Question
Two objects of masses m and 4m are moving toward each other along the x-axis with the same initial speeds v0 = 45.0 m/s. The object with mass m is traveling to the left, and the object with mass 4m is traveling to the right. They undergo an elastic glancing collision such that mass m is moving downward after the collision at right angles from its initial direction.
(a) Find the final speeds of the two masses.
(b) What is the angle from its initial direction of motion at which the mass 4m is scattered?
Answer
Let
v = final speed of mass m,
v' = final speed of mass 4m and
θ = angle at which mass 4m is scattered
By the law of conservation of linear momentum applied in the horizontal and vertical direction respectively,
(4m) * (45) - (m) * (45) = 4mv'cosθ
=> v'cosθ = 135/4 ... ( 1 ) and
4mv'sinθ = mv
=> v'sinθ = v /4 ... ( 2 )
Squarring and adding eqns. ( 1 ) and ( 2 ),
v'^2 = (135/4)^2 + (v/4)^2 ... ( 3 )
As the collison is elastic, kinetic energy is conserved
=> (1/2)m*(45)^2 + (1/2)(4m)*(45)^2 = (1/2)mv^2 + (1/2)(4m)v'^2
=> 10125 = v^2 + v'^2 ... ( 4 )
Putting the value of v'^2 from eqn. ( 3 ) in eqn. ( 4 ),
10125 = v^2 + (135/4)^2 + (v/4)^2
=> 162000 = 16v^2 + 18225 + v^2
=> 143775 = 17v^2
=> v = 92 m/s
Putting this value of v in eqn. ( 4 ),
v'^2
= 10125 - v^2
= 10125 - (92)^2
= 1661
=> v' = 40.8 m/s
Putting thee values of v and v' in equation ( 2 ),
sinθ = (92) / 4*(40.8)
=> θ = 34.3°.
Two objects of masses m and 4m are moving toward each other along the x-axis with the same initial speeds v0 = 45.0 m/s. The object with mass m is traveling to the left, and the object with mass 4m is traveling to the right. They undergo an elastic glancing collision such that mass m is moving downward after the collision at right angles from its initial direction.
(a) Find the final speeds of the two masses.
(b) What is the angle from its initial direction of motion at which the mass 4m is scattered?
Answer
Let
v = final speed of mass m,
v' = final speed of mass 4m and
θ = angle at which mass 4m is scattered
By the law of conservation of linear momentum applied in the horizontal and vertical direction respectively,
(4m) * (45) - (m) * (45) = 4mv'cosθ
=> v'cosθ = 135/4 ... ( 1 ) and
4mv'sinθ = mv
=> v'sinθ = v /4 ... ( 2 )
Squarring and adding eqns. ( 1 ) and ( 2 ),
v'^2 = (135/4)^2 + (v/4)^2 ... ( 3 )
As the collison is elastic, kinetic energy is conserved
=> (1/2)m*(45)^2 + (1/2)(4m)*(45)^2 = (1/2)mv^2 + (1/2)(4m)v'^2
=> 10125 = v^2 + v'^2 ... ( 4 )
Putting the value of v'^2 from eqn. ( 3 ) in eqn. ( 4 ),
10125 = v^2 + (135/4)^2 + (v/4)^2
=> 162000 = 16v^2 + 18225 + v^2
=> 143775 = 17v^2
=> v = 92 m/s
Putting this value of v in eqn. ( 4 ),
v'^2
= 10125 - v^2
= 10125 - (92)^2
= 1661
=> v' = 40.8 m/s
Putting thee values of v and v' in equation ( 2 ),
sinθ = (92) / 4*(40.8)
=> θ = 34.3°.
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