Worked exercise — The cat on the cart

Problem

A cart of $m_c = 4\;\text{kg}$ moves to the right at $5\;\text{m/s}$ relative to the road. On the cart, a cat of $m_g = 2\;\text{kg}$ moves at $-3\;\text{m/s}$ relative to the cart. Find:

1. the velocity of the cat relative to the road;
2. the velocity of the CM of cat+cart relative to the road;
3. the velocity of the cat and of the cart relative to the CM.

Solution

1) Velocity of the cat relative to the road. From the little arrow diagram, path cat → cart → road: $v_{g(s)} = v_{g(c)} + v_{c(s)} = -3 + 5 = 2\;\text{m/s}$

2) Velocity of the CM relative to the road. Weighted average of the velocities: $v_{\text{CM}(s)} = \frac{m_g v_{g(s)} + m_c v_{c(s)}}{m_g + m_c} = \frac{2\cdot 2 + 4\cdot 5}{6} = \frac{24}{6} = 4\;\text{m/s}$

3) Velocity in the CM frame. Each object has $v_{X(\text{CM})} = v_{X(s)} - v_{\text{CM}(s)}$: $v_{g(\text{CM})} = 2 - 4 = -2\;\text{m/s} \qquad v_{c(\text{CM})} = 5 - 4 = 1\;\text{m/s}$

Check. The momentum relative to the CM must be zero: $m_g v_{g(\text{CM})} + m_c v_{c(\text{CM})} = 2\cdot(-2) + 4\cdot 1 = 0 \;\checkmark$ The result confirms that, in the CM reference frame, the total momentum is always zero.

Links

Topics: Centre of mass Concepts: Centre of mass Skills: Changing reference frame · Calculating the centre of mass