Two skaters of masses 2m and 3m are initially at rest on frictionless ice. They push each other apart. Their relative speed after separation is 10 m/s. If the speed of the lighter skater is N m/s, find N.
Physics · Center of maas , Momentum and Collision · momentum and conservation of momentum
JEE Mains NEET JEE Advanced

Question

Two skaters of masses 2m and 3m are initially at rest on frictionless ice. They push each other apart. Their relative speed after separation is 10 m/s. If the speed of the lighter skater is N m/s, find N.

Physics · Center of maas , Momentum and Collision · momentum and conservation of momentum

G1 Physics · Accepted Answer

Let the lighter skater move with speed v₁ and the heavier skater with speed v₂ in the opposite direction. Since initial momentum is zero, 2m v₁ = 3m v₂. Their relative speed is v₁+v₂ = 10. From momentum conservation, v₁ = (3/2)v₂. Therefore (3/2)v₂ + v₂ = 10, so (5/2)v₂ = 10 and v₂ = 4 m/s. Hence v₁ = 6 m/s and N = 6.