Question

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 .50kg ball moves away at -14 m/s. what is the velocity of the other ball after the collision?

Answer 1

After using the conservation of momentum, the velocity of the 1.00 kg ball after the collision is calculated to be -2 m/s, meaning it continues to move in the westward direction.

Step-by-step explanation:

To solve for the velocity of the 1.00 kg ball after the collision, we must use the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision. We can set up the equation as follows:

Momentum before = Momentum after

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

3 kg·m/s - 12 kg·m/s = -7 kg·m/s + (1.00 kg * v)

-9 kg·m/s = -7 kg·m/s + 1.00 kg * v

v = (-9 kg·m/s + 7 kg·m/s) / 1.00 kg

v = (-2 kg·m/s) / 1.00 kg

v = -2 m/s

Therefore, the velocity of the 1.00 kg ball after the collision is -2 m/s, which means it continues to move in the -x direction, or westward if we assume east is positive.

Answer 2

Answer: -2 m/s

Step-by-step explanation: