Starting with an initial speed of 3.61 m/s at a height of 0.315 m, a 1.06-kg bal
ID: 2259846 • Letter: S
Question
Starting with an initial speed of 3.61 m/s at a height of 0.315 m, a 1.06-kg ball swings downward and strikes a 5.10-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 1.06-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 1.06-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 5.10-kg ball just after the collision. (d) How high does the 1.06-kg ball swing after the collision, ignoring air resistance? (e) How high does the 5.10-kg ball swing after the collision, ignoring air resistance?
Explanation / Answer
a) Ei = Ef
1/2 mv^2 + m gh = 1/2 mv^2
0.5*3.61^2 + 9.81*0.315 = 0.5*v^2
v=4.38 m/s
b)
conservation of momentum
1.06*4.38 = 1.06*x + 5.1*y
y = (1.06*4.38 - 1.06*x)/5.1
conservation of kinetic energy
1.06*4.38^2 = 1.06*x^2 + 5.1*y^2
1.06*4.38^2 = 1.06*x^2 + 5.1*((1.06*4.38 - 1.06*x)/5.1)^2
x=-2.87 m/s
so 2.87 m/s to the left
c)
y = (1.06*4.38 - 1.06*-2.87)/5.1=1.51 m/s to the right
d) conservation of energy of the ball
1/2 mv^2 = m g h
0.5*2.87^2 = 9.81*h
h= 0.5*2.87^2/9.81=0.420 m
e) conservation of energy of the 5.1 kg ball
1/2 mv^2 = m g h
0.5*1.51^2 = 9.81*h
h= 0.5*1.51^2/9.81=0.116 m
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