A 25.0-gram bullet enters a 2.25-kg watermelon with a speed of 220 m/s and exits the opposite side with a speed of 110 m/s. If the melon was originally at rest, then what speed will it have as the bullet leaves its opposite side?

3.67 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

mu+m'u' = mv+m'v' Equation 1

Where m = mass of the bullet, m' = mass of the watermelon, u = initial velocity of the bullet, u' = initial velocity of the watermelon, v =  final velocity of the bullet, v' = final velocity of the watermelon.

make v' the subject of the equation,

v' = (mu+m'u'-mv)/m' Equation 2

Given: m = 25 g = 0.025 kg, u = 220 m/s, m' = 2.25 kg, u' = 0 m/s ( at rest), v = 110 m/s.

Substitute into equation 2

v' =[ (0.025×220) +(2.25×0)+(0.025×110)]/2.25

v' = 3.67 m/s.

Hence the velocity of the watermelon = 3.67 m/s

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step-by-step explanation: