rightarrow f=\frac{30}{1}=30\ cm

The focal length of the mirror is 30 cm

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A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th
irina [24]

The angular momentum is defined as,

L=I\omega

Acording to this text we know for conservation of angular momentum that

L_i=L_f

Where L_iis initial momentum

L_f is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

L_i=mvl

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

L_f=(I_b+I_s)\omega

Substituting

L_f=(ml^2+\frac{10ml^2}{3})\omega

L_f=\frac{13}{10}ml^2w

m(0.65)l=\frac{13}{10}ml^2 \omega

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Applying conservative energy equation we have

\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'

\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)

Replacing the values and solving

l=\frac{13}{0.54g}

Substituting

l=\frac{13}{0.54(9.8)}

l=2.45cm

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