Two people are standing on a 2.2-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of air, like a hovercraft. One person throws a 6.1-kg ball to the other, who catches it. The ball travels nearly horizontally. Excluding the ball, the total mass of the platform and people is 124 kg. Because of the throw, this 124-kg mass recoils. How far does it move before coming to rest again?
Because there are no external horizontal forces, the center of mass of board + people + ball remains in the same place. Since the ball moves from one end to the other, the board and people on it must recoil in the opposite direction. Motion stops when the ball is caught.
Use those facts to calculate the motion.
To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the throw is equal to the total momentum after the throw.
Let's consider the initial momentum before the throw. The platform and people have a total mass of 124 kg and are initially at rest, so the initial momentum is zero (since the velocity is zero).
After the throw, the ball is in motion, and the platform and people will recoil in the opposite direction to conserve momentum.
Let's denote the mass of the person who throws the ball as m1, the mass of the person who catches it as m2, and the mass of the ball as m3.
The momentum of the ball just before it is caught is given by the equation P3 = m3 * v3, where v3 is the velocity of the ball. Since the ball is thrown horizontally, its vertical velocity is zero, so v3 only has a horizontal component.
According to the principle of conservation of momentum, the total momentum before the throw is equal to the total momentum after the throw. So we have:
0 = -(m1 + m2) * v + m3 * v3
Where v is the velocity at which the platform and people recoil, and it has the opposite direction to v3.
Now, we can solve for v:
v = (m3 * v3) / (m1 + m2)
Given that m1 = m2 = 124 kg (the total mass of the platform and people), and m3 = 6.1 kg (the mass of the ball), we can substitute these values into the equation:
v = (6.1 kg * v3) / (124 kg + 124 kg)
Simplifying the equation further:
v = (6.1 kg * v3) / 248 kg
Now, to calculate the distance the platform and people move before coming to rest, we can use the equation for uniform acceleration:
v^2 = u^2 + 2as
Where u is the initial velocity (equal to v), v is the final velocity (which is zero when the platform and people come to rest), a is the acceleration, and s is the distance.
Rearranging the equation:
s = (v^2 - u^2) / (2a)
Since the platform and people come to rest (v = 0), we can rewrite the equation as:
s = -u^2 / (2a)
Given that u = v, the equation becomes:
s = -v^2 / (2a)
Substituting the value of v we calculated earlier:
s = -[(6.1 kg * v3) / 248 kg]^2 / (2a)
Now, we need to determine the acceleration, which can be obtained by dividing the horizontal force on the platform and people by their total mass.
The only horizontal force acting on the platform and people is the force exerted by the ball when it is thrown. This force can be calculated using Newton's second law:
F = m * a
Knowing that F = (m1 + m2) * v (the force from the recoil), we have:
(m1 + m2) * v = (m1 + m2) * a
Dividing both sides of the equation by (m1 + m2):
v = a
So, the acceleration of the platform and people is equal to the velocity they recoil with.
Therefore, we can substitute v for a in the equation for distance:
s = -[(6.1 kg * v3) / 248 kg]^2 / (2v)
Finally, we have the equation for the distance the platform and people move before coming to rest again:
s = -[(6.1 kg * v3) / 248 kg]^2 / (2v)