a particle of mass m strikes elastically on a wall with velocity v, at an angle 60degree from the wall then magnitude of change in momentum of ball along the wall is
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Since momentum is vector quantity
so momentum toward the wall=mvcos30=mv√3/2
and momentum of ball while going back=-mv√3/2 since momentum remains conserved but direction is opposite
Momentum change=mv√3/2-(-mv√3/2)
2mv√3/2=mv√3
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so momentum toward the wall=mvcos30=mv√3/2
and momentum of ball while going back=-mv√3/2 since momentum remains conserved but direction is opposite
Momentum change=mv√3/2-(-mv√3/2)
2mv√3/2=mv√3
PLEASE MARK AS BRAINLIEST IF HELPFUL!!!
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