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zhuklara [117]
3 years ago
8

A billiard ball (ball #1) moving at 5.00 m/s strikes a stationary ball (ball #2) of the same mass. after the collision, ball #1

moves at a speed of 4.35 m/s. find the speed of ball #2 after the collision.
Physics
2 answers:
irakobra [83]3 years ago
8 0

Answer

Two billiard balls of equal mass are on a level, frictionless surface. The first ball is moving and collides with the second ball, which was stationary. After the collision, both balls are moving. What is a possible speed for the first ball after the collision? Need help

Explanation:

please

Rashid [163]3 years ago
4 0
If it is completely elastic, you can calculate the velocity of the second ball from the kinetic energy 
<span>v1 = velocity of #1 </span>
<span>v1' = velocity of #1 after collision </span>
<span>v2' = velocity of #2 after collision. </span>

<span>kinetic energy: v1^2 = v1' ^2 + v2' ^2 (1/2 and m cancel out) </span>
<span>5^2 = 4.35^2 + v2' ^2 </span>
<span>v2 = 2.46 m/s <--- ANSWER</span>
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A 20-kg child running at 1.4 m/s jumps onto a playground merry-go-round that has mass 180 kg and radius 1.6m. She is moving tang
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Answer:

ωf = 0.16 rad/s

Explanation:

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Moment of Inertia of the MGR = ½mr² = ½(180)1.6² = 230.4 kg•m²

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Conservation of angular momentum

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Read 2 more answers
A bead slides without friction around a loopthe-loop. The bead is released from a height 21.9 m from the bottom of the loop-the-
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Answer:

Part a)

v = 12.45 m/s

Part B)

F_n = 0.05 N

Explanation:

Part A)

As we know that the point A lies on the top of the loop

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mgH = \frac{1}{2}mv^2 + mg(2R)

so the speed at point A is given as

mg(H - 2R) = \frac{1}{2}mv^2

v = \sqrt{2g(H - 2R)}

v = \sqrt{2(9.81)(21.9 - 2\times 7)}

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Part B)

Now the force equation at point A is given as

F_n + mg = \frac{mv^2}{R}

F_n = \frac{mv^2}{R} - mg[/tex]

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QUESTION 36
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