Question

Elastic collisions in one dimension: a 620-g object traveling at 2.1 m/s collides head-on with a 320-g object traveling in the opposite direction at 3.8 m/s. if the collision is perfectly elastic, what is the change in the kinetic energy of the 620-g object?

a) it loses 0.23 j.
b) it gains 0.69 j.
c) it loses 0.47 j.
d) it loses 1.4 j.
e) it doesn't lose any kinetic energy because the collision is elastic.

Answer 1

The change in kinetic energy of the 620-g object in this perfectly elastic collision is -0.47 J.

Step-by-step explanation:

To determine the change in kinetic energy of the 620-g object, we can use the conservation of momentum and apply it to the elastic collision between the two objects. The equation for the conservation of momentum is:

m1v1 + m2v2 = m1v1f + m2v2f

where m1 and m2 are the masses of the objects, v1 and v2 are their initial velocities, and v1f and v2f are their final velocities. We can solve for v1f to find the final velocity of the 620-g object. Then, we can calculate the change in kinetic energy using the equation:

Change in KE = (1/2) m (vf2 - vi2)

where m is the mass of the object, vf is the final velocity, and vi is the initial velocity.

Using these equations, we find that the change in kinetic energy of the 620-g object is -0.47 J (option c).