A 12.0 g bullet is fired horizontally into a 108 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring of constant 144 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 82.0 cm, what was the speed of the bullet at impact with the block?
What equation would I have use to solve the question??
What is the stored energy in the spring at max compression? What Veloicity does that equal on the block/bullet system. Waht momentum is that? Then, finally that momentum is the mometnum of the bullet...solve for bullet speed.
I am so cunfused? Can you still use the PEi+PEs+KE = PEf+PEs+KE?
No. You cant use that. conservation of energy does not apply when the bullet hits the block.
You can use conservation of energy at
spring: block, but that is all.
Then how would I set it up?
To solve this question, you can use the principle of conservation of momentum and the concept of elastic collisions. The equation you would need to use is the conservation of linear momentum equation:
m1v1 + m2v2 = (m1 + m2)V
Where:
m1 = mass of the bullet
v1 = initial velocity of the bullet
m2 = mass of the block
v2 = initial velocity of the block (zero since it is at rest)
V = final velocity of the combined bullet-block system
Since the bullet becomes embedded in the block, the final velocity of the system after the collision can be assumed to be the same for the bullet and the block.
After finding the value of V, you can use this information to calculate the initial velocity of the bullet.