# Answer to Question #12287 in Mechanics | Relativity for jim murillo

Question #12287

an 11.7 kg object is at rest on a perfectly frictionless surface when it is struck head on by a 3.9 kg object moving at 18 m/s. if the collision is perfectly elastic, what is the speed of the 11.7 kg object?

Expert's answer

Let's use the momentum conservation law:

m1·u1 + m2·u2 = m1·v1 + m2·v2.& (1)

Here m1 is the mass of the first object (11.7 kg), m2 is the mass of the second object (3.9 kg), u1 and u2 are initial velocities of the first and the second objects respectfully and v1 and v2 are their velocities after collision.

The equation expressing conservation of kinetic energy is

m1·u1² + m2·u2² = m1·v1² + m2·v2².& (2)

Using expressions (1) and (2) we'll get

v1 = (m1-m2)·u1/(m1+m2) + 2·m2·u2/(m1+m2),

or, substituting given masses and initial velocities

v1 = (11.7-3.9)·0/(11.7+3.9) + 2·3.9·18/(11.7+3.9) = 9 m/s.

So, 11.7 kg object obtain speed 9 m/s.

m1·u1 + m2·u2 = m1·v1 + m2·v2.& (1)

Here m1 is the mass of the first object (11.7 kg), m2 is the mass of the second object (3.9 kg), u1 and u2 are initial velocities of the first and the second objects respectfully and v1 and v2 are their velocities after collision.

The equation expressing conservation of kinetic energy is

m1·u1² + m2·u2² = m1·v1² + m2·v2².& (2)

Using expressions (1) and (2) we'll get

v1 = (m1-m2)·u1/(m1+m2) + 2·m2·u2/(m1+m2),

or, substituting given masses and initial velocities

v1 = (11.7-3.9)·0/(11.7+3.9) + 2·3.9·18/(11.7+3.9) = 9 m/s.

So, 11.7 kg object obtain speed 9 m/s.

## Comments

## Leave a comment