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)?.
Additional: I know its a conservation of momentum problem, Newtons 3rd law, but i don't know how to solve it or what equation to use.
To solve this problem, we can use the principle of conservation of momentum. According to Newton's third law, the total momentum before the collision is equal to the total momentum after the collision.
Let's first convert the given velocities to meters per second (m/s) for consistent units.
The car is traveling at a speed of 44 miles per hour (mi/h). To convert to meters per second, we multiply by a conversion factor of 0.44704 (1 mile = 1.60934 kilometers = 1609.34 meters; 1 hour = 3600 seconds):
Car velocity = 44 mi/h * 0.44704 m/s per mi/h = 19.58736 m/s.
The bug is traveling in the opposite direction, and we want to find its velocity. We'll call it v_bug.
The initial momentum is the sum of the momentum of the car and the momentum of the bug, which is given by:
Initial momentum = (mass of car x velocity of car) + (mass of bug x velocity of bug)
Final momentum is the momentum of the car after being slowed down by the bug, which is given by:
Final momentum = (mass of car x final velocity of car)
Since the collision is one-dimensional (the bug's motion is opposite to the car's), we can set up the equation as:
Initial momentum = Final momentum
(mass of car x velocity of car) + (mass of bug x velocity of bug) = (mass of car x final velocity of car)
Substituting the given values:
(880 kg x 19.58736 m/s) + (7.9 mg x v_bug) = (880 kg x 19.58736 m/s - 1 mi/h x 0.44704 m/s per mi/h)
Notice that we converted the mass of the bug to kilograms by dividing it by 1000 since 1 mg = 0.001 kg.
We can now solve for v_bug:
(880 kg x 19.58736 m/s) + (7.9 x 10^(-6) kg x v_bug) = (880 kg x 19.58736 m/s - 1 x 0.44704 m/s)
Simplifying this equation will give us the velocity of the bug.