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slava [35]
3 years ago
14

If Roger dropped pages instead of mongoer, you would have would have to wait longer to catch them than when you were catching ma

ngoes Why would have to wait longer? ​
Physics
1 answer:
slamgirl [31]3 years ago
8 0
I’m confused retype your question please
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The maximum tension the wire can withstand without breaking is 300 N . A 0.800 kg projectile traveling horizontally hits and emb
topjm [15]

Answer:

Let's investigate the case where the cable breaks.

Conservation of angular momentum can be used to find the speed.

\vec{L}_1 = \vec{L}_2\\\vec{L}_1 = m\vec{v_0} \\\vec{L}_2 = I\vec{\omega}\\

The projectile embeds itself to the ball, so they can be treated as a combined object. <u>The moment of inertia of the combined object is equal to the sum of the moment of inertia of both objects. </u>

I = I_{projectile} + I_{ball}\\I = mr^2 + mr^2\\I = 2mr^2

where r is the length of the cable.

<u>After the collision, the ball and the projectile makes a circular motion because of the cable.</u> So, the force (tension) in circular motion is

F = \frac{mv^2}{r}

The relation between linear velocity and the angular velocity is

v = \omega r

So,

F = \frac{m(\omega r)^2}{r} = m\omega^2 r = 300\\mv_0r  =I\omega\\\\\omega = mv_0r/I\\300 = m(\frac{mv_0r}{I})^2r = m(\frac{mv_0r}{2mr^2})^2r = m(\frac{v_0}{2r})^2r = \frac{mv_0^2r}{4r^2} = \frac{mv_0^2}{4r}\\300 = \frac{0.8v_0^2}{4r}\\1500 = v_0^2/r\\v_0 = \sqrt{1500r}

As can be seen, the maximum velocity for the projectile without breaking the cable is \sqrt{1500r}, where r is the length of the cable.

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