A block A of mass 3.50kg moving east with a speed of 3.40 m/s colloids head on w

Question

A block A of mass 3.50kg moving east with a speed of 3.40 m/s colloids head on with a second block 13 of mass 5.60kg moving west with a speed of 2.15 m/s. After collision block A moves west with a speed of 0.12 m/s. a) What is magnitude and direction of the velocity of block B after collision? b) How much Impulse is experienced by the Block B? e) If the collision time is 0.0556 second, What is average force experienced by the block A during collision? d) Is this collision elastic or inelastic? Justify your answer.

Explanation / Answer

(taking east as +ve then west will be -ve)

a) Applying momentum conservation for this colllision,

(3.50 x 3.40) + ( 5.60 x -2.15) = (3.50 x -0.12) + ( 5.60 v)

v = 0.05 m/s

b) Impulse = change in momentum

= 5.60 ( - 2.15 - (0.05)) = - 12.32 kg m/s

magnitude of impulse = 12.32 kg m/s

c) Impulse =F x t

12.32 =F x 0.0556

F = 221.58 N

d) initial energy = 3.50 x 3.40^2 /2 +    5.60 x 2.15^2 /2   = 33.17 J

energy after collision = 3.50 x 0.12^2 /2   + 5.60x0.05^2/2 = 0.0322 J

initial energy is not equal to final energy/

so this is not elastic collision.

this is an Inelastic collision.

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