A man of mass m1 = 76.0 kg is skating at v1 = 9.00 m/s behind his wife of mass m
ID: 2012300 • Letter: A
Question
A man of mass m1 = 76.0 kg is skating at v1 = 9.00 m/s behind his wife of mass m2 = 55.0 kg, who is skating at v2 = 3.00 m/s. Instead of passing her, he inadvertently collides with her. He grabs her around the waist, and they maintain their balance.(a) Sketch the problem with before-and-after diagrams, representing the skaters as blocks. (Do this on paper. Your instructor may ask you to turn in this work.)
(b) Is the collision best described as elastic, inelastic, or perfectly inelastic?
Why?
(c) Write the general equation for conservation of momentum in terms of m1, v1, m2, v2, and final velocity vf. (Do this on paper. Your instructor may ask you to turn in this work.)
(d) Solve the momentum equation for vf. (Use the following as necessary: m1, v1, m2, v2. Do not substitute numerical values, use variables only.)
vf =
(e) Substituting values, obtain the numerical value for vf, their speed after the collision.
couple vf = m/s
man vf = m/s
wife vf = m/s
Explanation / Answer
A) In the before picture the two blocks are moving in the same direction with one moving faster and coming up from behind the other. In the after, they are moving together as one object, stuck together.
There is an animation that is a moving representation of this in the Perfectly Inelastic Collision section of this page: http://en.wikipedia.org/wiki/Inelastic_collision
B) This is a perfectly inelastic collision because they stick together after colliding.
C) m1v1 + m2v2m = (m1 + m2)vf
D) By rearranging the above equation, we simply divide both sides by (m1 + m2). Therefore,
E) To find vf, simply plug in your given values for m1, m2, v1, and v2. You find that vf = 6.48 m/s.
The vf is the same for the husband, wife, and both together, because they stuck together in the inelastic collision.
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