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3 years ago

You have a mass of 60-kg, and you are facing your friend who has a mass of 100 kg. You skate towards

Physics

1 answer:

White raven [17]3 years ago

4 0

Answer: My initial velocity is 5 m/s.

Explanation:

In this case, momentum can be conserved.

initial momentum = final momentum

Since both the bodies come to rest after collision,

Final momentum = 0

Let my velocity be v, and mass, m1 = 60 kg

Friend's mass, m2 = 100 kg

Friend's velocity, v2 = 3 m/s

Intial momentum = m1v + m2v2

= 60v + 300

Conserving momentum,

60v + 300 = 0

v= -5 m/s

( Negative sign indicates that me and my friend are moving in opposite directions that is towards each other)

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3 years ago

The solid right-circular cylinder of mass 500 kg is set into torque-free motion with its symmetry axis initially aligned with th

aivan3 [116]

Answer:

30.95°

Explanation:

We need to define the moment of inertia of cylinder but in terms of mass, that equation say,

$A=\frac{1}{12}m(3r^2+l^2)$

Replacing the values we have,

$A=\frac{1}{12}(500)(3(0.5)^2+(2)^2)$ $A=197.9kg.m^2$

At the same time we can calculate the mass moment of intertia of cylinder but in an axial way, that is,

$c=\frac{1}{2}mr^2$

$c=\frac{1}{2}(500)(0.5)^2$

$c=62.5kg.m^2$

Finally we need to find the required angle between the fixed line a-a (I attached an image )

$\Phi = 2tan^{-1}\sqrt{\frac{(\frac{A}{cos\gamma})^2-A^2}{c^2}}$

Replacing the values that we have,

$\Phi = 2tan^{-1}\sqrt{\frac{(\frac{197.9}{cos5\°})^2-197.9^2}{62.5^2}}$

$\Phi = 2tan^{-1}(\sqrt{0.076634})$

$\Phi = 2tan^{-1}(0.2768)$

$\Phi = 2(15.47)$

$\Phi = 30.95\°$

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4 years ago