Unit 2: Conservation of Momentum in 2 Dimensions?

Question

Two ice skaters undergo a collision, after which their arms are intertwined and they have a common velocity of 0.85 m/s [27° S of E]. Before the collision, one skater of mass 71 kg had a velocity of 2.3 m/s [12° N of E], while the other skater had a velocity of 1.9 m/s [52° S of W]. What is the mass of the second skater?
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Answer 1

we make use of conservation of momentum



x momentum: before collision = 71kg x 2.3 cos 12 m/s - M (1.9 cos 52)



the minus sign indicates a negative momentum since it is directed in the -x direction; M is the mass of the second skater



x momentum after collision = (71+M)(0.85 cos 27)



the x momentum is unchanged with collision, so we have

159.7-1.17M= 53.8+0.76M

solving for M gives M=54.9kg



we can use this value of M to verify momentum conservation in the y direction:



71x2.3sin12 -54.9x1.9sin 52 = -(71+54.9)x0.85sin 27

-48.2= -48.5...so apart from some round off error, we have verified that this mass satisfies both the x and y momentum conditions

Answer 2

the web page (below) is titled : The Derivation of E=mc2 introductory paragraph: The Derivation of E=mc2 Perhaps the most famous equation of all time is E = mc2. The equation is a direct result of the theory of special relativity, but what does it mean and how did Einstein find it? In short, the equation describes how energy and mass are related. Einstein used a brilliant thought experiment to arrive at this equation, which we will briefly review here. followed by discussion and formulae!

Answer 3

Use conservation of momentum



components in x-dir:

m1*v1x + m2*v2x = (m1 + m2)*v3x

m2 = m1*(v3x - v1x)/(v2x - v3x)



v1x = 2.3*cos(12)

v2x = -1.9*cos(52)

v3x = .85*cos(27)



m2 = 56 kg