Ball A has a mass of 8kg. Ball B has twice the mass of ball A. The two balls collide and stick together in a perfectly inelastic collision. After the collision the combined balls are at rest. If the velocity of ball A before the collision was 18m/s, what was the velocity of ball B before the collision?
To solve this problem, we can use the principle of conservation of momentum. In an isolated system, the total momentum before an event (such as a collision) is equal to the total momentum after the event.
In this case, we can assume that ball A and ball B are the only objects involved in the collision, and the system is isolated.
The equation for conservation of momentum is:
(mass of ball A)(velocity of ball A before collision) + (mass of ball B)(velocity of ball B before collision) = (mass of combined balls)(velocity of combined balls after collision)
Let's assign variables to the given values:
Mass of ball A = 8 kg
Mass of ball B = 2 * (mass of ball A) = 2 * 8 kg = 16 kg
Velocity of ball A before collision = 18 m/s
Velocity of combined balls after collision = 0 m/s (since the combined balls are at rest)
Plugging these values into the equation, we get:
(8 kg)(18 m/s) + (16 kg)(velocity of ball B before collision) = (24 kg)(0 m/s)
Simplifying the equation:
144 kg m/s + 16 kg (velocity of ball B before collision) = 0 kg m/s
Subtracting 144 kg m/s from both sides:
16 kg (velocity of ball B before collision) = -144 kg m/s
Dividing both sides by 16 kg:
velocity of ball B before collision = -144 kg m/s / 16 kg
Calculating:
velocity of ball B before collision = -9 m/s
Therefore, the velocity of ball B before the collision was -9 m/s. Note the negative sign indicates the direction of motion, which means that ball B was moving in the opposite direction of ball A.