A 2.0kg bag is held by a string to the ceiling as shown in the diagram below. A 10-g bullet travelling at 200m/s strikes the stationary bag. The height of the bad after the collision is 10cm. Assuming there is no friction, determine the speed, in metres per second of that bag after the collision.
To determine the speed of the bag after the collision, we can apply the law of conservation of momentum.
1. First, let's calculate the momentum of the bullet before the collision. Momentum is calculated by multiplying the mass of an object by its velocity.
Momentum of the bullet before collision = mass of bullet × velocity of bullet
Mass of bullet = 10 g = 0.01 kg
Velocity of bullet = 200 m/s
Momentum = 0.01 kg × 200 m/s
2. Next, we need to calculate the momentum of the bag after the collision. Since the bullet gets embedded in the bag, their momenta combine. Let's assume the final velocity of the bullet-bag system is V.
Momentum of the bag after collision = mass of bag × velocity of bag
Mass of bag = 2.0 kg
Velocity of bag = V
Momentum of the bullet-bag system after collision = (0.01 kg + 2.0 kg) × V
3. According to the law of conservation of momentum, the total momentum before the collision should equal the total momentum after the collision.
Therefore, we can write:
Momentum before collision = Momentum after collision
0.01 kg × 200 m/s = (0.01 kg + 2.0 kg) × V
Simplifying the equation, we have:
2 = 2.01V
Dividing both sides by 2.01, we get:
V = 2 / 2.01
V ≈ 0.995 m/s
So, the speed of the bag after the collision is approximately 0.995 m/s.