A 340-g air track cart traveling at 1.25 m/s suddenly collides elastically with a stationary 300-g cart.What is the speed of the 300-g cart just after the collision?
A)0.0781 m/s
B)1.33 m/s
C)0.625 m/s
D)1.25 m/s
E)0.664 m/s

Question

Quizplus · Accepted Answer

In an elastic collision, both momentum and kinetic energy are conserved. The formula for the final velocity of the second cart (v2') in an elastic collision is given by v2' = [(2*m1*v1) + (m2 - m1)*v2] / (m1 + m2), where m1 and v1 are the mass and velocity of the first cart before the collision, m2 and v2 are the mass and velocity of the second cart before the collision (v2 = 0 since it's stationary), and v2' is the velocity of the second cart after the collision. Plugging in the given values: m1 = 0.34 kg, v1 = 1.25 m/s, m2 = 0.3 kg, and v2 = 0, we find v2' = 1.33 m/s.

A 340-G Air Track Cart Traveling at 1