A 2-kg ball of putty moving to the right has a head-on
inelastic collision with a 1-kg putty ball moving to the left.
If the combined blob doesn’t move just after the collision,
what can we conclude about the relative speeds of the
balls before they collided?
Conclusion: m1*V1 = m2*V2.
2*V1 = 1*V2
V2 = 2V1
To solve this problem, we can use the principle of conservation of momentum.
The momentum before the collision is given by the sum of the individual momenta of the two balls. Let's denote the velocity of the 2-kg ball as v1 and the velocity of the 1-kg ball as v2. Since the 2-kg ball is moving to the right and the 1-kg ball is moving to the left, we can write the momenta as:
Momentum of the 2-kg ball = (mass of the 2-kg ball) * (velocity of the 2-kg ball) = 2 kg * v1
Momentum of the 1-kg ball = (mass of the 1-kg ball) * (velocity of the 1-kg ball) = -1 kg * v2
Here, we take the velocity of the 1-kg ball as negative since it is moving in the opposite direction.
Now, according to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces are acting on the system:
Momentum before collision = Momentum after collision
So, the sum of the individual momenta after the collision is zero:
2 kg * v1 + (-1 kg * v2) = 0
Simplifying this equation, we have:
2v1 - v2 = 0
Since the combined blob doesn't move just after the collision, we can conclude that the relative speeds of the balls before the collision were such that the magnitude of the velocity of the 2-kg ball was twice the magnitude of the velocity of the 1-kg ball. In other words, the 2-kg ball was moving twice as fast as the 1-kg ball, but in the opposite direction.
To determine the relative speeds of the balls before the collision, we need to analyze the conservation of momentum during the collision. The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision in an isolated system.
In this case, we have a head-on inelastic collision. In an inelastic collision, the objects stick together or deform upon impact, and some kinetic energy is lost. Since the combined blob doesn't move just after the collision, we can conclude that the total momentum before the collision is zero.
To find the relative speeds of the balls before the collision, we first need to calculate their individual momenta. The momentum (p) of an object is given by the product of its mass (m) and velocity (v), so p = m * v.
Let's assume that the positive direction is to the right and the negative direction is to the left. We can assign the following values:
Mass of the first ball (m1) = 2 kg
Mass of the second ball (m2) = 1 kg
Since the combined blob doesn't move just after the collision, the total momentum before the collision is zero. Hence:
Total momentum before the collision = (m1 * v1) + (m2 * v2) = 0
Since m2 is negative (moving to the left), we can rewrite the equation as:
(2 kg * v1) + (1 kg * (-v2)) = 0
Simplifying the equation, we get:
2v1 - v2 = 0
Therefore, we can conclude that the relative speed of the first ball (v1) is equal to the negative of the relative speed of the second ball (v2).