A 11 kg block moving at 3 m/s catches up to a 20 block moving with a speed of 1 m/s . After they collide?

FIXED (was originally using equation for inelastic collision)

Conservation of energy. Momentum and kinetic energy are conserved in elastic collisions.

Momentum before = momentum after (elastic collision)

(m1*u1) + (m2*u2) = (m1*v1) + (m2*v2)

(11*3) + (20*1) = (11*v1) + (20*v2)

53 = (11*v1) + (20*v2)

Because you have 2 unknowns, you will need to use the conservation of Kinetic energy law (elastic collision)

½(m1*u1²) + ½(m2*u2²) = ½(m1*v1²) + ½(m2*v2²)

½(11*3²) + ½(20*1²) = ½(11*v1²) + ½(20*v2²)

119 = (11*v1²) + (20*v2²)

Now, we have 2 equations and 2 unknowns. Solve for v2 in terms of v1 in the first equation, and plug the value into the second equation for v2. This will give you the answer the value of v1. Use that to find v2.

NOTE:

m = mass, u = initial velocity, v = final velocity

Use these equations and solve the system, you should get the velocities after. The guy above me is doing it as an inelastic collision, it is wrong.

if we assume m1 (mass) and v1 (velocity) for first mass , m2 and v2 for second mass ,we have :

m1 v1i + m2 v2i = m1 v1f + m2 v2f

and

1/2 m1 v1i2 + 1/2 m2 v2i2 = 1/2 m1 v1f2 +1/2 m2 v2f2

i : initial

f : final