A car (mass 880 kg) is traveling with a speed of 44 mi/h. A bug (mass 7.9 mg) is traveling in the opposite direction. What speed would the bug need in order to slow down the truck by one mile per hour in a big bug splash (i.e., final speed 43 mi/h)? I don't know how to approach the problem.
To solve this problem, we can use the concept of conservation of momentum.
Conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision, assuming no external forces act on the system.
The formula for momentum is given by:
Momentum = mass * velocity
Since the car and the bug are traveling in opposite directions, we'll consider their velocities as negative quantities.
Given:
Mass of the car (m1) = 880 kg
Velocity of the car (v1) = -44 mi/h
Mass of the bug (m2) = 7.9 mg = 7.9 x 10^(-6) kg
Final velocity of the car after the collision (v1f) = -43 mi/h
Final velocity of the bug after the collision (v2f) = ?
Initial momentum = final momentum
Using the formula for momentum, we can write the equation as:
(m1 * v1) + (m2 * v2) = (m1 * v1f) + (m2 * v2f)
Substituting the given values and rearranging the equation to solve for v2f, we get:
(880 kg * -44 mi/h) + (7.9 x 10^(-6) kg * v2) = (880 kg * -43 mi/h) + (7.9 x 10^(-6) kg * v2f)
Now, we can solve for v2f by rearranging the equation:
v2f = [(880 kg * -44 mi/h) + (7.9 x 10^(-6) kg * v2) - (880 kg * -43 mi/h)] / (7.9 x 10^(-6) kg)
Simplifying the equation, we get:
v2f ≈ [(880 kg * -44 mi/h) - (880 kg * -43 mi/h)] / (7.9 x 10^(-6) kg) + v2
v2f ≈ [(-880 * 44 + 880 * 43) / (7.9 x 10^(-6) kg)] + v2
Now, we can substitute the values and calculate the equivalent speed of the bug (v2f).
Please note that the formula used assumes an elastic collision, where kinetic energy is conserved.