13.3 NewtonÕs Second Law for Particle Systems

413

However, you might argue that the external forces acting on a spinning ball are different because of air resistance: It is the interaction with the air that causes a spinning ball to move sideways, which is called curving the ball. This is a valid objection. The motion of the center of mass is determined by the external forces acting on the object. But, if we can neglect the effects of air resistance, the motion of the center of mass of a rod when you throw it is the same when it is rotating as when it is not rotating. This may be surprising, but it is a result of NewtonÕs second law for particle systems.

13.3.1 Example: Ballistic Motion with an Explosion

Problem: A projectile is Þred from the ground. Its initial velocity in the horizontal direction is v0. When it reaches its maximum height of h, a charge is set off, splitting the projectile into two equal parts. One part moves forward with the velocity v1. Find the trajectory of each of the parts, and the trajectory of their center of mass. You may neglect air resistance.

Solution: We have illustrated the process in Fig. 13.8. The process has three stages. In stage one, the projectile moves to its maximum height only under the inßuence of gravity. In the second stage, the explosion takes place, after which the projectile is split into two projectiles. In the third stage, each part propagates to the ground only affected by gravity.

		v0						vA	vB
						A		B
	15
]	10
[M
y
	5
	0	5	10	15	20	25	30	35	40
	0
					x [M ]

Fig. 13.8 A projectile explodes at the top of its path, splitting into two equal pieces. We track the position of each part until they hit the ground

414	13 Multiparticle Systems

Model: First, let us address the explosion. During the explosion, each part of the projectile is subject to large forces. But the only external force acting on the parts is gravity. Consequently, there is no external horizontal force acting on the system. The horizontal momentum is therefore conserved at all times. In particular, the momentum is the same immediately before and immediately after the explosion. We use conservation of momentum to determine the horizontal velocities of the parts after the explosion.

Before the explosion, the horisontal momentum of the system is:

p0 = mv0 ,

(13.38)

and after the collision, the horizontal momentum is:

p1 = m A vA + m B vB =	m		m		(13.39)
		vA +		vB .
	2		2

Conservation of momentum gives:

11

v0 =		vA +		vB ,	(13.40)
	2		2

where vB = v1 is the velocity of part B after the explosion, and vA is the velocity of part A, which we need to Þnd:

vA = 2v0 − vB = 2v0 − v1 .

(13.41)

We therefore know the initial conditions for the motion in the third stage. Each part is affected by gravity alone: G A = −m A g j, GB = −m B g j.

Finding the motion of part B: First, we Þnd the motion of part B. NewtonÕs second law in the x -direction gives:

Fx = 0 = m B aB .

(13.42)

The velocity in the x -direction is constant. The position is therefore:

x B (t ) = x B (t0) + vB t = v0t ,

(13.43)

where we have placed the origin at the ground directly below the explosion. Therefore x B (t0) = 0.

The motion in the y-direction corresponds to the motion of a falling object, hence:

1		gt 2 .	(13.44)
yB (t ) = h −
2

13.3 NewtonÕs Second Law for Particle Systems

415

Finding the motion of part A: Similarly, we Þnd the position of part A:

x A (t ) = vA t = (2v0	1		gt 2 .	(13.45)
	− v1) t , yA (t ) = h −
	2

We notice that the motion in the y-direction is the same for the two parts, which is as expected. The two parts will therefore strike the ground at the same time.

Finding the motion of the center of mass: We use these results to Þnd the center of mass:

R =	i i mi	.	(13.46)
	mi ri

We Þnd the x - and y-components independently:

X =	i	mi	=	m2 + m2	= 2 (x A + x B ) =		2 ((2v0 − v1) t + v1t ) = v0t .
	i mi xi			m2 x A + m2 x B	1		1
							(13.47)

This is what we get if we apply NewtonÕs second law for a particle system. There are no external horizontal forces acting on the system, therefore the horizontal component of the center of mass moves with constant velocity:

Fext = −mg j = mA Ax = 0 X = v0t .

(13.48)

We see that NewtonÕs second law for particle systems gives the same result as when NewtonÕs second law is applied to each object.

Similarly, we Þnd the y-position of the center of mass:

Y =	i mi			=				m2 + m2					2		(yA + yB )						(13.49)
			i mi yi				m2 yA + m2				yB		1
		1		1		2			1			2				1			2
=			h −		gt			+ h −			gt				= h −			gt		,	(13.50)
	2			2						2							2

We could also have found this directly from NewtonÕs second law for particle systems:

Fext = mA = −mg j Ay = −g Y (t ) = Y (t0) −	1		gt 2 = h −	1		gt 2 .
				2
	2
				(13.51)

Analyze: We have demonstrated that we can Þnd the motion of the center of mass either by calculating the motion of each of the parts of the system, or we can Þnd the motion of the center of mass by applying NewtonÕs second law for particle systems directly. The results are of course the same. However, there are many questions