Answer
given,
mass of the dog = 29.6 Kg
speed in northward = 2.90 m/s
mass of cat = 7.70 Kg
speed of the cat in eastward = 3.76 m/s
Mass of owner = 76.7 Kg
Momentum of dog
P_y = m₁ x u₁
P_y = 29.6 x 2.9
P_y = 85.84 Kg.m/s
P_x = m₂ x u₂
P_x = 7.7 x 3.76
P_x =28.952 Kg.m/s
magnitude of momentum
[tex]P = \sqrt{P_x^2+P_y^2}[/tex]
[tex]P = \sqrt{28.952^2+85.84^2}[/tex]
P = 90.59 Kg.m/s
[tex]\theta = tan^{-1}(\dfrac{P_y}{P_x})[/tex]
[tex]\theta =tan^{-1}(\dfrac{85.84}{28.952})[/tex]
[tex]\theta =71.36^0[/tex]
Answer:
u= 1.17 m/s
θ = 71.21 °
Explanation:
Given that
m₁=29.6 kg ,u₁=2.9 j m/s
m₂ = 7.7 kg ,u₂=3.76 i m/s
m= 76.7 kg
The speed of the mass 76.7 kg = u m/s
The linear momentum
m₁ u₁+m₂u₂ = m u
29.6 x 2.9 j + 7.7 x 3.76 i = 76.7 u
76.7 u = 28.95 i +85.84 j
u =0.377 i +1.11 j
The magnitude of the velocity u
[tex]u=\sqrt{0.377^2+1.11^2}\ m/s[/tex]
u= 1.17 m/s
The direction θ ( angle from eastward direction)
[tex]tan \theta= \dfrac{1.11}{0.377}[/tex]
θ = 71.21 °
