Starting with an initial speed of 5.57 m/s at a height of 0.185 m, a 2.37-kg bal

Question

Starting with an initial speed of 5.57 m/s at a height of 0.185 m, a 2.37-kg ball swings downward and strikes a 4.22-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 2.37-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 2.37-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.22-kg ball just after the collision. (d) How high does the 2.37-kg ball swing after the collision, ignoring air resistance? (e) How high does the 4.22-kg ball swing after the collision, ignoring air resistance?

Explanation / Answer

a) Given; b) m1=2.37kg M2=4.22kg Vi=5.557m/s (Vi)=0 h=0.185m Step one: (a) For m1(2.37kg)Find (E) the energy amount to get Vf at h=0 The total mechanical energy: E=K+U k=1/2*m*V^2 U=m*g*h E= (1/2*m1*Vi^2) + (m1*g*h) E= (1/2*2.37*5.557^2) + (2.37*9.8*0.185) E=40.89J Next solve for Vf by using K=1/2mv^2, Units for K(Joules) since E= Joules E= (1/2*m1*Vf^2) 40.89=(1/2*2.37*Vf^2) Let me know if u want me to show work? Vf=5.874 (rite before collision for 2.37kg) Step Two: (b) Since we are assuming that its "elastic collision " There is a special formula that is used to find the magnitude and the direction. The final value from the Magnitude will be either positive or negative establishing the direction. ' Vi*(m1-M2)/(m1+M2)=(Vf) Since the direction of Vi is down a minus is used. -5.557*(2.37-4.22)/(2.37+4.22)=(Vf) Vf= 1.56 (after collision for 2.37 kg) we got positive answer so the direction is up now. Meaning it bounce up. I am still working on (c),(d), (e) I be right back. let know if you want to see more steps..

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