Question 14.4: Ball B, of mass mB, is suspended from a cord of length l att......
Ball B, of mass m_B, is suspended from a cord of length l attached to cart A, of mass m_A, which can roll freely on a frictionless horizontal track. If the ball is given an initial horizontal velocity \text v_0 while the cart is at rest, determine (a) the velocity of B as it reaches its maximum elevation, (b) the maximum vertical distance h through which B will rise. (It is assumed that v_0^2 < 2gl.)
The impulse-momentum principle and the principle of conservation of energy will be applied to the cart-ball system between its initial position 1 and position 2, when B reaches its maximum elevation.
Velocities Position 1: \left( \text v _A\right)_1=0 \quad\quad\quad\left( \text v _B\right)_1=\text v _0 (1)
Position 2: When ball B reaches its maximum elevation, its velocity \left(\text v _{B / A}\right)_2 relative to its support A is zero. Thus, at that instant, its absolute velocity is
\left(\text v _B\right)_2=\left(\text v _A\right)_2+\left(\text v _{B / A}\right)_2=\left(\text v _A\right)_2 (2)
Impulse-Momentum Principle. Noting that the external impulses consist of W _A t, W _B t, and Rt, where R is the reaction of the track on the cart, and recalling (1) and (2), we draw the impulse-momentum diagram and write
\sum m\text v _1+\sum E\text x t~ I m p _{1 \text y 2}=\Sigma m \text v _2\overset{+}{\text y}~x~ components:~~~~~~~m_B v_0=\left(m_A+m_B\right)\left(v_A\right)_2
which expresses that the linear momentum of the system is conserved in the horizontal direction. Solving for \left(v_A\right)_2:
\left(v_A\right)_2=\frac{m_B}{m_A+m_B} v_0 \quad\quad\left(\text v _B\right)_2=\left(\text v _A\right)_2=\frac{m_B}{m_A+m_B} v_0 ~\text yConservation of Energy
Position 1. Potential Energy: V_1=m_A g l
Kinetic Energy: T_1=\frac{1}{2} m_B v_0^2
Position 2. Potential Energy: V_2=m_A g l+m_B g h
Kinetic Energy: T_2=\frac{1}{2}\left(m_A+m_B\right)\left(v_A\right)_2^2
T_1+V_1=T_2+V_2: \quad \frac{1}{2} m_B v_0^2+m_A g l=\frac{1}{2}\left(m_A+m_B\right)\left(v_A\right)_2^2+m_A g l+m_B g h
Solving for h, we have
h=\frac{v_0^2}{2 g}-\frac{m_A+m_B}{m_B} \frac{\left(v_A\right)_2^2}{2 g}or, substituting for \left(v_A\right)_2 the expression found above,
h=\frac{v_0^2}{2 g}-\frac{m_B}{m_A+m_B} \frac{v_0^2}{2 g} \quad\quad\quad h=\frac{m_A}{m_A+m_B} \frac{v_0^2}{2 g}Remarks. (1) Recalling that v_0^2<2 g l, it follows from the last equation that h < l; we thus check that B stays below A as assumed in our solution.
(2) For m_A \gg m_B, the answers obtained reduce to\left( \text v _B\right)_2=\left(\text v _A\right)_2=0 and h=v_0^2 / 2 g; B oscillates as a simple pendulum with A fixed. For m_A \ll m_B, they reduce to \left( \text v _B\right)_2=\left(\text v _A\right)_2=\text v _0 and h = 0; A and B move with the same constant velocity \text v _0.
