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Neporo4naja [7]
3 years ago
9

A projectile of mass 1.800 kg approaches a stationary target body at 4.800 m/s. The projectile is deflected through an angle of

60.00° and its speed after the collision is 2.900 m/s. What is the magnitude of the momentum of the target body after the collision?
Physics
1 answer:
Artyom0805 [142]3 years ago
5 0

Answer:

P = 5.22 Kg.m/s

Explanation:

given,

mass of the projectile = 1.8 Kg

speed of the target = 4.8 m/s

angle of deflection = 60°

Speed after collision = 2.9 m/s

magnitude of momentum after collision = ?

initial momentum of the body = m x v

                                                  = 1.8 x 4.8 = 8.64 kg.m/s

final momentum after collision

momentum along x-direction

P_x = m v cos θ

P_x = 1.8 x 2.9 x  cos 60°

P_x = 2.61 kg.m/s

momentum along y-direction

P_y = m v sin θ

P_y = 1.8 x 2.9 x  sin 60°

P_y = 4.52 kg.m/s

net momentum of the body

P = \sqrt{P_x^2 + P_y^2}

P = \sqrt{4.52^2 + 2.61^2}

 P = 5.22 Kg.m/s

momentum magnitude after collision is equal to P = 5.22 Kg.m/s

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A 2.55 kg block starts from rest on a rough inclined plane that makes an angle of 60 degree with the horizontal. the coefficient
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Explanation:

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