Question

Question 1 In the figure, two particles are launched from the origin of the coordinate system at time t = 0. Particle 1 of mass my - 5.20 g is shot directly along the x axis (on a frictionless floor), where it moves with a constant speed of 10.4 m/s. Particle 2 of mass m2 - 3.80 g is shot with a velocity of magnitude 17.0 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? (a) Number Units (b) Number i + Units (c) Number j Units

Answer 1

Given

m1 = 5.20 g

m2 = 3.80 g

v_2,x = 10.4 m/s

v₂ =17.0 m/s

v² = u² + 2*a*s

where:

v = final velocity, u = initial velicty

a = acceleration, s = distance/height

v²_2,y = u²_2,y + 2*g*y

(a)

at maximum height, final velocity v_2,y = 0 m/s

-u²_2y = -2*g*y_max

particle 2 always stays directly above particle 1

v_{2,x =} v_1,x

u²₂,_y = v²₂ - v²_2,x

u²₂,_y = 17² – 10.4²

u₂,_y = 13.7 m/s

-u²_2y = -2*g*y_max

13.7² = 2*9.81*y_max

y_max = 9.63 m/s

H_max = (m₂*y_max)/(m₁+m₂)

H_max = (3.8*9.63)/(5.2+3.8)

H_max = 4.06 m

(b)

v_com = (m₁v₁ + m₂v₂ ) / (m₁ + m₂)

v_com,y = 0 ; v_com,x = (m₁v_1,x + m₂v_2,x ) / (m₁ + m₂)

v_{2,x =} v_1,x

v_com,x = v_1,x (m₁ + m₂ ) / (m₁ + m₂)

v_com,x = v_1,x

v_com,x = 10.4 m/s i

v_com = 10.4 i + 0 j m/s

(c)

a_com = (m₁a₁ + m₂a₂ ) / (m₁ + m₂)

a_com = (0 + m₂g) / (m₁ + m₂)

a_com = (3.8*9.81) / (3.8+ 5.2)

a_com = - 4.14 m/s/s

-ve sign is because a_com is downward

a_com = 0 i + (-4.14) j m/s²