Question 14.SP.5: Ball B, with a mass of mB, is suspended from a cord with a l......

STRATEGY: You are asked about the velocity of the system at two different positions, so use the principle of work and energy for the cart−ball system. You will also use the impulse–momentum principle, because momentum is conserved in the x direction.

MODELING and ANALYSIS: For your system, choose the ball and the cart and model them as particles.

Velocities.

Position 1 :$\qquad\qquad(\mathbf{v}_{A})_{1}=0\qquad\qquad(\mathbf{v}_{B})_{1}=\mathbf{v}_{0}$             (1)

Position 2: When ball B reaches its maximum elevation, its velocity $(v_{B/A})_{2}$ relative to its support A is zero (Fig. 1). Thus, at that instant, its absolute velocity is

$(\mathbf{v}_{B})_{2}=(\mathbf{v}_{A})_{2}+(\mathbf{v}_{B/A})_{2}=(\mathbf{v}_{A})_{2}$             (2)

Impulse–Momentum Principle. The external impulses consist of${W}_{A}t,{W}_{B}t,\mathrm{~and~}{R}t,$ where R is the reaction of the track on the cart. Recalling Eqs. (1) and (2), draw the impulse–momentum diagram (Fig. 2) and write

$\Sigma m\mathbf{v}_{1}+\Sigma\mathrm{Ext~Imp}_{1\rightarrow2}=\Sigma m\mathbf{v}_{2}$

$\underrightarrow{+}$x components: $\qquad \qquad m_{B}\nu_{0}=(m_{A}+m_{B})(\nu_{A})_{2}$

This expresses that the linear momentum of the system is conserved in the horizontal direction. Solving for $(v_{A})_{2}$, you have

$(\nu_{A})_{2}=\frac{m_{B}}{m_{A}+m_{B}}\nu_{0}\qquad(\mathbf{v}_{B})_{2}=(\mathbf{v}_{A})_{2}=\frac{m_{B}}{m_{A}+m_{B}}\nu_{0}$ → ◂

Conservation of Energy. The system is shown in Fig. 3 in the two positions. Define your datum at the location of B in position 1 (although you could also choose to place it at A). You can now calculate the kinetic and potential energies in the two positions:

Position 1. Potential Energy: $\qquad V_{1}=m_{A}g l$

Kinetic Energy: $\qquad T_{1}={\frac{1}{2}}m_{B}\nu_{0}^{2}$

Position 2. Potential Energy: $\qquad V_{2 }=m_{A}g l+m_{B}g h$

Kinetic Energy: $\qquad T_{2}={\textstyle{\frac{1}{2}}}(m_{A}+m_{B})(\nu_{A})_{2}^{2}$

Substituting these into the conservation of energy gives

$T_{1}+V_{1}=T_{2}+V_{2}\colon \qquad{\textstyle{\frac{1}{2}}}m_{B}v_{0}^{2}+m_{A}g l={\textstyle{\frac{1}{2}}}(m_{A}+m_{B})(\nu_{A})_{2}^{2}+m_{A}g l+m_{B}g h$

Solving for h, you have

$h=\frac{\nu_{0}^{2}}{2g}-\frac{m_{A}+m_{B}}{m_{B}}\frac{(\nu_{A})_{2}^{2}}{2g}$

or substituting $(v_{A})_{2}$ from above, you have

$h=\frac{\nu_{0}^{2}}{2g}-\frac{m_{B}}{m_{A}+m_{B}}\frac{\nu_{0}^{2}}{2g}\qquad h=\frac{m_{A}}{m_{A}+m_{B}}\frac{\nu_{0}^{2}}{2g}$  ◂

REFLECT and THINK: Recalling that $v_{0}^{2}$< 2gl, it follows from the last equation that h < l; this verifies that B stays below A, as assumed in the solution. For $m_{A} ≫ m_{B}$, the answers reduce to $(v_{B})_{2} = (v_{A})_{2}= 0$ and h = $v_{0}^{2}$/2g; B oscillates as a simple pendulum with A fixed. For $m_{A} ≪ m_{B}$, they reduce to $(v_{B})_{2}= (v_{A})_{2} = v_{0}$ and h = 0; A and B move with the same constant velocity $v_{0}$.