Two pucks collide on an air hockey table, then bounce off each other in opposite directions. Each puck has a mass of 0.2 kg. Just before the collision, puck 1 has an initial velocity of +10 m/s, and puck 2 has an initial velocity of −25 m/s. Just after the collision, puck 1 has a velocity of −25 m/s. What is the velocity of puck 2 just after the collision? Group of answer choices−40 m/s−10 m/s+10 m/s+40 m/s
Question
Two pucks collide on an air hockey table, then bounce off each other in opposite directions. Each puck has a mass of 0.2 kg. Just before the collision, puck 1 has an initial velocity of +10 m/s, and puck 2 has an initial velocity of −25 m/s. Just after the collision, puck 1 has a velocity of −25 m/s. What is the velocity of puck 2 just after the collision? Group of answer choices−40 m/s−10 m/s+10 m/s+40 m/s
Solution
To solve this problem, we need to use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
The total momentum before the collision is the sum of the momentum of puck 1 and puck 2. Momentum is calculated as mass times velocity.
Momentum of puck 1 before collision = mass of puck 1 * velocity of puck 1 = 0.2 kg * 10 m/s = 2 kgm/s Momentum of puck 2 before collision = mass of puck 2 * velocity of puck 2 = 0.2 kg * -25 m/s = -5 kgm/s
So, total momentum before collision = 2 kgm/s - 5 kgm/s = -3 kg*m/s
The total momentum after the collision is also the sum of the momentum of puck 1 and puck 2.
We know that the velocity of puck 1 after the collision is -25 m/s.
So, momentum of puck 1 after collision = mass of puck 1 * velocity of puck 1 = 0.2 kg * -25 m/s = -5 kg*m/s
Let's denote the velocity of puck 2 after the collision as V. So, the momentum of puck 2 after the collision is mass of puck 2 * V = 0.2 kg * V.
The total momentum after the collision is therefore -5 kg*m/s + 0.2 kg * V.
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. So, we can set up the following equation:
-3 kgm/s = -5 kgm/s + 0.2 kg * V
Solving this equation for V gives us:
V = (-3 kgm/s + 5 kgm/s) / 0.2 kg = 10 m/s
So, the velocity of puck 2 just after the collision is +10 m/s.
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