In the experiment shown above, you are testing the \"elasticity\" of a collision
ID: 1495070 • 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
for elastic collision, let the velocity of the second sphere after collision be v
velocity of first sphere just before collision = u
velocity of the first sphere after collision = v'
from momentom conservation
initial momentum = fibnal momentum
0.7u = 0.7v' + 1v ... (1)
now to find u
ke at bottom of the string for the mass = PE of the mass when raised to height 0.25m [conservation of energy]
0.5*0.7*u^2 = 0.7*9.8*0.25
u = 0.5916 m/s
0.7u = 0.7v' + v
0.4141 = 0.7v' + v ... (2)
Now from energy conservation for perfectly el;astic collision
0.5*0.7*0.5916^2 = 0.5*0.7(v')^2 + 0.5*1*v^2
0.24499 = 0.7(v')^2 + (0.4141 - 0.7v)^2 = 0.7(v')^2 + 0.17147 + 0.49(v')^2 - 0.57974v'
1.19(v')^2 - 0.57974v' - 0.07352 = 0
v' = 0.5916 m/s, -0.10443 m/s
v = -2*10^-5 m/s, 0.487201 m/s
now, v has to be positive, so
v = 0.487201 m/s, v' = -0.10443 m/s
let the sphere land at distance x from the table, after time t
1.3 = 0.5*g*t^2
t = 0.515 s
x = vt = 0.2509 m