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An object of mass 500 g moving with a speed of 10 m s^(-1) collides with another object of mass 250 g moving with a speed of 2 m s^(-1) in the opposite direction. After collision both the objects move in the same direction with common velocity. Write the proper order of steps to find out their common velocity after collision. <br> (a) Calculate the total momentum of two bodies before collision as m_(1) u_(1) + m_(2) u_(2). <br> (b) Write down the given values of m_(1) and m_(2) in SI system <br> (c ) Write the expression for their total momentum after collision as (m_(1) + m_(2)) v where v is their common velocity. <br> (d) Assign proper signs to the initial velocities u_(1) and u_(2). <br> (e) Using law of conservation of momentum Equate the above two expressions and determine the value of v as (m_(1) u_(1) + m_(2) u_(2))/(m_(1) + m_(2))

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Answer Text

b d a c e b d c e a a d b c e c d e b a

Answer :
A
Solution :
Given `m_(1) = 500 g = (1)/(2) kg, m_(2) = 250 g = (1)/(4)kg (b)` <br> `u_(1) = 10 m s^(-1), u_(2) = -2 m s^(-1) (d)` <br> Momentum before collision, `m_(1) u_(1) + m_(2) u_(2)` <br> `=(1)/(2) xx 10 + (1)/(4) xx (-2), 5 - (1)/(2) = (9)/(2) Kg m s^(-1)` <br> Momentum after collision, `(m_(1) + m_(2)) v = (3)/(4) v (c )` <br> From law of conservation of momentum, <br> `(9)/(2) = (3)/(4)v implies v = 6m s^(-1) (e)`

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