Answered

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8. An object of mass 13 kg going to the right witha speed of 26 m/s collides with a(n) 10 kg object at rest.After collision the 13 kg object moves with a speed of 11 m/smaking an angle of 20 degree with the horizontal-right.Calculate the x-component of the velocity of the 10 kgobject after collision. (1 point)A. 07.069 m/sB. O 17.629 m/sC. O 3.745 m/sD. O 11.865 m/sE. O 20.362 m/s

8 An Object Of Mass 13 Kg Going To The Right Witha Speed Of 26 Ms Collides With An 10 Kg Object At RestAfter Collision The 13 Kg Object Moves With A Speed Of 11 class=

Sagot :

8)Answer: 20.362 m/s

Explanation:

According to the law of conservation of momentum,

Total momentum before collision = total momentum after collision

Total momentum before collision = m1u1 + m2u2

where

m1 is the mass of the object moving to the right

u1 is its velocity

m2 is the mass of the object at rest

u2 is its velocity

From the information given,

m1 = 13

u1 = 26

m2 = 10

u2 = 0

Total momentum before collision = 13 x 26 + 10 x 0 = 338

Total momentum after collision = m1v1cosθ1 + m2v2cosθ2

where

θ1 is the angle made by m1 with the horizontal

θ2 is the angle made by m2 with the horizontal

v1 is the final velocity of m1 after collision

v2 is the final velocity of m2 after collision

From the information given,

v1 = 11

θ1 = 20

Total momentum = 13 x 11cos20 + 10v2cosθ2 = 134.376 + 10v2cosθ2

Thus,

338 = 134.376 + 10v2cosθ2

10v2cosθ2 = 338 - 134.376 = 203.624

v2cosθ2 = 203.624/10

v2cosθ2 = 20.362 m/s

The x component of the velocity of the 10kg ball after collision is 20.362 m/s

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