Question

Two balls undergo a perfectly elastic head-on collision, with one ball initially at rest. If the incoming ball has a speed of 200 m/s, what are the final speed and direction of each ball if a. The incoming ball is much more massive than the stationary ball? b. The stationary is much more massive that the incoming ball?

Answer 1

a. First ball 200 m/s and second ball 400 m/s both in same direction

b. First ball 200 m/s opposite direction and second ball 0 m/s at rest

Step-by-step explanation:

Now for elastic collision there is a formula including mass and velocity of first and second body.

Now,

m1 = mass of first body

m2 = mass of second body

u1 = initial velocity of first body

u2 = initial velocity of second body

v1= final velocity of first body

v2 = final velocity of second body

Formula for final velocities of both bodies are

v1 = [(m1-m2)/m1+m2)]*u1 + [(2m2)/(m1+m2)]*u2

v2 = [(2m1)/(m1+m2)]*u1 + [(m2-m1)/m2+m1)]*u2

Now according to given data

u1 = 200 m/s

u2 = 0 m/s

For a)

consider m2 = 0 kg (in comparison, first ball is much massive, so the mass of second ball is negligible)

Now by putting values of m2, u1 and u2, we get

v1 = u1

v2 = 2u1

Since u1 = 200 m/s

v1 = 200 m/s

v2 = 400 m/s and both will move in same direction that is from where first body was coming towards second body.

For b)

Consider m1 = 0 Kg for same assumption

Now by putting values of m1, u1 and u2, we get

v1 = -u1

v2 = 0

So v1 = 200 m/s in opposite direction after collision with massive body

v2 = 0 m/s which means that massive body will not move after collision and will be at rest but smaller body will move in opposite direction after collision in double speed.