A 50.0 g ball is fired horizontally with initial speed v0 toward a 100 g ball that is hanging motionless from a 1.10 m -long string. The balls undergo a head-on, perfectly elastic collision, after which the 100 g ball swings out to a maximum angle θmax = 50.0 ∘. What is v0?
To find the initial speed (v0) of the 50.0 g ball, we can use conservation of energy and momentum principles.
First, let's assume that the positive x-direction is in the horizontal direction that the ball is fired.
Conservation of momentum:
Momentum is conserved in an elastic collision. Since the 100 g ball is initially at rest, its momentum (P2) after the collision will be zero. The momentum before the collision (P1) is given by:
P1 = m1 * v1
where m1 is the mass of the 50.0 g ball and v1 is its velocity.
Conservation of energy:
The potential energy stored in the 100 g ball after swinging to a maximum angle θmax can be converted into kinetic energy.
The potential energy at the maximum angle is given by:
PE = m2 * g * h
where m2 is the mass of the 100 g ball, g is the acceleration due to gravity, and h is the height of the ball's displacement.
The initial kinetic energy before the collision is given by:
KE = 0.5 * m1 * v1^2
Since it is an elastic collision, the total energy before and after the collision remains the same.
KE + PE = 0.5 * m1 * v1^2 + 0
Substituting the values, we get:
m2 * g * h = 0.5 * m1 * v1^2
Rearranging the equation, we have:
v1^2 = (2 * m2 * g * h) / m1
Finally, to find v1, we take the square root of both sides:
v1 = sqrt((2 * m2 * g * h) / m1)
Substituting the given values:
m1 = 50.0 g = 0.050 kg
m2 = 100 g = 0.100 kg
h = 1.10 m
θmax = 50.0°
g = 9.8 m/s^2
v1 = sqrt((2 * 0.100 kg * 9.8 m/s^2 * 1.10 m) / 0.050 kg)
Calculating the above expression will give us the initial velocity (v1) of the 50.0 g ball.