Question

A hockey puck of mass m₁=155 g slides from left to right with an initial velocity of 21.5 m/s. It collides head on with a second puck of the same mass, m₂=m₁, moving in the opposite direction with velocity -25.5 m/s. They collide elastically head-on. After the collision, the velocity of m₂ is:

Answer 1

initial momentum of puck 1 =mass*veloicty=0.155*21.5=3.3325 kg.m/s

initial momentum of puck 2 =mass*veloicty=0.155*(-25.5)=-3.9525 kg.m/s

then total initial momentum of the system=3.3325-3.9525=-0.62 kg.m/s

initial kinetic energy of puck 1=0.5*mass*speed^2=0.5*0.155*21.5^2=35.8243 J

initial kinetic energy of puck 2=0.5*mass*speed^2=0.5*0.155*25.5^2=50.3943 J

then total initial kinetic energy of the system=35.8243+50.3943=86.2186 J

let after the collision speed of m1 is v1 and speed of m2 is v2.

then final total momentum=m1*v1+m2*v2

=0.155*(v1+v2)

as total momentum is conserved (there is no external forces present)

total initail momentum=total final momentum

==>-0.62=0.155*(v1+v2)

==>v1+v2=-4

==>v1=-v2-4...(1)

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

(total mechanical energy is conserved, but as there is no change in potential energy between before collision and after collision, total kinetic energy is also conserved)

==>total final kinetic energy=initial total kinetic energy

==>0.5*0.155*(v1^2+v2^2)=86.2186

==>v1^2+v2^2=1112.498

using expression for v1 from equation 1,

(-4-v2)^2+v2^2=1112.498

==>16+v2^2+8*v2+v2^2=1112.498

==>2*v2^2+8*v2-1096.498=0

solving for v2, we get

v2=21.5 m/s

or v2=-25.5 m/s

as v2=-25.5 m/s is the velocity before colision, it can be discarded.

hence the solution for v2 is 21.5 m/s

hence m2 will move with a veloicty of 21.5 m/s after collison.