In Figure 9-43, two particles are launched from the origin of the coordinate sys

Question

In Figure 9-43, two particles are launched from the origin of the coordinate system at time t = 0. Particle 1 of mass m1 = 6.40 g is shot directly along the x axis (on a frictionless floor), where it moves with a constant speed of 10.2 m/s. Particle 2 of mass m2 = 3.80 g is shot with a velocity of magnitude 25.0 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?

Explanation / Answer

since m2 always stays directly above particle 1 during its flight,means x component of velocity of m2 will be equal to velocity of m1.Let be the angle with horizontal

25 cos = 10.2

cos=10.2 /25 =

= 66degree with the horizontal

velocity of m2 in y component = 25 sin=22.83

max height of m2 = vy2 / 2g = 26.06

max height of com= m2*y2+m1*y1 / (m1+m2)

= 3.8*26.06 + m1*0/(3.8+6.4)

= 9.708 m

acceleration of m2 in x component will be 0 and it will have y component = g =10

acceleration of com in x component = 0

acceleration of com in y component = m1*a1 +m2*a2/(m1+m2) = 6.4*0+3.80*10 /(6.4+3.80) = 3.72 downward

hence acceleration of com = -3.72j

velocity of com in x component = m1v1 + m2v2 / (m1+m2) velocities are x component of both particle

= 6.4*10.2+3.8*10.2/(6.4+3.8) = 10.2

velocity of com in y component will be 0 as both the particle have 0 velocity in y component.

hence the velocity of com = 10.2 i

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