<h3>Explanation: M1(mass of the train) = 12000kg</h3><h3>U1(initial velocity of the train) = 2.2m/s</h3><h3>M2(mass of boxcars) = 25000kg</h3><h3>U2(initial velocity of the boxcars which is at rest) = 0</h3><h3>V = common velocity of train and boxcars since the collision was inelastics.</h3><h3>M1U1+M2U2 = (M1+M2)V</h3><h3>12000kg×2.2m/s+25000kg×0m/s = (12000kg+25000kg)V</h3><h3>⇒26400 + 0 = 37000V</h3><h3>V = 0.7135m/s</h3><h3 />
Full solution calculation can be found in the attachment below.
From the principle of conservation of linear momentum, the sum of momentum before collision equals the sum of momentum after collision.
Before collision only the train had momentum. After the collision the train and the boxcars stick together and move as one body. The initial momentum of the train is now shared with the boxcars as they move together as one body. The both move with a common velocity v.
See the attachment below for the solution calculation.
The butterfly takes a vertical perpendicular path equivalent to 9m and travels a horizontal distance of 17m. The net path between the two is equivalent to that of the hypotenuse, so we will apply the Pythagorean theorem.
Therefore the magnitude of the butterfly's displacement is 9m