Starting with an initial speed of 5.00 m/s at a height of 0.300 m, a 1.50 kg ball swings downward and strikes a 4.60kg ballthat is at rest, as the drawing shows. a. using the principle of conservation of mechanicalenergy,find the speed of the 1.50 kg ball just before impact b. assuming that the collision is elastic, find the velocities( magnitude and direction ) of both balls just after thecollision c. how high does each abll swing after the collision, ignoringair resistance?

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pattererX · Accepted Answer

Let m=1.5kg, v1 the initial speed of m, v2 the speed of mbefore impact, v3 speed after impactBy conservation of energy12m×v12+m×g×h=12×m×v22  before impact andv22=v12+2×g×h=25+2×9.8×.3=30.88 and v2=5.56m/s Let M=1.5, V the speed of M after impact and v3 the speed ofm after impactSince the collision is elastic12×m×v22=12×m×v32+12×M×V2Rewrite this as m imes (v22−v32)=M×V2 (I)By conservation of momentum m×v2=m×v3+M×VRewrite this as m×(v2−v3)=M×V (II)Divide equation (I) by equation (II) and get v2+v3=VSubstitute V in equation (II) and m×v2−m×v3=M×v2+M×v3Solve for v3 and get v3=v2×m−MM+m It&#039;sworth noting that if M=m then v3=0v3=−5.56×3.16.1=−2.83 m/s and V=5.56m/s−2.83m/s=2.73m/sNow you can use 12×m×v32=m×g×h and 12×M×V2=×g×HTo solve for the height that each ballrises.

Starting with an initial speed of 5.00 m/s at a height of0.300 m, a 1.50 kg ball swings downward and strikes a 4.60kg ballthat is at rest, as the draw

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