Question

A ball of mass 0.440 kg moving east (+x direction) with a speed of 3.60 m/s collides head-on with a 0.260 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?

Answer 1

let speed of first ball is v1 m/s and speed of second be v2 m/s after collision along +ve x direction.

as the collision is elastic,

momentum is conserved.

so momentum before collision=momentum after collision

==>0.44*3.6+0.26*0=0.44*v1+0.26*v2

==>0.44*v1+0.26*v2=1.584

==>v1=-0.591*v2+3.6....(1)

as the collision is elastic in nature, total kinetic energy is also conserved.

initial kinetic energy=final kinetic energy

==>0.5*0.44*3.6^2=0.5*0.44*v1^2+0.5*0.26*v2^2

==>0.22*v1^2+0.13*v2^2=2.8512

using v1=-0.591*v2+3.6

==>0.22*(-0.591*v2+3.6)^2+0.13*v2^2=2.8512

==>0.22*(0.591^2*v2^2+3.6^2-4.2552*v2)+0.13*v2^2=2.8512

==>0.20684*v2^2-0.936144*v2=0

==>v2=4.526 m/s

using expression for v1:

v1=-0.591*v2+3.6=0.925 m/s

hence both the balls will move in +ve x direction

with first ball moving with 0.925 m/s and second ball moving with 4.526 m/s