In Figure 9-45, two particles are launched from the origin of the coordinate sys
ID: 1341309 • Letter: I
Question
In Figure 9-45, two particles are launched from the origin of the coordinate system at time t = 0. Particle 1 of mass m is shot directly along the x axis (on a frictionless floor), where it moves with a constant speed v. Particle 2 of mass 3m is shot with a velocity of magnitude 2v at an upward angle such that it always stays directly above particle 1 during its flight. (a) What is the maximum height Hmax reached by the com of the two-particle system? What are the (b) x and (c) y components of the velocity as well as the (d) x and (e) ycomponents of the acceleration of the com when the com reaches Hmax? Express your answer in terms of the variables given and g
9 FIG. 9-45 Problem 14.Explanation / Answer
use this example to solve this problem
two particles are launched from the origin of the coordinate system at time t = 0. Particle 1 of mass m1 = 3.90 g is shot directly along the x axis (on a frictionless floor), where it moves with a constant speed of 12.8 m/s. Particle 2 of mass m2 = 4.30 g is shot with a velocity of magnitude 17.1 m/s, at an upward angle such that it always stays directly above particle 1 during its flight. (a) What is the maximum height Hmax 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 Hmax
As particle 2 stays directly above particle 1, it has the same horizontal velocity.
Let x1, x2, be horizontal co-ordinates at time t for the two masses, and y1, y2 be vertical co-ordinates, a be the angle above the horizontal at which mass 2 is fired. Let x3, y3 be the co-ordinates of the centre of mass.
x1 = 12.8t
y1 = 0
x2 = 17.1t cos(a)
y2 = 17.1t sin(a) - 9.81t^2 / 2
Equating x1 and x2:
cos(a) = 12.8 / 17
a = 41.15 deg.
x3 = 12.8t
y3 = (3.90y1 + 4.30y2) / (3.90 + 4.30)
= 4.30 * (17.1t sin(41.15) - 9.81t^2 / 2) / 8.20
= 5.901t - 2.572t^2 ...(1)
y3 is maximum when:
5.901 - 5.144t = 0
t = 1.147 sec.
(a)
Substituting t = 1.147 sec. in (1), the maximum height is:
3.38m.
(b)
The vertical velocity is 0, and therefore the total velocity is:
12.8 * 1.147 = 14.7 m/s horizontally.
(c)
The acceleration is constant at:
(9.81 * 4.30) / 8.2
= 5.14 m/s^2 downwards
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.