A 0.075-kilogram dart strikes a 2-kilogram stationary block of wood. If the dart sticks in the wood and the dart and wood fly off together at 1.5 meters per second, what was the approximate original speed of the dart?
.075 v = (.075+2.000)1.5
so
v = 3/.075 = 40 m/s approximately
To find the approximate original speed of the dart, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces act on the system.
The momentum of an object is the product of its mass and velocity. In this case, we have a dart with a mass of 0.075 kilograms and an unknown initial velocity, colliding with a stationary block of wood with a mass of 2 kilograms. After the collision, the dart sticks in the wood, and they fly off together with a velocity of 1.5 meters per second.
Using the conservation of momentum formula:
(mass of dart * initial velocity of dart before collision) + (mass of wood * initial velocity of wood before collision) = (total mass * final velocity after collision)
Plugging in the given values:
(0.075 kg * initial velocity of dart) + (2 kg * 0 m/s) = (2.075 kg * 1.5 m/s)
Now we can solve for the initial velocity of the dart:
0.075 kg * initial velocity of dart = (2.075 kg * 1.5 m/s)
initial velocity of dart = (2.075 kg * 1.5 m/s) / 0.075 kg
initial velocity of dart ≈ 41 m/s
Therefore, the approximate original speed of the dart was 41 meters per second.