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A 35-gram stainless steel ball on a track is moving at a velocity of 9 m/s. on the same track, a 75-gram stainless steel ball is moving at a velocity of –7 m/s. after colliding, the 35-gram ball moves at a velocity of –15 m/s. assume there is no net force on the system. what is the velocity of the 75-gram ball after the collision? round to the nearest tenth. –4.2 m/s –0.5 m/s 0.5 m/s 4.2 m/s



Answer :

The velocity of second ball after collision is 4.2m/s

Given:

mass of first ball, m1 = 35g

mass of second ball, m2 = 75g

velocity of first ball before collision, u1 = 9m/s

velocity of second ball before collision, u2 = -7m/s

velocity of first ball after collision, v1 = -15m/s

velocity of second ball after collision, v2 = ? (To Find)

As a result, the equation of the law of conservation of momentum is as follows: m1 u1 + m2 u2 represents the total momentum of particles A and B before the collision, and m1 v1 + m2 v2 represents the total momentum of particles A and B after the collision.

Now it is given that there is no net force on the system of two balls

According to the Law of Conservation of Momentum so, here momentum is constant before and after collision:

P1 = P2

m1u1 + m2u2 = m1v1 + m2v2

35g × 9m/s + 75g × (-7m/s) = 35g × (-15m/s) + 75g × v2

v2 = (35g × 9m/s + 75g × (-7m/s) - 35g × (-15m/s))/75g

v2 = 4.2m/s

Therefore, the velocity of second ball after collision is 4.2m/s

Learn more about Law of Conservation of Momentum here

brainly.com/question/7538238

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