Answer to Question #250977 in Physics for Pranali

Question

Two 18.0 kg ice sleds are placed a short distance apart, one directly behind the other, as shown in the figure. A 3.96 kg cat, initially standing on sled 1, jumps across to sled 2 and then jumps back to sled 1. Both jumps are made at a horizontal speed of 4.30 m/s relative to the ice. What is the final velocity of sled 2? (Assume the ice is frictionless.) What is the final velocity of sled 1?

Answer 1

When the cat lands on sled 2 from sled 1, the conservation of momentum gives

$$m_cv_c=(m_s+m_c)u,\\\space\\ u=v_c\frac{m_c}{m_s+m_c}=0.775\text{ kg m/s}.$$

Then, ignoring the motion of sled 2, the cat jumps at 4.3 m/s relative to the ice:

$$(m_s+m_c)u=m_cv_c+m_sv_{s2f},\\\space\\\\\space\\ v_{s2f}=\frac{(m_s+m_c)u-m_cv_c}{m_s},\\\space\\ v_{s2f}=\frac{(18+3.96)0.775-3.96·(-4.3)}{18}=1.89\text{ m/s}.$$

When the cat jumped from sled 1 to sled 2, the speed of sled 1 became

$$0=m_cv_c-m_sv_s,\\\space\\ v_s=v_c\frac{m_c}{m_s}=0.946\text{ m/s}.$$

Then, while sled 1 was moving to the left with negative speed of 0.946 m/s, the cat moving at 4.3 m/s landed on sled 1. Conservation of momentum gives:

$$-m_sv_{s}-m_cv_c=-(m_c+m_s)v_{s1f},\\\space\\ v_{s1f}=\frac{m_sv_s+m_cv_c}{m_c+m_s},\\\space\\ v_{s1f}=\frac{18·0.946+3.96·4.3}{4.3+18}=1.53\text{ m/s}.$$