What is the speed of the resulting ball of clay in m/s?
A 50.0 g ball of clay traveling east at 9.00 m/s collides and sticks together with a 60.0 g ball of clay traveling north at 1.00 m/s. What is the speed of the resulting ball of clay in meters per second (m/s)?
2 Answers
Since they stick together, this is a "perfectly inelastic collision."
The formula for inelastic collisions is:
(m1)(v1)i + (m2)(v2)i = (m1 + m2)vf
You're going to have to run through the equation twice, once using the north/south speeds for velocity, and once using the east/west speeds.
Let's start with the north/south speeds.
(50.0)(0) + (60.0)(1) = (50.0 + 60.0)(vf)
0 + 60 = (110)(vf)
vf = .545 m/s to the north
That's just the north/south velocity. We're going to solve for the east/west velocity now:
(50.0)(9) + (60.0)(0) = (50.0 + 60.0)(vf)
450.0 = (110)(vf)
vf = 4.09 m/s to the east
Now we're going to use the Pythagorean Theorem to find the resultant vector:
a^2 + b^2 = c^2
(4.09)^2 + (.545)^2 = c^2
17.03 = c^2
c = 4.13 m/s
now you need to find the angle:
inverse sin of (opp/hyp) will give us the angle:
inverse sin of .545/4.13 = 7.6 degrees north of east
So the answer is:
4.13 m/s at 7.6 degrees north of east
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