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A mass m moves with a velocity v and collides inelastically with another identical mass. After collision the 1st mass moves with velocity in a direction perpendicular to the initial direction of motion. Find the speed of the second mass after collision. [AIEEE 2005]
- a)v
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A mass m moves with a velocity v and collides inelastically with anoth...
Let mass A moves with velocity v and collides inelastically with mass B, which is at rest.
According to problem mass A moves in a perpendicular direction and let the mass B moves at angle θ with the horizontal with velocity v.
Initial horizontal momentum of system
(before collision) = mv ....(i)
Final horizontal momentum of system
(after collision ) = mVcosθ.................(ii)
From the conservation of horizontal linear momentum mv
= mvcosθ ⇒ v = Vcosθ .....(iii)
Initial vertical momentum of system (before collision) is zero.
Final vertical momentum of system mv/√3 − mV sinθ
From the conservation of vertical linear momentum
According to problem mass A moves in a perpendicular direction and let the mass B moves at angle θ with the horizontal with velocity v.
Initial horizontal momentum of system
(before collision) = mv ....(i)
Final horizontal momentum of system
(after collision ) = mVcosθ.................(ii)
From the conservation of horizontal linear momentum mv
= mvcosθ ⇒ v = Vcosθ .....(iii)
Initial vertical momentum of system (before collision) is zero.
Final vertical momentum of system mv/√3 − mV sinθ
From the conservation of vertical linear momentum
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A mass m moves with a velocity v and collides inelastically with anoth...
Let mass A moves with velocity v and collides inelastically with mass B, which is at rest.
According to problem mass A moves in a perpendicular direction and let the mass B moves at angle θ with the horizontal with velocity v.
Initial horizontal momentum of system
(before collision) = mv ....(i)
Final horizontal momentum of system
(after collision ) = mVcosθ.................(ii)
From the conservation of horizontal linear momentum mv
= mvcosθ ⇒ v = Vcosθ .....(iii)
Initial vertical momentum of system (before collision) is zero.
Final vertical momentum of system mv/√3 − mV sinθ
From the conservation of vertical linear momentum
According to problem mass A moves in a perpendicular direction and let the mass B moves at angle θ with the horizontal with velocity v.
Initial horizontal momentum of system
(before collision) = mv ....(i)
Final horizontal momentum of system
(after collision ) = mVcosθ.................(ii)
From the conservation of horizontal linear momentum mv
= mvcosθ ⇒ v = Vcosθ .....(iii)
Initial vertical momentum of system (before collision) is zero.
Final vertical momentum of system mv/√3 − mV sinθ
From the conservation of vertical linear momentum
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A mass m moves with a velocity v and collides inelastically with another identical mass. After collision the 1st mass moves with velocity in a direction perpendicular to the initial direction of motion. Find the speed of the second mass after collision. [AIEEE 2005]a)vb)c)d)Correct answer is option 'C'. Can you explain this answer?
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