The drawing shows a collision between two pucks on an air-hockey table. Puck A h

Question

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.28 kg and is moving along the x axis with a velocity of 6.60 m/s. It makes a collision with puck B, which has a mass of 0.56 kg and is initially at rest. After the collision, the two pucks fly apart with angles as shown in the drawing (? = 50° and ? = 30°). Find the final speed of puck B.

Apparently this is how you do it.. But I keep getting the wrong answer.

U have to separate it into components. M1(6.6)= m1vcos50 +m2vcos30 x-component.

Y-component 0=m1vsin50 - m2vsin30

This is a 2 system equations so u plug in all the numbers that were given and then see what variables r unknown. Those unknown variable u isolate in one of the equations preferably the y-component equation and then u substitute the isolated variable in the x-component and u solve for the variable that u dont know. It should be m1 final velocity

Explanation / Answer

0.28(6.6)= 0.28 * V1*cos50 +0.56 *V2cos30 in x direction momentum conservation

0=0.28* V1*sin50 - 0.56*V2sin30

V1  = 3.35 m/s

V2 = 2.5669 m/s

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