Question

The figure shows two 24.3 kg ice sleds that are placed a short distance apart, one directly behind the other. A 3.90 kg cat initially standing on one sled jumps to the other one and then back to the first. Both jumps are made at a speed of 3.11 m/s relative to the ice. What are the final speeds of (a) the first sled and (b) the other sled?

Answer 1

mass of ice sleds m₁ = 24.3 kg;

mass of cat m₂ = 3.9 kg;

speed of cat v₂ = 3.11 m/s; (both times);

(a) Calculate speed of ice sleds:

apply conservation of momentum

m₂v₂ = m₁v₁;

v₁ = 0.499 m/s. (speed of first sled when cat jumped first time to second sled)

when cat again jumped to came back on the first sled then it will be inelastic collision, therefore

m₂v₂ + m₁v₁ = v(m₂ + m₁);

v = (3.9x3.11 + 24.3x0.499)/(24.3 + 3.9);

= 0.86 m/s ( final speeds of the first sled)

(b) speed of second sled (v) when cat jumped first time to second sled

m2v2 + 0 = (m1 + m2)v

v = 3.9x3.11/(3.9 + 24.3);

v = 0.43 m/s

now final speed of second sled;

m₂v₂ = m₁v₁;

v₁ = 0.499 m/s