a)final velocity of the person and cart relative to theground b)frictional force

Question

a)final velocity of the person and cart relative to theground b)frictional force acting on the person while he is slidingacross the top surface of the cart c)how long does the frictional force act on the person? d)find the charge in the momentum of the person and the changein the momentum of the cart e)displacement of the person relative to the ground while heis sliding on the cart f) displacement of the cart relative to the ground while theperson is sliding g)change in kinetic energy of the person? h)change in kinetic energy of the cart i)Explain why the answers to g and h differ. What kind ofcollision is this and what accounts for the loss of mechanicalenergy?

Explanation / Answer

Initial momentum of person = 69 kg * 4.10 m/s = 282.9 kgm/s Initial momentum of cart = 0 The final momentum of person + cart = (69kg + 138 kg)V2 a) According to conservation of momentum             Pp1 + Pc1 = Ppc             282.9 kg m/s + 0 = (69kg + 138 kg) V2 therefore V2 = ( 282.9 kg m/s ) / (69kg + 138 kg)   = 1.366 m/s b) The frictional force acting on the person        f = k N =k mg = 0.40 * 69 kg * 9.8 m/s2 =270.48 N c) we know that change in momentum is equal to impulse             I = f t = mv2 - mv1             t = (mv2 - mv1) / - f                               t = (mv2 - mv1) / - f                  

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