Consider the head-on collision of two identical bowling balls, each with mass 5

Question

Consider the head-on collision of two identical bowling balls, each with mass 5 kg (see figure). Ball A with velocity = m/s strikes ball B_1 which was at rest. Then ball A stops and ball B moves with the same velocity that ball A had initially. Choose a system consisting only of ball A. What is the momentum change of the system during the collision? Delta system = kg . m/s What is the momentum change of the surroundings? Delta surroundings = kg . m/s Choose a system consisting only of ball B. What is the momentum change of the system during the collision? Delta system = kg . m/s What is the momentum change of the surroundings? Delta surroundings = kg . m/s Choose a system consisting of both balls. What is the momentum change of the system during the collision? Delta system = kg . m/s What is the momentum change of the surroundings? Delta surroundings = kg . m/s

Explanation / Answer

a)
SInce A is sytem,
delta P system = mA* (VfA - ViA)
= 5*(0 - <6,0,0>)
= <-30,0,0> Kgm/s

SInce B is surrounding,
delta P surrounding= mB* (VfB - ViB)
= 5*(<6,0,0> - 0)
= <30,0,0> Kgm/s

b)
SInce B is sytem,
delta P system = mB* (VfB - ViB)
= 5*(<6,0,0> - 0)
= <30,0,0> Kgm/s

SInce A is surrounding,
delta P surrounding= mA* (VfA - ViA)
= 5*(0 - <6,0,0>)
= <-30,0,0> Kgm/s

c)
If A and B are both system, momentum must be conserved because lost in momentum of A = gain in momentum of B

delta P system = 0 kgm/s
delta P surrounding= 0 kgm/s

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