Question

A 10 kg mass slides on a frictionless surface with speed 10 m/s, and has an elastic collision with a 2.0 kg mass. Assume both masses remain along a straight line before and after the collision. (a) If the 10 kg mass is at rest after the collision, what was the initial velocity of the 2.0 kg mass? What was the final velocity of the 2.0 kg mass? (b) If the 2.0 kg mass is at rest after the collision, what was the initial velocity of the 2.0 kg mass? What was the final velocity of the 10 kg mass?

Answer 1

part a:

as the collision is elastic, both momentum and kinetic energy will be consersrved.

we will take velocity to the right as positive and velocity to the left as negative.

initial velocity of 10 kg mass=10 m/s

initial velocity of 2 kg mass=v1 m/s (let)

final velocity of 10 kg mass=0 m/s

let final velocity of 2 kg mass be v2 m/s.

conserving momentum:

10*10+2*v1=10*0+2*v2

==>v2=50+v1...(1)

conserving kinetic energy (as potential energy is not changing)

0.5*10*10^2+0.5*2*v1^2=0.5*10*0^2+0.5*2*v2^2

==>1000+2*v1^2=2*(50+v1)^2=5000+200*v1+2*v1^2

==>200*v1=-4000

==>v1=-20 m/s

then v2=50+v1=30 m/s

so initial velocity of 2 kg mass was 20 m/s towards the 10 kg mass and final velocity was 30 m/s, along the original direction of motion of 10 kg mass.

part b:

let final velocity of 10 kg mass be v2 m/s.

final velocity of 2 kg mass=0 m/s

conserving mometum:

10*10+2*v1=10*v2

==>v1=5*v2-50

consefving kinetic energy,

0.5*10*10^2+0.5*2*v1^2=0.5*10*v2^2+0.5*2*0^2

==>1000+2*(5*v2-50)^2=10*v2^2

==>1000+50*v2^2+5000-1000*v2=10*v2^2

==>40*v2^2-1000*v2+6000=0

==>v2=10 m/s

then v1=0 m/s

so initial velocity of 2 kg mass is 0 m/s and final velocity of 10 kg mass is 10 m/s