8. A 45.0 kg student sits at rest on a new class rolling chair. A friend tosses the student a 5.0 kg backpack at 4m/s to the left. The student freaks out and bats the backpack away moving the student backwards across the class at a speed of 0.5m/s to the left. What is the backpack's final velocity? Type square

Question

8. A 45.0 kg student sits at rest on a new class rolling chair. A friend tosses the student a 5.0 kg backpack at 4m/s to the left. The student freaks out and bats the backpack away moving the student backwards across the class at a speed of 0.5m/s to the left. What is the backpack's final velocity? Type square

Answer 1

To solve this problem, we can use the principle of conservation of momentum. The total momentum of the system before the backpack is tossed must be equal to the total momentum of the system after the backpack is tossed.Let's denote the following variables:- m1 = mass of the student = 45.0 kg- m2 = mass of the backpack = 5.0 kg- v1i = initial velocity of the student = 0 m/s (at rest)- v1f = final velocity of the student = -0.5 m/s (to the left)- v2i = initial velocity of the backpack = 4 m/s (to the left)- v2f = final velocity of the backpack (unknown)According to the conservation of momentum:(m1 * v1i) + (m2 * v2i) = (m1 * v1f) + (m2 * v2f)Substituting the given values:(45.0 kg * 0 m/s) + (5.0 kg * 4 m/s) = (45.0 kg * -0.5 m/s) + (5.0 kg * v2f)Simplifying the equation:0 + 20 kg*m/s = -22.5 kg*m/s + 5.0 kg * v2fSolving for v2f:20 kg*m/s + 22.5 kg*m/s = 5.0 kg * v2f42.5 kg*m/s = 5.0 kg * v2fv2f = 42.5 kg*m/s / 5.0 kgv2f = 8.5 m/sTherefore, the backpack's final velocity is 8.5 m/s to the right.

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