In the experiment shown above, you are testing the \"elasticity\" of a collision
ID: 1495093 • 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 1.0 kg is placed at the edge of a table of height 1.3 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 0.25 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?
Explanation / Answer
v1' = 2gh = 0.25*19,6 = 2.21 m/sec
momentum of the first sphere, p1' = m1*v1' = 0.7*2.21 = 1.55kgm/s
and initial kinetic energy; E1' = 1/2*m1*v1'2 = 1/2 * 0.7 * 2.21 * 2.21 = 1.71J
after the collision, both p1' and E1' are conserved
so, m1v1 + m2v2 = 1.55 (1)
and 1/2m1v12 + 1/2m2v22 = 1.71
or m1v12 +m2v22 = 1.71*2 = 3.42 (2)
from equation (1) ; v1 = [1.55 - m2v2]/m1
substituting the value of v1 in (2), we get
[1.55 - m2v2]2/m1 + m2v22 =3.42
substituing the values of m1 and m2 ,we obtain
[1.55 - v2]2/0.7 + v22 =3.42
so, v2 =1.83 m/s
time taken for fall ; t = sqrt(2h/g) = sqrt ( 2*1.3/9.8) =0.51secs
and the position where the second sphere lands; x2 = v2*t = 1.83*0.51 = 0.93m
and distance