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A ball of mass 3.0 kg, moving at 2 m/s eastward, strikes head-on a ball of mass 1.0 kg that is moving at 2 m/s westward. The balls stick together after the impact. What are the magnitude and direction of the velocity of the combined mass after the collision?

User Fqdn
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Answer:

To solve this problem, we can use the law of conservation of momentum, which states that the total momentum of a closed system remains constant before and after a collision.

Let's first find the initial momentum of the system before the collision:

p = m1v1 + m2v2

where p is the total momentum, m1 and v1 are the mass and velocity of the first ball, and m2 and v2 are the mass and velocity of the second ball.

p = (3.0 kg)(2 m/s) + (1.0 kg)(-2 m/s) [since the second ball is moving in the opposite direction]

p = 2 kg m/s eastward

Now, since the balls stick together after the collision, we can consider them as a single object with a combined mass of 4.0 kg. Let's use the law of conservation of momentum again to find the final velocity of the combined mass:

p = mv

where p is the total momentum (which is the same before and after the collision), m is the combined mass of the two balls, and v is the final velocity of the combined mass.

2 kg m/s eastward = (4.0 kg) v

v = 0.5 m/s eastward

Therefore, the magnitude of the velocity of the combined mass after the collision is 0.5 m/s, and the direction is eastward.

User Mark Stickley
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