What is the speed of the resulting ball of clay in m/s?

Question

A 50.0 ball of clay traveling east at 9.00 collides and sticks together with a 60.0 ball of clay traveling north at 1.00 .

Raymond Moen · Accepted Answer

Since they stick together, this is a "perfectly inelastic collision."The formula for inelastic collisions is:(m1)(v1)i + (m2)(v2)i = (m1 + m2)vfYou'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 northThat'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 eastNow 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^217.03 = c^2c = 4.13 m/snow 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 eastSo the answer is:4.13 m/s at 7.6 degrees north of east

Raymond Moen · Answer

Ball Of Clay

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What is the speed of the resulting ball of clay in m/s?

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