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IRINA_888 [86]
3 years ago
6

Billiard ball A (0.35 kg) is struck such that it moves at 10 m/s toward a second identical ball (Ball B) initially at rest. Afte

r the collision, Ball A continues to move in the same direction at 2 m/s. What is the magnitude of the velocity for Ball B after the collision?
Physics
1 answer:
timurjin [86]3 years ago
3 0

Answer:

Explanation:

We shall apply law of conservation of momentum during the collision of ball A and B .

Total momentum before collision of A and B = .35 x 10 = 3.5 kg m/s

Let the velocity of B after collision be v .

Total momentum after collision  = .35 x 2 + .35v

According to law of conservation of momentum

.35 x 2 + .35v  = 3.5

.35 v = 2.8

v = 8 m /s .

The direction of B will be same as direction of A .

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Explain how balanced and unbalanced forces effect an objects motion differently
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Balanced Forces acting on an object will not change the object's motion. Unbalanced Forces acting on an object will change the change the object's motion.
5 0
3 years ago
A 244.0 N block is at rest on a flat, frictionless table. A hooked cable applies an upward force of 24.0 N on the block. What is
blagie [28]

Answer:

268N

Explanation:

The upward force acting on the block are the reaction and the hooked table..

The total normal force acting = normal reaction + 24N

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4 0
3 years ago
Assume it takes 5.00 minutes to fill a 50.0-gal gasoline tank. (1 U.S. gal = 231 in.3).
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7 0
4 years ago
Which formula describes acceleration?<br><br> m/s^2<br><br><br> m/s<br><br><br> s/m<br><br><br> m2
Aleks04 [339]

Answer:

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3 0
3 years ago
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