A 1.6-kg block is attached to the end of a 2.0-m string to form a pendulum. The pendulum is released from rest when the string is horizontal. At the lowest point of its swing when it is moving horizontally, the block is hit by a 10-g bullet moving horizontally in the opposite direction. The bullet remains in the block and causes the block to come to rest at the low point of its swing. What was the magnitude of the bullet's velocity just before hitting the block?
The kinetic energy = the original potential energy, so KE = mgh = .5mv^2
Solve for velocity to get 6.26 m/s for the block
Since Vf = 0, use inelastic equation to get m2v2=-m1v1 and solve for v2
v2 = 1540 m/s
To solve this problem, we need to apply the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
Step 1: Find the initial momentum of the block and bullet system.
The block is released from rest, so its initial velocity is 0. The bullet is moving horizontally with an unknown velocity, let's call it v. The mass of the bullet is 10 g, which is equal to 0.01 kg.
The initial momentum is given by:
Initial momentum = (mass of block)(velocity of block) + (mass of bullet)(velocity of bullet)
Initial momentum = (1.6 kg)(0) + (0.01 kg)(v)
Initial momentum = 0 + 0.01v
Initial momentum = 0.01v
Step 2: Find the final momentum of the block and bullet system.
After the collision, the block comes to rest. Therefore, its final velocity is 0. The bullet is embedded in the block and also comes to rest.
The final momentum is given by:
Final momentum = (mass of block)(velocity of block) + (mass of bullet)(velocity of bullet)
Final momentum = (1.6 kg)(0) + (0.01 kg)(0)
Final momentum = 0 + 0
Final momentum = 0
Step 3: Apply the conservation of momentum principle.
According to the conservation of momentum principle, the initial momentum is equal to the final momentum.
0.01v = 0
v = 0/0.01
v = 0
Therefore, the magnitude of the bullet's velocity just before hitting the block was 0 m/s.
To find the magnitude of the bullet's velocity just before hitting the block, we can use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.
1. First, let's calculate the initial momentum of the bullet-block system before the collision.
The block is at the lowest point of its swing, so its velocity is zero. The bullet is moving horizontally in the opposite direction, so its initial velocity can be represented as -vb (negative vb). The mass of the bullet is 10 grams, which is 0.01 kg.
Momentum before collision = (mass of bullet) x (velocity of bullet) + (mass of block) x (velocity of block)
= (0.01 kg) x (-vb) + (1.6 kg) x 0
= -0.01 vb
2. Next, let's calculate the final momentum of the bullet-block system after the collision. The block comes to rest at the lowest point of its swing, so its velocity is 0. The bullet remains in the block, so their final velocity will be the same.
Momentum after collision = (mass of bullet + mass of block) x (final velocity)
= (0.01 kg + 1.6 kg) x 0
= 0
3. Since momentum is conserved, we can equate the initial momentum to the final momentum and solve for the magnitude of the bullet's velocity.
-0.01 vb = 0
vb = 0
This means that the magnitude of the bullet's velocity just before hitting the block is 0. The bullet has no initial velocity in this scenario.