In the experiment shown above, you are testing the \"elasticity\" of a collision
ID: 1492331 • Letter: I
Question
In the experiment shown above, you are testing the "elasticity" of a collision between two metal spheres. One sphere with mass 0.7 kg is attached to a string (assume the string is massless and of constant length), forming a pendulum. The second sphere with mass 0.2 kg is placed at the edge of a table of height 1.4 m. When the first sphere is hanging straight down, it just touches the second sphere and is right above the table.
To run the experiment, you pull back the pendulum until the first sphere is a height 0.65 m above the table. You release the pendulum, the first sphere swings down and collides with the second sphere. You then measure how far from the table the second sphere lands on the ground, xf.
If the collision between the spheres is totally elastic, where would the second sphere land, xf?
Give your answer in meters to at least three digits. Do not include the units in your answer.
You will not be graded on the number of significant digits you provide. Providing at least three digits.
Explanation / Answer
for the first sphere
potential energy U = m*g*h
at the lowest point KE of the first obhet K = 0.5*m*u1^2
KE = PE
u1 = sqrt(2*g*h) = sqrt(2*9.8*0.65) = 3.57 m/s
ELASTIC COLLISION
m1 = 0.7 kg m2 = 0.2kg
speeds before collision
u1 = 3.57 m/s u2 = 0 m/s
speeds after collision
v1 = ? v2 =
initial momentum before collision
Pi = m1*u1 + m2*u2
after collision final momentum
Pf = m1*v1 + m2*v2
from momentum conservation
total momentum is conserved
Pf = Pi
m1*u1 + m2*u2 = m1*v1 + m2*v2 .....(1)
from energy conservation
total kinetic energy before collision = total kinetic energy after collision
KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2
KEf = 0.5*m1*v1^2 + 0.5*m2*v2^2
KEi = KEf
0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2 .....(2)
solving 1&2
we get
v1 = [ ((0.7-0.2)*3.57) + (2*0.2*0) ] /(0.7+0.2) = 9.8 m/s
v2 = ( ((0.2-0.7)*0) + (2*0.7*3.57) ) /(0.7+0.2)
v2 = 5.55 m/s
after for 2 nd sphere
along vertical initial velocity voy = 0
acceleration ay = -g
displacement y = -14 m
from equation of motion
y = voy*t + 0.5*g*t^2
-14 = -0.5*9.8*t^2
t = 1.69 s
along horizantal
xf = v2*t = 5.55*1.69 = 9.38 m <<<<----answer