Working out help? (1 Viewer)

BLIT2014 · Jul 15, 2013

Q)momentum of 20 kg m s-1 N.10° E and momentum of 20 kg m s-1 W.10°. N.

BLIT2014 · Jul 15, 2013

anyone?

BLIT2014 · Jul 15, 2013

28kg m/s N35W?

BLIT2014 · Jul 15, 2013

Is that answer correct?Thats what it came with?

SpiralFlex · Jul 15, 2013

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$I will draw you a diagram. Edit: Updated diagram for clarity.$
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$Step 1: Resolve each of the momentum components in the$\; x \;$and$\; y \;$direction.$
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$Step 2:$

$Let's consider N$10$E first. We have in the$\; x\; $direction$\; 20\cos 80 \; $kgm/s$,\; y \;$direction$\; 20\sin 80\; $kgm/s$

$Now consider W$10$N. We have in the$\; x \;$direction$ \;-20\cos 10 \;$kgm/s, \;$ y \;$direction$\; 20\sin 10\; $kgm/s$

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$Step 3:$

$Now adding each components together,$

$x\; $direction:$\; (20\cos 80 - 20\cos 10) \;$kgm/s$$

$y \;$direction:$\; (20\sin 80 + 20\sin 10) \;$kgm/s$$

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$Step 4:$

$Now you can use Pythagora's theorem to find the resultant momentum.$

$\rho_{R} = \sqrt{(20\cos 80 - 20\cos 10)^2+(20\sin 80 + 20\sin 10)^2} \; $kgm/s$$

$I will leave the direction for you to find yourself.$

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Parvee · Jul 15, 2013

Never Entertain Sexy Women? wot?

SpiralFlex · Jul 15, 2013

Parvee said:
Never Entertain Sexy Women? wot?

My perverted year 2 friend taught me that when we learnt bearings.

Edit: But my year 2 friend has a point, you should never entertain them.

BLIT2014 · Jul 15, 2013

thank you spiral

Working out help? (1 Viewer)

BLIT2014

The pessimistic optimist.

BLIT2014

The pessimistic optimist.

BLIT2014

The pessimistic optimist.

BLIT2014

The pessimistic optimist.

SpiralFlex

Well-Known Member

Parvee

(╯°□°）╯︵ ┻━┻

SpiralFlex

Well-Known Member

BLIT2014

The pessimistic optimist.

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