Two balls have masses 44 kg and 86 kg. The
44 kg ball has an initial velocity of 80 m/s (to
the right) along a line joining the two balls
and the 86 kg ball is at rest. The 86 kg
ball has in initial velocity of −24 m/s. The
two balls make a head-on elastic collision with
each other.
44 kg
80 m/s
86 kg
−24 m/s
What is the final velocity of the 44 kg ball?
To find the final velocity of the 44 kg ball after the head-on elastic collision, we can use the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision if no external forces are acting on the system. Mathematically, this can be expressed as:
Total initial momentum = Total final momentum
In this case, the total initial momentum is the sum of the individual momenta of the two balls, and the total final momentum is also the sum of the individual momenta of the two balls after they collide.
Let's assign variables to the quantities involved:
m1 = mass of the 44 kg ball
v1i = initial velocity of the 44 kg ball
m2 = mass of the 86 kg ball
v2i = initial velocity of the 86 kg ball
Using the principle of conservation of momentum, we can write the equation for the total initial momentum:
m1 * v1i + m2 * v2i = total initial momentum
Substituting the given values:
m1 = 44 kg
v1i = 80 m/s
m2 = 86 kg
v2i = -24 m/s
44 kg * 80 m/s + 86 kg * (-24 m/s) = total initial momentum
Simplifying the equation:
3520 kg*m/s - 2064 kg*m/s = total initial momentum
1456 kg*m/s = total initial momentum
Now, let's determine the total final momentum. Since the collision is elastic, the total final momentum remains the same as the total initial momentum. Therefore, the total final momentum is also 1456 kg*m/s.
Let's assign a variable to the final velocity of the 44 kg ball after the collision:
v1f = final velocity of the 44 kg ball
The momentum of the 44 kg ball after the collision is given by:
m1 * v1f
And the momentum of the 86 kg ball after the collision is given by:
m2 * v2f, where v2f is the final velocity of the 86 kg ball
Using the equation for the total final momentum, we can write:
m1 * v1f + m2 * v2f = total final momentum
Substituting the given values and the calculated total final momentum:
44 kg * v1f + 86 kg * v2f = 1456 kg*m/s
Now, we have one equation with two unknowns (v1f and v2f). However, since the collision is head-on and elastic, the relative velocity between the two balls before and after the collision remains the same. In other words:
v1i - v2i = v1f - v2f
Substituting the given values:
80 m/s - (-24 m/s) = v1f - v2f
80 m/s + 24 m/s = v1f - v2f
104 m/s = v1f - v2f
Now, we have two equations with two unknowns. We can solve these equations simultaneously to find the final velocities of both balls.
44 kg * v1f + 86 kg * v2f = 1456 kg*m/s
v1f - v2f = 104 m/s
By manipulating these equations algebraically, we can solve for v1f:
From the second equation, we have:
v1f = v2f + 104 m/s
Substituting this expression for v1f into the first equation:
44 kg * (v2f + 104 m/s) + 86 kg * v2f = 1456 kg*m/s
44 kg * v2f + 4576 kg*m/s + 86 kg * v2f = 1456 kg*m/s
Combining like terms:
130 kg * v2f + 4576 kg*m/s = 1456 kg*m/s
130 kg * v2f = -3120 kg*m/s
Dividing both sides by 130 kg:
v2f = -24 m/s
Now, substituting this value of v2f back into the expression for v1f:
v1f = v2f + 104 m/s
v1f = -24 m/s + 104 m/s
v1f = 80 m/s
Therefore, the final velocity of the 44 kg ball after the head-on elastic collision is 80 m/s.
You know the momentum is the same before and after
and
you know the kinetic energy is the same before and after
u = final velocity of 44 kg ball
v = final velocity of 86 kg ball
44*80 + 86* -24 = 44 u + 86 v
(1/2)44(80^2)+(1/2)86(24^2) =(1/2)44u^2+(1/2)86v^2
solve for u
Since they have not told you the final velocity of at least one ball, you are going to have to use conservation of momentum AND kinetic energy to come up with answers. (Conservation of kinetic energy may not be a good assumption for some kinds of balls that do not bounce well, such as squash balls).
The math gets rather tedious, and I will leave it up to you. You can probably find the solution formula online.
A convenient way to solve this type of problem is to use a coordinate system that moves with the center of mass. In that system, each ball will simply reverse direction and keep the same speed. Then transform speeds back to lab coordinates.