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 m1 = 3.80 g is shot directly along the x axis (on a frictionless floor), where it moves with a constant speed of 11.1 m/s. Particle 2 of mass m2 = 2.50 g is shot with a velocity of magnitude 22.7 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?

** Part b and c are in vector componet form. They must be calculated as i and j seperated. **

Explanation / Answer

a)

Center of mass : COM = ( m1*r1 + m2*r2 )/(m1 + m2)

Now, r1 = (11.1*t) i

and , r2 = (11.1 i + y j )*t + 0.5*(-9.8j)*t^2

y = sqrt(22.7^2 - 11.1^2) = 19.8 m/s

= (11.1*t) i + (y*t - 4.9*t^2) j

So, COM = ( 3.8*(11.1*t ) i + 2.5*( (11.1*t) i + (yt - 4.9*t^2) j) ) / (3.8 + 2.5)

= ( (69.9*t) i + 2.5*(yt - 4.9*t^2) j ) / (6.3)

= (11.1*t) i + (0.397*(yt - 4.9*t^2)) j

So, height heached , h = 0.397*(yt - 4.9*t^2)

= 0.397*(19.8*t - 4.9*t^2)

Now, for maximum height , dh/dt = 0

So, 19.8 - 9.8*t = 0

So, t = 2.02 s

So, Hmax = 0.397*(19.8*2.02 - 4.9*2.02^2) = 7.94 m

b)

velocity , v = d(COM)/dt = 11.1 i + (0.397*(19.8 - 9.8*2.02)) j

= 11.1 i + 0 j

acceleration , a = dv/dt = 0 i + 0.397*(-9.8 j)

= 0 i - 3.89 j

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