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A ball of mass 143 g is thrown with an initial velocity of 25 m/s at an angle of 30° with the horizontal direction. Ignore air resistance. What is the momentum of the ball after 0.6 s? (Do this problem by finding the components of the momentum first and then constructing the magnitude and direction of the momentum vector from the components. Find the magnitude in kg · m/s and the direction in degrees counterclockwise from the +x-axis. Assume the +x-axis points horizontally to the right.)

Respuesta :

Answer:

P=1.53 i +1.92 j kg.m/s

P=2.45  kg.m/s

α = 51.34

Explanation:

Given that

m=123 g = 0.123 Kg

U= 25 m/s

θ=30°

t= 0.6 s

This is the case of projectile motion

So the horizontal component of velocity U =  u cosθ

u = 25 cosθ

u = 25 cos 30°

u=21.65 m/s

The vertical component of velocity U = U sinθ

Vo= U sinθ

Vo= 25 sin 30°

Vo = 12.5 m/s

We know that horizontal component of velocity of ball will remain same.So the horizontal component of momentum

Px= m u

Px= 0.143 x 12.5 kg.m/s

Px=1.53 kg.m/s

The vertical component of ball after 0.6 s

V= Vo- g t

V= 21.65 - 10 x 0.6 m/s

V= 15.65 m/s

Py= m V

Py= 0.123 x 15.65 kg.m/s

Py=1.92 kg.m/s

P=1.53 i +1.92 j kg.m/s

Magnitude P

[tex]P=\sqrt{1.53^2+1.92^2}\ kg.m/s[/tex]

P=2.45  kg.m/s

Direction

[tex]tan\alpha =\dfrac{P_y}{P_x}[/tex]

α = 51.34° (measured from x direction)

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