A high-flying hawk of mass mh = 1.2 kg spots a sparrow of mass m2 = 0.400 kg fly

Question

A high-flying hawk of mass mh = 1.2 kg spots a sparrow of mass m2 = 0.400 kg flying low to the ground with a horizontal velocity vs = 6.0 m/s in the + x direction. The hawk goes into a downward vertical dive, attaining a velocity vh = 20.0 m/s in the ? y direction before grabbing the sparrow out of the air.

(a) What is the x-component vx’ of the velocity of the two-bird system immediately after the collision? Don’t forget the sign! (b) What is the y-component vy’ of the velocity of the two-bird system immediately after the collision? Don’t forget the sign! (c) What is the speed v’ of the two-bird system immediately after the collision?

Explanation / Answer

Here ,

mass of hawk , mh = 1.2 Kg

m2 = 0.4 Kg

vs = 6 i m/s

vh = -20 j m/s

a) Using conservation of momentum is x direction

Vx' * (m2 + m1) = vs * m2

Vx' * (1.2 + 0.4) = 6 * 0.40

solving Vx'

Vx' = 1.5 m/s

x-component vx’ of the velocity of the two-bird system immediately after the collision is 1.5 m/s

b)

Using conservation of momentum in y direction

Vy' * (m1 + m2) = m2 * vh

Vy' * (1.2 + 0.4) = -20 j * 1.2

Vy' = -15 j m/s

he y-component vy’ of the velocity of the two-bird system immediately after the collision is -15 m/s

c)

speed , v' = sqrt(15^2 + 1.5^2)

v' = 15.1 m/s

the speed v' of the two bird system immediately after the collision is 15.1 m/s

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