A 220 g cart moves on an air track (a one-dimensional system in which motion is friction free) at 1.3 m/s. It collides with and sticks to another cart of mass 550 g, which was moving in the opposite direction at 0.80 m/s before the collision. What is the velocity of the composite cart after the collision? (Answer: 0.2m/s Direction)
Answer is provided above. Please show all work on how to get the answer.
To find the velocity of the composite cart after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
The momentum (p) of an object can be calculated by multiplying its mass (m) by its velocity (v). Mathematically, the momentum is given as p = m * v.
Let's denote the mass and velocity of the first cart (220 g) as m1 and v1, respectively. Similarly, let the mass and velocity of the second cart (550 g) be denoted as m2 and v2, respectively. After the collision, the carts stick together and move as a single composite cart. Let's denote the mass and velocity of the composite cart as m and v.
Using the principle of conservation of momentum, we can write the equation as:
(m1 * v1) + (m2 * v2) = (m * v)
Now, let's substitute the given values into the equation:
(m1 * v1) + (m2 * v2) = (m * v)
((220 g) * (1.3 m/s)) + ((550 g) * (-0.8 m/s)) = ((220 g + 550 g) * v)
Simplify the equation:
(286 g * m/s) + (-440 g * m/s) = (770 g * v)
-154 g * m/s = 770 g * v
Now, isolate the velocity (v) by dividing both sides of the equation by 770 g:
v = (-154 g * m/s) / 770 g
Simplify the equation:
v = -0.2 m/s
Since velocity has both magnitude and direction, the answer should be v = -0.2 m/s. The negative sign indicates that the composite cart is moving in the opposite direction compared to the initial motion of the two carts.
Therefore, the velocity of the composite cart after the collision is 0.2 m/s in the opposite direction.