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A 2kg ball traveling at 3.5 m/s collides with a 3 kg traveling in the same direction at 2.5 m/s. If the 3 kg block moves at 5.0 m/s in its original direction after the collision, what is the final speed of the 2 kg block?​

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answer: the final speed of the 2 kg ball is 0.25 m/s.

Step-by-step explanation:

To solve this problem, we can use the law of conservation of momentum, which states that the total momentum of a system before a collision is equal to the total momentum after the collision.

The momentum of an object is defined as the product of its mass and velocity:

momentum = mass x velocity

So, the total momentum before the collision can be calculated as:

total momentum before = (mass of ball 1 x velocity of ball 1) + (mass of ball 2 x velocity of ball 2)

total momentum before = (2 kg x 3.5 m/s) + (3 kg x 2.5 m/s)

total momentum before = 7 kg m/s + 7.5 kg m/s

total momentum before = 14.5 kg m/s

After the collision, the 3 kg ball moves at 5.0 m/s in its original direction. Let's assume that the 2 kg ball moves at a final velocity of v.

Using the law of conservation of momentum, we can write:

total momentum after = (mass of ball 1 x final velocity of ball 1) + (mass of ball 2 x final velocity of ball 2)

total momentum after = (2 kg x v) + (3 kg x 5.0 m/s)

total momentum after = 2v kg m/s + 15 kg m/s

Since the total momentum before the collision is equal to the total momentum after the collision, we can set these two expressions equal to each other:

total momentum before = total momentum after

14.5 kg m/s = 2v kg m/s + 15 kg m/s

Solving for v, we get:

v = (14.5 kg m/s - 15 kg m/s) / 2 kg

v = -0.25 m/s

Since the final velocity cannot be negative, we know that the 2 kg ball is moving in the opposite direction after the collision. So, we can take the absolute value of v to find the final speed of the ball:

final speed = |v| = |-0.25 m/s| = 0.25 m/s

Therefore, the final speed of the 2 kg ball is 0.25 m/s.

User Kostix
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2 votes

To solve this problem, we can use the law of conservation of momentum, which states that the total momentum of a system before a collision is equal to the total momentum of the system after the collision.

The equation for conservation of momentum is:

m1v1 + m2v2 = m1v1' + m2v2'

Where:

m1 = mass of object 1 (2 kg)

v1 = velocity of object 1 before collision (3.5 m/s)

m2 = mass of object 2 (3 kg)

v2 = velocity of object 2 before collision (2.5 m/s)

v1' = velocity of object 1 after collision (unknown)

v2' = velocity of object 2 after collision (5.0 m/s)

Plugging in the given values, we get:

(2 kg)(3.5 m/s) + (3 kg)(2.5 m/s) = (2 kg)(v1') + (3 kg)(5.0 m/s)

Simplifying, we get:

7 + 7.5 = 2v1' + 15

14.5 = 2v1'

v1' = 7.25 m/s

Therefore, the final speed of the 2 kg block after the collision is 7.25 m/s.

User Duy Tran
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