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A 0.50 kg ball traveling at 6.0 m/s due east collides head on with a 1.00 kg ball traveling in the opposite direction at -12.0 m/s. After the collision, the 0.50 kg ball moves away at -14 m/s. Find the velocity of the second ball after the collision.

User Anfab
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Answer: The velocity of the second ball after the collision is -2 m/s due east.

Step-by-step explanation:

We can use the law of conservation of momentum to solve this problem. According to this law, the total momentum of a system before a collision is equal to the total momentum of the system after the collision, provided that no external forces act on the system.

The momentum p of an object is given by the product of its mass m and velocity v: p = mv.

Before the collision, the momentum of the system is:

p_initial = m_1 * v_1 + m_2 * v_2

where m_1 and v_1 are the mass and velocity of the first ball, m_2 and v_2 are the mass and velocity of the second ball.

Plugging in the given values, we get:

p_initial = (0.50 kg)(6.0 m/s) + (1.00 kg)(-12.0 m/s) = -6.0 kg*m/s

After the collision, the momentum of the system is:

p_final = m_1 * v'_1 + m_2 * v'_2

where v'_1 and v'_2 are the velocities of the first and second ball after the collision.

We are given that the first ball moves away at -14 m/s after the collision, so:

v'_1 = -14 m/s

We can now use the conservation of momentum to solve for v'_2:

p_initial = p_final

m_1 * v_1 + m_2 * v_2 = m_1 * v'_1 + m_2 * v'_2

Plugging in the given values, we get:

(0.50 kg)(6.0 m/s) + (1.00 kg)(-12.0 m/s) = (0.50 kg)(-14 m/s) + (1.00 kg)(v'_2)

Solving for v'_2, we get:

v'_2 = -2 m/s

Therefore, the velocity of the second ball after the collision is -2 m/s due east.

User TjDillashaw
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