Two objects of masses m and 3m are moving toward each other along the x-axis wit
ID: 2184278 • Letter: T
Question
Two objects of masses m and 3m are moving toward each other along the x-axis with the same initial speed vo. the object with mass m is traveling to the left. and the object with mass 3m is traveling to the right. they undergo an elastic glancing collision such that m is moving downward after the collision at right angles from its initial direction.
a) find the final speeds of the two objects
b) what is the angle? at which the object with mass 3m is scattered.
Explanation / Answer
(a) mass m: - 91.8 m/s mass 4m: - 41.2 m/s (b) Scattering angle of mass 4m is 33.6° with its initial direction Solution: v = final speed of mass m, v' = final speed of mass 4m and ? = angle at which mass 4m is scattered By law of conservation of linear momentum applied in the horizontal and vertical direction respectively, (4m) * (45) - (m) * (45) = 4mv'cos? => v'cos? = 135/4 ( 1 ) and 4mv'sin? = mv => v'sin? = v /4 ( 2 ) Squarring adding eqns. ( 1 ) and ( 2 ), v'^2 = (135/4)^2 + (v/4)^2 ( 3 ) By the law of conservation of kinetic energy, (1/2)m*(45)^2 + (1/2)(4m)*(45)^2 = (1/2)mv^2 + (1/2)(4m)v'^2 => 10125 = v^2 + v'^2 ( 4 ) Putting the value of v'^2 from eqn. ( 3 ) in eqn. ( 4 ), 10125 = v^2 + (135/4)^2 + (v/4)^2 => 162000 = 16v^2 + 18225 + v^2 => 143275 = 17v^2 => v = 91.8 m/s Putting this value of v in eqn. ( 4 ), v'^2 = 10125 - v^2 = 10125 - (91.8)^2 = 1698 => v' = 41.2 m/s Putting thee values of v and v' in equation ( 2 ), sin? = (91.8) / 4*(41.2) => ? = 33.6°
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