Starting with an initial speed of 5.57 m/s at a height of 0.406 m, a 1.20-kg bal

Question

Starting with an initial speed of 5.57 m/s at a height of 0.406 m, a 1.20-kg ball swings downward and strikes a 4.81-kg ball that is at rest. (a) Using the principle of conservation of mechanical energy, find the speed of the 1.20-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 1.20-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.81-kg ball just after the collision. (d) How high does the 1.20-kg ball swing after the collision, ignoring air resistance? (e) How high does the 4.81-kg ball swing after the collision, ignoring air resistance?

Explanation / Answer

(a) the small ball has potential and kinetic energy initially. It loses potential energy as it swings down and this is converted to kinetic. Total energy at the top = mgh +(1/2)mv^2 = 1.20*9.8*0.406J + (1/2)1.20*5.57^2J = 23.4J Total energy at the bottom is the same, since conserved, but now none is potential (1/2)mv^2 = 23.4 v^2 = 23.4*2/m = 38.98m^2/s^2 v = SQRT(38.98m^2/s^2) = 6.24 m/s Note that a small change in velocity is a relative large change in kinetic energy. This is why a hurricane can do way more damage when the wind speed is just 10mph faster. In an elastic collision, both momentum and kinetic energy are conserved m1V1o = m1V1 + m2V2 (1/2)m1V1o^2 = (1/2)m1V1^2 + (1/2)m2V2^2 cancelling out the (1/2)s and substituting in values 1.2*6.24 = 1.2V1 + 4.81V2 1.2*6.24^2 = 1.2V1^2 + 4.81V2^2 (7.49 - 1.2V1)/4.81 = V2 substitute this in for V2 in the KE eqn 46.779 = 1.2V1^2 + 4.81*{(7.49 - 1.2V1)/4.81}^2 1.50V1^2 - 3.74V1 - 35.1 = 0 V1^2 - 2.49V1 - 23.4 = 0 quadratic formula [2.49 +- sqrt{2.49^2 - 4*1*(-23.4)}]/2 = v1 = -3.75 or +6.24 +6.24 would be if the balls missed and the 1.2kg kept going -3.75 is - 3.75 m/s; it means that the ball bounced backwards. (c) Now let's find the speed of the big ball (7.49 - - (1.2)3.75)/4.81 = 2.49 so the big ball is going at 2.49m/s in the original direction of the small ball. Let's check our work momentum before was 1.2kg*6.24m/s does this equal 4.81kg*2.49m/s - 1.2*3.75 Yes, momentum is conserved! Now let's check KE Does (1/2)1.2kg(6.24m/s)^2 = (1/2)1.2kg(-3.75m/s)^2 + (1/2)4.81kg(2.49m/s)^2 Yes, kinetic energy is also conserved! Note that energy doesn't have a direction. The small ball and large ball are going in opposite directions, but we still add their KEs. (d) small ball will go up until it has no more KE all of KE will be PE (1/2)mv^2 = mgh h = (1/2) v^2/g = (1/2)(-3.75m/s)^2/(9.8m/s^2) = 0.718m (I got scared when I saw that this is higher than the height it swung from, but it started with some kinetic energy) (e) h = (1/2) v^2/g = (1/2)(2.49m/s)^2/(9.8m/s^2) = 0.317m

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