Starting with an initial speed of 6.31 m/s at a height of 0.314 m, a 2.73-kg bal

Question

Starting with an initial speed of 6.31 m/s at a height of 0.314 m, a 2.73-kg ball swings downward and strikes a 4.92-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 2.73-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 2.73-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.92-kg ball just after the collision. (d) How high does the 2.73-kg ball swing after the collision, ignoring air resistance? (e) How high does the 4.92-kg ball swing after the collision, ignoring air resistance? (a) Number Units (b) Number Units (c) Number Units (d) Number Units (e) Number Units

Explanation / Answer

a.) using conservation of energy

mgh + 0.5mv^2 = 0.5 mV^2

9.81*0.314 + 0.5 * 6.31 ^2 = 0.5 V^2

V = 6.78 m/s { velocity of ball just before impact

b.) using conservation of momentum

m1v1 + m2v2 = m1V1 + m2V2 ( V1 and V2 is velocity after collision od 2.73 kg and 4.92 kg ball respectively }

2.73*6.78 + 0 = 2.73*V1 + 4.92*V2 -------------1

as collision is elastic

V1- V2 = v2 -v1

V1-V2 = -6.78 ------------2

solving 1 and 2

V1 = - 1.94 m/s

V2 = 4.833 m/s

velocity 0f 2.73 kg ball is (V1) = - 1.94 m/s

c.)velocity of 4.92 Kg (V2) = 4.833 m/s

d.) using law of motion

V^2 = u^2 + 2as

0 = 1.94^2 - 2* 9.81*S

S = 0.1918 m

e.) V^2 = U^2 + 2as

0 = 4.833^2 - 2 *9.81* S

S = 1.19 m

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