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AshleyBcdd
 3 years ago
Ina shoots a large marble (Marble A, mass: 0.08 kg) at a smaller marble (Marble B, mass: 0.05 kg) that is sitting still. Marble A was initially moving at a velocity of 0.5 m/s, but after the collision it has a velocity of –0.1 m/s. What is the resulting velocity of marble B after the collision? Be sure to show your work for solving this problem along with the final answer.
AshleyBcdd
 3 years ago
Ina shoots a large marble (Marble A, mass: 0.08 kg) at a smaller marble (Marble B, mass: 0.05 kg) that is sitting still. Marble A was initially moving at a velocity of 0.5 m/s, but after the collision it has a velocity of –0.1 m/s. What is the resulting velocity of marble B after the collision? Be sure to show your work for solving this problem along with the final answer.

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AshleyBcdd
 3 years ago
Best ResponseYou've already chosen the best response.0@amistre64 @SheldonEinstein

SheldonEinstein
 3 years ago
Best ResponseYou've already chosen the best response.1I know this, wait!

SheldonEinstein
 3 years ago
Best ResponseYou've already chosen the best response.1@AshleyBcdd prefer to draw a diagram of it first, I mean that of condition

SheldonEinstein
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1351258288836:dw

mayank06
 3 years ago
Best ResponseYou've already chosen the best response.1initial linear momentum=final linear momentum so 0.08*0.5=0.1*0.08+0.05*v v=0.96

SheldonEinstein
 3 years ago
Best ResponseYou've already chosen the best response.1Now, what we have is : the condition I mentioned above in the form of a diagram is of before collision... Now here is the condition of the marbles AFTER collision in the form of a diagram... dw:1351258449229:dw

SheldonEinstein
 3 years ago
Best ResponseYou've already chosen the best response.1now using the law of conservation of momentum : Momentum before collision = Momentum after collision @AshleyBcdd are you getting what I am telling here?

SheldonEinstein
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large{m_A u_A + m_B u_B = m_A v_A + m_B v_B }\] \[\large{0.08 kg\times (0.05 ms^{1}) + 0.05 kg \times 0 ms^{1} = 0.08 kg \times 0.01 ms^{1} \\+ 0.05 kg \times v_B }\]

SheldonEinstein
 3 years ago
Best ResponseYou've already chosen the best response.1you will get : 0.096 after solving for v_B

AshleyBcdd
 3 years ago
Best ResponseYou've already chosen the best response.0I'm sorry i was away, yes i understand! thankyou so much :)

SheldonEinstein
 3 years ago
Best ResponseYou've already chosen the best response.1You're welcome....
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