13. A block with a mass m1 = 2.0-kg moves with a speed of v1 = 8.0 m/s on a fric

Question

13. A block with a mass m1 = 2.0-kg moves with a speed of v1 = 8.0 m/s on a frictionless table approaching a second block with a mass m2 = 6.0-kg and a speed of 2.0 m/s.

(i) If this is a collision where after the event the two blocks are deformed into one large mass, then what is the velocity (speed and direction) of that mass?

(ii) If instead this is a collision where the two blocks collide and conserve kinetic energy in the process, each moving away at its own velocity, then what is the final velocity of each of the blocks?

Explanation / Answer

2*8+6*(-2)=(2+6)(velocity after collision of combined system)

=> v=0.5 meter per sec{======> direction}

PART B)

as K.E CONSERVES THIS IS A ELASTIC COLLISION

[AS THER IS NO DEFORMATION]

BY APPLYING CINSERVATION OF MOMENTUM AND THE FACT THAT K.E CONSERVES WE GET

VELOCITY OF MASS 2KG=14 MT PER SEC =====>(DIRECTION EMPHASIS)

VELOCITY OF MASS 6KG=4 MT PER SEC <=====(DIRECTION EMPHASIS)

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