Jake has a mass of 88.0 kg and is zooming along in a 110.0-kg bumper car at 9.4 m/s. He bumped Allen's car, which is sitting at rest. Allen has a mass of 88.0 kg. After the elastic collision, Jake continues ahead with a speed of 3.2 m/s. How fast is Allen's car bumped across the floor?
To solve this problem, we can use the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision.
The momentum (p) of an object is calculated by multiplying its mass (m) by its velocity (v). Mathematically, momentum can be represented as p = mv.
Let's denote Jake's initial velocity as v1 (before the collision) and his final velocity as v1' (after the collision). Similarly, Allen's initial velocity is 0 m/s since his car is at rest, and his final velocity is v2 (after the collision).
Using the law of conservation of momentum, we can write the equation:
(Mass of Jake × Jake's initial velocity) + (Mass of Allen × Allen's initial velocity) = (Mass of Jake × Jake's final velocity) + (Mass of Allen × Allen's final velocity)
(88.0 kg × v1) + (110.0 kg × 0 m/s) = (88.0 kg × 3.2 m/s) + (88.0 kg × v2)
Simplifying the equation:
88.0 kg × v1 = 88.0 kg × 3.2 m/s + 88.0 kg × v2
Since the mass of Jake cancels out on both sides:
v1 = 3.2 m/s + v2
Now, we have two equations:
(1) v1 = 3.2 m/s + v2 (from above)
(2) (88.0 kg × v1) + (110.0 kg × 0 m/s) = (88.0 kg × 3.2 m/s) + (88.0 kg × v2
We can solve this system of equations to find the value of v2, which is Allen's final velocity.
First, substitute the value of v1 from equation (1) into equation (2):
(88.0 kg × (3.2 m/s + v2)) + (110.0 kg × 0 m/s) = (88.0 kg × 3.2 m/s) + (88.0 kg × v2)
Simplifying:
(88.0 kg × 3.2 m/s) + (88.0 kg × v2) + (110.0 kg × 0 m/s) = (88.0 kg × 3.2 m/s) + (88.0 kg × v2)
Now, cancel out the common terms:
(88.0 kg × v2) = (88.0 kg × 3.2 m/s)
Finally, divide both sides of the equation by 88.0 kg:
v2 = 3.2 m/s
Therefore, Allen's car is bumped across the floor at a speed of 3.2 m/s.