I have three questions. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bottle, and the bottle flies forward at 25 m/s. How fast is the ball traveling after hitting the bottle?

2. Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of woof is sitting on a frictionless surface, so it moves freely when the bullet hits it). The wood block is initially at rest. The bullet is traveling 300 m/s when it hits the wood block and sticks inside it. Now the bullet and the woof block move together as one object. How fast are they traveling?

3. A big league hitter attacks a fastball. The ball has a mass of 0.16 kg. It is pitched at 38 m/s. After the player hits the ball, it is now traveling 44 m/s in the opposite direction. The impact lasted 0.002 seconds. How big of a force did the ballplayer put on that ball?

Question

Leonardo Nelson

Physics

23 November, 07:27

I have three questions. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bottle, and the bottle flies forward at 25 m/s. How fast is the ball traveling after hitting the bottle?

2. Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of woof is sitting on a frictionless surface, so it moves freely when the bullet hits it). The wood block is initially at rest. The bullet is traveling 300 m/s when it hits the wood block and sticks inside it. Now the bullet and the woof block move together as one object. How fast are they traveling?

3. A big league hitter attacks a fastball. The ball has a mass of 0.16 kg. It is pitched at 38 m/s. After the player hits the ball, it is now traveling 44 m/s in the opposite direction. The impact lasted 0.002 seconds. How big of a force did the ballplayer put on that ball?

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Answers (1)

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Markus Wu · Answer 1

1) The total momentum of the system has to be conserved to satisfy the principle of conservation of momentum. Before the ball hits the bottle, the momentum of the system is 0.4 x 18 = 7.2 kg m/s

The momentum of the bottle after being hit is 0.2 x 25 = 5 kg m/s

So the momentum of the ball now is 7.2 - 5 = 2.2 kg m/s

Hence its velocity is 2.2/0.4 = 5.5 m/s

2) The momentum of the wood and the bullet is equal to the momentum of the bullet before it hits the wood = 0.04 x 300 = 120 kg m/s

Hence (0.5 + 0.04) * v = 120 kg m/s, v = 120 kg m/s / 0.54 kg = 222.2 m/s

3) Force = rate of change of momentum = 0.16 (38+44) / 0.002 = 6560 N