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 resista

nce?

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. Steps four and five, to answer (d) and (e), which states to find the height after the collision for both of our massess. So lets find a equation that we will use for both of them. Well since we already used the conservation of energy of mechanics, the kinematics equations should work. Since time was not given then the equation V^2= Vi^2 - 2g(yf-yi) this equation is also already derived to a simpliar set up; h=(Vi^2)/(2*g) h= (1.56^2)/(2*9.8) (d) h=0.12416 m This is for the height of mass ball 2.37kg. For part (e) it is no w plug and chuck because we already have our equation. V^2= Vi^2 - 2g(yf-yi) h=(Vi^2)/(2*g) h= (1.122^2)/(2*9.8) (e) h=0.06422 m This is for the heavier mass ball of 4.22kg As you are going through the steps make sure u double checked your values because I was am not 100% on mine. I sorry but matters what matters is that you fully understand how to do the problems. Let me know is I need to explain or need to show more steps.

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