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Answer: a) 0.78 m/s b) 1.57 m/s
Explanation:
M = father's mass
m = son's mass = M/3
V = father's initial speed
v = son's initial speed
(1/2)MV^2 = (1/2)*(1/2)*m v^2
M*V^2 = (1/2)(M/3)v^2
V^2/v^2 = 1/4
V = v/2
Second equation:
(1/2)M*(V + 1.4)^2 = (1/2)m*v^2
= (1/2)*(M/3)*(3V)^2
cancel out the M's and (1/2)'s
(V + 1.4)^2 = 3V^2
V^2 + 2.8V + 1.96 = 3V^2
V^2 -1.4V -0.98 = 0
V^2 = 0.98/0.4 = 2.45
V = 1.57
Answer:
Explanation:
There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.
momentum of first piece = .320 kg x 2 m/s
= 0.64 kg m/s along x -axis.
momentum of second piece = .355 kg x 1.5 m/s
= 0.5325 kg m/s along y- axis .
Let the velocity of third piece be v and it is making angle of θ with x -axis .
Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ
vertical component of its velocity = .100 kg x v sinθ = .1 v sinθ
For making total momentum in the plane zero
.1 v cosθ = 0.64 kg m/s
.1 v sinθ = 0.5325 kg m/s
Dividing
Tanθ = .5325 / .64 = .83
θ = 40⁰.
The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.
