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Karo-lina-s [1.5K]

3 years ago

8

6.If a 250. gram cart moving to the right with a velocity of .31 m/s collides inelastically with a 500. gram cart traveling to t

he left with a velocity of -0.22 m/s, what is the total momentum of the system before the collision

2 answers:

spin [16.1K]3 years ago

5 0

Answer:

The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

Explanation:

Given that,

Mass of the cart, m = 250 g = 0.25 kg

Initial velocity of the cart, u = 0.31 m/s (due right)

Mass of another cart, m' = 500 g = 0.5 kg

Initial velocity of the another cart u' = -0.22 m/s (due left)

Let p is the total momentum of the system before the collision. It is given by :

$p=mu+m'u'\\\\p=0.25\times 0.31+0.5\times (-0.22)\\\\p=-0.0325\ kg-m/s$

So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

miv72 [106K]3 years ago

3 0

Answer:

Explanation:

mass of I cart, m = 250 g = 0.25 kg

initial velocity of I cart, u = 0.31 m/s

mass of II cart, m' = 500 g =0.5 kg

initial velocity of II cart, u' = - 0.22 m/s

Total momentum before collision = m x u + m' x u'

= 0.25 x 0.31 - 0.5 x 0.22

= 0.0775 - 0.11

= - 0.0325 kg m/s

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A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

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a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

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Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

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$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

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$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

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$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

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$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

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Since h(t) =0, it means the rocket is falling towards the ground, when it gets to the ground when it's at rest the height h(t) = 0m

{Note the rocket is launched from the height of the Cliff so that would be the initial height of the rocket,h0}

2.Substituting into h(t) = -16t2 + vt + h0

We have;

0 = -16t2 + 116t + 101=> 16t2 -116t -101=0

Using formula method for solving quadratic equation we have;

t = -(-116)+_√[(-116)^2 -( 4× 16 ×-101]/ (2× 16)

t = [116 +_(141.1382)]/32

t = (116 -141.1382)/32 or (116 +141.1382)/32

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-0.8s or 8.0s to the nearest tenth.

Now time cannot be negative in real life situation hence the time is 8.0s

Note : the general equation of a quadratic equation with variable t is given below;

at2 + bt + c=0

Formula method for quadratic equation is :

t =( -b+_√[(b^2 -( 4× a×c)] ) / (2× a)

6 0

3 years ago