A 2.0 kg melon is balanced on a circus performer's head. An archer shoots a 50.0 g arrow at the melon with a speed of 30 m/s. The arrow passes through the melon and emerges with a speed of 18 m/s. Find the speed of the melon as it flies off the performer's head.

The law of conservation of momentum states that for a system of colliding bodies, the total momentum before the collision is equal to the total momentum after the collision, provided there is no net external force acting on the bodies.

To find the speed of the melon, we use the formula from the law of conservation of momentum.

Formula:

mu+m'u' = mv+m'v'............ Equation

Where:

m = mass of the melon
u = initial velocity of the melon
m' = mass of the arrow
u' = initial velocity of the arrow
v = final velocity of the melon
v' = final velocity of the arrow.

From the question,

Given:

m = 2 kg
m' = 50 g = 0.05 kg
u = 0 m/s (at rest)
u' = 30 m/s
v' = 18 m/s.

Substitute these values into equation 1

(2×0)+(0.05×30) = (0.05×18)+(2×v)
0+1.5 = 0.9+2v
2v = 1.5-0.9
2v = 0.6
v = 0.3 m/s.

Hence, The speed of the melon as it flies off the performer's head is 0.3 m/s.

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