In the experiment shown below, you are testing the \"elasticity\" of a collision
ID: 2305236 • Letter: I
Question
In the experiment shown below, you are testing the "elasticity" of a collision between two metal spheres. One sphere with mass m1 is attached to a string (assume the string is massless and of constant length), forming a pendulum. The second sphere with mass m2 is placed at the edge of a table, such that m2 is a height h above the floor.
When the first sphere is hanging straight down, it just touches the second sphere and the center of the two spheres have the same height. You can assume the spheres have the same radius.
To run the experiment, you pull back the pendulum until the first sphere, m1, is a height yi above its lowest point during the swing. You release the pendulum, m1swings down, and the two spheres collide. You then measure how far from the table the second sphere lands on the ground, d. d is measured from the edge of the able in the horizontal direction (along the floor).
m1 = 0.550 kg
m2 = 0.666 kg
h = 0.93 m
yi = 0.443 m
d = 0.728 m
How much energy is lost during the collision?
Give your answer in Joules 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 will avoid an answer being counted as incorrect due to rounding errors.
Explanation / Answer
for initial speed of m1 (just before collision)
m g yi = m v1i^2 /2
v1i = sqrt(2 x 9.8 x 0.443) = 2.947 m/s
and v2i = 0
after collision,
in vertical, yf - y0 = v0y t + ay t^2 /2
0 - 0.93 = 0 - 9.8 t^2 /2
t = 0.4357 sec
v2f = d/t = 1.671 m/s
Applying momentum conservation for collision,
m1 v1i + m2 v2i = m1 v1f + m2 v2f
(0.550 x 2.947) + 0 = (0.550 v1f) + (0.666 x 1.671)
v1f = 0.9235 m/s
Ki = m1 v1i^2 /2 + m2 v2i^2 /2 = 2.3883 J
Kf = m1 v1f^2 /2 + m2 v2f^2 /2 = 1.16435 J
energy lost = Ki - Kf = 1.224 J