In the figure, two particles are launched from the origin of the coordinate syst

Question

In the figure, two particles are launched from the origin of the coordinate system at time t = 0. Particle 1 of mass m_1 = 5.90 g is shot directly along the x axis (on a frictionless floor), where it moves with a constant speed of 12.0 m/s. Particle 2 of mass m_2 = 2.70 g is shot with a velocity of magnitude 18.9 m/s, at an upward angle such that it always stays directly above particle 1 during its flight. (a) What is the maximum height H_max reached by the com of the two-particle system? In unit-vector notation, what are the (b) velocity and (c) acceleration of the com when the com reaches H_max

Explanation / Answer

m1 = 5.9 g, m2 = 2.7 g, v1 = 12 m/s, v2 = 18.9 m/s

If the particle 2 remains overhead of particle 1 at all instances then the horizontal velocity of both the particles must be same.

v2cos= v1

18.9cos =12

= 50.6 degrees

Now initially both particles are projected from origin, so initially com is also at origin.

The net horizontal velocity of com is, (and it remains constant at all instances)

(vcom)x = m1v1x+m2v2x/(m1+m2)

= (5.9*12 +0)/(5.9+2.7) = 8.23 m/s

(vcom)y = (5.9*0 +2.7*18.9)/(5.9+2.7)

= 5.93 m/s

Now there is no horizontal force acting on any of the particle of the system so net horizontal force (and acceleration )is zero.However a particle 2 has vertical acceleration (but particle 1 has none)

(acom)x = 0

So like vertical velocity the vertical acceleration(pointing downwards) of the com will be (it will be constant)

(acom)y = (5.9*0 +2.7*9.8)/(5.9+2.7)

(acom)y = 3.08 m/s^2

so maximum height attained by com will be

(Hcom)max = (vcom)y^2/2*acomy

= 5.93^2/(2*3.08)

= 5.71 m

The vertical velocity of com becomes zero at top point.

vcom = 8.23 i + 0 j m/s

acom = 0 i -3.08j m/s^2

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.