Question

Henry is at the state fair. At one booth he throws a 0.75kg ball forward with a velocity of 25m/s in order to hit a 0.2kg bottle sitting on a shelf. When the ball makes contact the bottle goes flying forward at 39.5m/s. What is the velocity of the ball after it hits the bottle?

Answer 1

Using the conservation of momentum, the ball's velocity after it hits the bottle is calculated to be 14.47 m/s, assuming there are no external forces acting on the ball and bottle system other than their interaction.

Step-by-step explanation:

To determine the velocity of the ball after it hits the bottle at the state fair, we can use the principle of conservation of momentum. In an isolated system, the total momentum before collision is equal to the total momentum after collision. The formula for momentum (p) is p = mv, where m is mass and v is velocity.

The initial momentum of the system can be calculated as the momentum of the ball plus the momentum of the bottle before they collide. In this case, the bottle is initially at rest, so its momentum is zero. Therefore, the initial momentum is just pball initial = mball x vball initial.

After the collision, the ball and the bottle move with different velocities, but the total momentum of the system must remain the same. So, we have

• Initial momentum of the ball: pball initial = (0.75 kg)(25 m/s) = 18.75 kg·m/s,
• Momentum of the bottle after collision: pbottle final = (0.2 kg)(39.5 m/s) = 7.9 kg·m/s.

Using conservation of momentum:

ptotal initial = pball final + pbottle final

18.75 kg·m/s = (0.75 kg)vball final + 7.9 kg·m/s

Solving for vball final gives us:

vball final = (18.75 kg·m/s - 7.9 kg·m/s) / 0.75 kg

vball final = 14.47 m/s

Therefore, after hitting the bottle, the ball's velocity is 14.47 m/s.