Ball B, with a mass of mB, is suspended from a cord with a length l attached to cart A, with a mass of mA, that can roll freely on a frictionless horizontal track. If the ball is given an initial horizontal velocity v0 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. (Assume v02< 2gl .)
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 :(vA)1=0(vB)1=v0 (1)
Position 2: When ball B reaches its maximum elevation, its velocity (vB/A)2 relative to its support A is zero (Fig. 1). Thus, at that instant, its absolute velocity is
(vB)2=(vA)2+(vB/A)2=(vA)2 (2)
Impulse–Momentum Principle. The external impulses consist ofWAt,WBt, and Rt, 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
Σmv1+ΣExt Imp1→2=Σmv2
+x components: mBν0=(mA+mB)(νA)2
This expresses that the linear momentum of the system is conserved in the horizontal direction. Solving for (vA)2, you have
(νA)2=mA+mBmBν0(vB)2=(vA)2=mA+mBmBν0 → ◂
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: V1=mAgl
Kinetic Energy: T1=21mBν02
Position 2. Potential Energy: V2=mAgl+mBgh
Kinetic Energy: T2=21(mA+mB)(νA)22
Substituting these into the conservation of energy gives
T1+V1=T2+V2:21mBv02+mAgl=21(mA+mB)(νA)22+mAgl+mBgh
Solving for h, you have
h=2gν02−mBmA+mB2g(νA)22
or substituting (vA)2 from above, you have
h=2gν02−mA+mBmB2gν02h=mA+mBmA2gν02 ◂
REFLECT and THINK: Recalling that v02< 2gl, it follows from the last equation that h < l; this verifies that B stays below A, as assumed in the solution. For mA≫mB, the answers reduce to (vB)2=(vA)2=0 and h = v02/2g; B oscillates as a simple pendulum with A fixed. For mA≪mB, they reduce to (vB)2=(vA)2=v0 and h = 0; A and B move with the same constant velocity v0.