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prisoha [69]
4 years ago
5

If two billiard balls collide, what happens to the systems overall momentum ?

Physics
1 answer:
andreyandreev [35.5K]4 years ago
4 0
Remains same. Here according to law of conservation of momentum, the system have momentum conserved and remains constant.
Given by the popular equation,
m1v1=m2v3
m1=Mass of billiard ball 1
m2= Mass of the 2nd ball
v1,v2 are the velocity of the ball 1,2 respectively.
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<h3><u>Answer;</u></h3>

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3 years ago
A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm ons tionless bearings. Two 500 g blocks fall from above, hit the tum ble s
Verdich [7]

There are mistakes in the question.The correct question is here

A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?

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Explanation:

Given data

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where I is moment of inertia

w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\   So\\I_{turntable}=M\frac{r^{2} }{2}\\I_{turntable}=2*(\frac{(0.2/2)}{2} )\\ I_{turntable}=0.01 \\And\\I_{block1}=I_{block2}=mr^{2}\\I_{block1}=I_{block2}=(0.5)*(0.2/2)^{2} \\ I_{block1}=I_{block2}==0.005\\so\\w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\w^{i}=\frac{0.01*(10.47)}{0.005+0.005+0.01} \\w^{i}=5.235 rad/s\\w^{i}=5.235*(60/2\pi )\\w^{i}=50 rpm

7 0
3 years ago
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