a body of mass M strikes a wall at an angle 60 with velocity V elastically the change in momentum is
a MV
b MV/2
c -2MV
d ZERO
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let the angle that velocity vector makes with the wall surface be 30 deg.
The velocity vector makes 60 deg. with the perpendicular to the wall at the point of contact.
The horizontal and vertical components before the collision are:
V Cos60° , V Sin 60°
After the rebound, the components are :
- V Cos 60° and V Sin 60°
Only the horizontal component changes. So change in momentum is M *2 V Cos 60° , so that is M * V.
The velocity vector makes 60 deg. with the perpendicular to the wall at the point of contact.
The horizontal and vertical components before the collision are:
V Cos60° , V Sin 60°
After the rebound, the components are :
- V Cos 60° and V Sin 60°
Only the horizontal component changes. So change in momentum is M *2 V Cos 60° , so that is M * V.
kvnmurty:
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sir,i am sorry to say,but it seems that your answer has errors...
according to my knowledge of physics,if initial velocity is u n final velocity is v,along the impulsive tangent,the wall...the velocity components become, usin30+vsin30....now since the collision is elastic,this means that,e=1...so v=u=V....hence,then ucos30 is balanced by vcos30...and along the wall velocity is 2Vsin30....so,momentum=M*2Vsin30=MV
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0
Answer: (a) mv
Explanation: Given, mass = m, velocity = v, therefore, momentum =?
We know that, momentum, P = mass x velocity
Therefore, P = mv
Thus, option (a) mv is correct
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