3 years ago

PLEASE HELP ASAPPPPPPPPP

Physics

1 answer:

Svetlanka [38]3 years ago

8 0

60kg would be the answer

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Two particles with masses m and 3m are moving toward each other along the x axis with the same initial speeds Vi. Particle m is

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Answer:

$v1 = \sqrt{(2)}Vi$

$v2 = \sqrt{(2/3)}Vi$

ANGLE is 35.3 degree celcius

Explanation:

Given data:

mass m and 3m

initial speed Vi

particle with mass m is moving toward left while particle with mass 3m is moving toward right

By using conservation of momentum :

$mVi + 3m(-Vi) = mv1 +3mv2$

$-2mVi = m(v1 + 3v2)$

$-2Vi = v1 + 3v2$

conservation of energy :

$m(Vi^2) + 3m(-Vi^2) = mv1^2 + 3mv2^2$

$4mVi^2 = m(v1^2+3v2^2)$

$4Vi^2 = v1^2+3v2^2$

After collision, particle with mass m moves at right angles, thus by considering conservation of momentum in x & y direction,

x direction : $-2mVi = 3m.v2i$

$-2Vi = 3v2i$

y direction : $0 = m(v1)j+3m(v2)j$

$-v1j = 3v2j$

subsitute these value in energy conservation

$v1 = \sqrt{(2)}Vi$

$v2 = \sqrt{(2/3)}Vi$

$angle = tan^{-1}(\frac{\sqrt{(2)}}{2}) =$ 35.3 degree from x-axis

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Explanation

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v = w r

Angular velocity and frequency are related by

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a = w² r

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f = 2.30 rev / s (2π rad / 1 rev) = 14.45 rad / s

.r = 10.3 cm = 0.103 m

Let's calculate

a = 4π² 14.45² 0.103

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