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Question: Two putty balls of equal mass moving with equal velocity in mutually perpendicular directions, stick...

Two putty balls of equal mass moving with equal velocity in mutually perpendicular directions, stick together after collision. If the balls were initially moving with a velocity of 452ms−145\sqrt{2}ms^{- 1} each, the velocity of their combined mass after collision is

A

452ms−145\sqrt{2}ms^{- 1}

B

45ms−145ms^{- 1}

C

90ms−190ms^{- 1}

D

22.52ms−122.5\sqrt{2}ms^{- 1}

Answer

45ms−145ms^{- 1}

Explanation

Solution

Initial momentum P⃗=m452i^+m452j^\vec { P } = m 45 \sqrt { 2 } \hat { i } + m 45 \sqrt { 2 } \hat { j }⇒ ∣P⃗∣=m×90| \vec { P } | = m \times 90

Final momentum 2m×V2 m \times V

By conservation of momentum 2m×V=m×902 m \times V = m \times 90

∴V=45 m/sV = 45 \mathrm {~m} / \mathrm { s }