Starting with an initial speed of 5.93 m/s at a height of 0.462 m, a 1.76-kg bal
ID: 2280699 • Letter: S
Question
Starting with an initial speed of 5.93 m/s at a height of 0.462 m, a 1.76-kg ball swings downward and strikes a 4.82-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 1.76-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 1.76-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.82-kg ball just after the collision. (d) How high does the 1.76-kg ball swing after the collision, ignoring air resistance? (e) How high does the 4.82-kg ball swing after the collision, ignoring air resistance?
Starting with an initial speed of 5.93 m/s at a height of 0.462 m, a 1.76-kg ball swings downward and strikes a 4.82-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 1.76-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 1.76-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.82-kg ball just after the collision. (d) How high does the 1.76-kg ball swing after the collision, ignoring air resistance? (e) How high does the 4.82-kg ball swing after the collision, ignoring air resistance?Explanation / Answer
A) So the first thing you need to find in the speed after collision. This can be found using SQRT Vo^2+2gHo. So square root of initial velocity squared plus two times gravity times height. Which for you would be. SQRT (5.93^2)+(2)(9.8)(0.462)=6.65m/s.
B) The next equation is ((m1-m2)/(m1+m2))*(Vo1) <---you just found
So for you it would be ((1.76-4.82)/(1.76+4.82))*(6.65)= -3.093m/s
C) The equation is ((2)m1)/(m1+m2)*(Vo1)
or ((2)(1.76)/(1.76+4.82))*(6.65)=3.56m/s
D) its asking for height after impact Vo2^2/2(g) is used
or -3.093^2/2(9.8)=0.488m
E) Same thing Vo2^2/2(g)
or 3.56^2/2(9.8)=0.647m
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