Elastic Collisions (continued) Practice 1. A0.015 kg marble sliding to the right at 22.5cm/s on a frictionless surface makes an elastic head-on collision with a 0.015 kg marble moving to the left at 18.0cm/s After the collision, the first marble moves to the left at 18.0cm/s a. Find the velocity of the second marble after the collision. b. Verify your answer by calculating the total kinetic energy before and after the collision.

To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.<br /><br />a. To find the velocity of the second marble after the collision, we can use the conservation of momentum equation:<br /><br />Initial momentum = Final momentum<br /><br />Momentum before collision = Momentum after collision<br /><br />(0.015 kg) * (22.5 cm/s) + (0.015 kg) * (-18.0 cm/s) = (0.015 kg) * (-18.0 cm/s) + (0.015 kg) * (v2)<br /><br />Simplifying the equation:<br /><br />0.3375 kg·cm/s - 0.27 kg·cm/s = -0.27 kg·cm/s + 0.015 kg * v2<br /><br />0.0675 kg·cm/s = -0.27 kg·cm/s + 0.015 kg * v2<br /><br />0.3375 kg·cm/s = 0.27 kg·cm/s + 0.015 kg * v2<br /><br />0.0675 kg·cm/s = 0.015 kg * v2<br /><br />v2 = 4.5 cm/s<br /><br />Therefore, the velocity of the second marble after the collision is 4.5 cm/s.<br /><br />b. To verify the answer, we can calculate the total kinetic energy before and after the collision.<br /><br />Total kinetic energy before collision:<br /><br />KE_initial = (1/2) * (0.015 kg) * (22.5 cm/s)^2 + (1/2) * (0.015 kg) * (18.0 cm/s)^2<br /><br />KE_initial = 0.1916 J + 0.1215 J = 0.3131 J<br /><br />Total kinetic energy after collision:<br /><br />KE_final = (1/2) * (0.015 kg) * (18.0 cm/s)^2 + (1/2) * (0.015 kg) * (4.5 cm/s)^2<br /><br />KE_final = 0.1215 J + 0.0176 J = 0.1391 J<br /><br />Since the total kinetic energy before and after the collision is approximately the same (0.3131 J ≈ 0.1391 J), the answer is verified.