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OpenStudy (anonymous):

A 6kg block, sliding to the east across a horizontal frictionless surface with a momentum of 30 kg*m/s strikes an obstacle. The obstacle exerts an impulse of 10N*s to the west on the block. What is the speed of the block after the collision?

OpenStudy (lastdaywork):

Use - |Initial momentum| - |impulse| = |Final momentum|

OpenStudy (anonymous):

So 30kg*m/s - 10 kg*m/s= 20N*s?

OpenStudy (anonymous):

nvm I got it. divide by mass. Thank you.

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