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.
(a) Calculate the total momentum of two bodies before collision as `m_(1) u_(1) + m_(2) u_(2)`.
(b) Write down the given values of `m_(1)` and `m_(2)` in SI system
(c ) Write the expression for their total momentum after collision as `(m_(1) + m_(2))` v where v is their common velocity.
(d) Assign proper signs to the initial velocities `u_(1)` and `u_(2)`.
(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))`

Options

A. b d a c e
B. b d c e a
C. a d b c e
D. c d e b a

Answer

A

Explanation

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