Question

the diagram shows a compressed spring between two carts that are initially at rest. cart x has a mass of 4 kg and cart y has a mass of 2 kg. a thread tied to a hook on each cart holds the carts together. when the thread is cut, the spring is released and the carts move apart. how does the speed of cart y compare to the speed of cart x after the thread is cut? a the speed of cart y is half the speed of cart x. b the speed of cart y is the same as the speed of cart x. c the speed of cart y is two times the speed of cart x. d the speed of cart y is four times the speed of cart x.

Explanation:

Step1: Apply law of conservation of momentum

The initial momentum of the system is 0 since the carts are at rest ($p_{i}=0$). Let the mass of cart X be $m_X = 4$ kg, mass of cart Y be $m_Y=2$ kg, velocity of cart X be $v_X$ and velocity of cart Y be $v_Y$. According to the law of conservation of momentum $p_{i}=p_{f}$, so $0 = m_Xv_X+m_Yv_Y$.

Step2: Solve for the ratio of velocities

From $0 = m_Xv_X+m_Yv_Y$, we can get $m_Xv_X=-m_Yv_Y$. Then $\frac{v_Y}{v_X}=-\frac{m_X}{m_Y}$. Substitute $m_X = 4$ kg and $m_Y = 2$ kg into the equation, we have $\frac{v_Y}{v_X}=-\frac{4}{2}=- 2$. The negative sign just indicates the direction. In terms of speed (magnitude of velocity), $v_Y = 2v_X$.

Answer:

C. The speed of cart Y is two times the speed of cart X.