Study the material on collisions, kinetic energy, momentum, and momentum conserv

Question

Study the material on collisions, kinetic energy, momentum, and momentum conservation in your text. Pay particular attention to totally inelastic collisions. Study the instructions for this experiment. Work the following problems. They are similar to those you will encounter in the experiment. Consider a scenario during which a 2.00 kg mass (A) moving at 2.00 m/s strikes another 2.00 kg mass (B) which is at rest. Both are confined to move along the same straight line. Three possible outcomes of the collision that conserve momentum are: they stick together and move off at 1.00 m/s. they bounce off one another and move off in the same direction, one (A) at 0.50 m/s and the other (B) at 1.50 m/s. (A) comes to rest and (B) moves off at 2.00 m/s. In each case the momentum before the collision is: (2.00 kg)(2.00 m/s) = 4.00 kg-m/s. In each of the three cases above show that momentum is conserved by finding the total momentum after the collision

Explanation / Answer

here,

case a)

mass of each mass , m = 2 kg

velocity after collison , v = 1 m/s

P(after) = m*v + m*v

P(after ) = 2*1 + 2*1

P(after) = 4 kg m/s

hence the momentum is conserved

case b)

mass of each mass , m = 2 kg

velocity after collison of block (a) , va = 0.5 m/s

velocity after collison of block (b) , vb = 1.5 m/s

P(after) = m*va + m*vb

P(after ) = 2*0.5 + 2*1.5

P(after) = 4 kg m/s

hence the momentum is conserved

case c)

mass of each mass , m = 2 kg

velocity after collison of block (a) , va = 0 m/s

velocity after collison of block (b) , vb = 2 m/s

P(after) = m*va + m*vb

P(after ) = 2*0 + 2*2

P(after) = 4 kg m/s

hence the momentum is conserved

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