A 11.0 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 151 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 90.0 cm, what was the speed of the bullet at impact with the block?
To find the speed of the bullet at impact with the block, we can use the principle of conservation of momentum and the law of conservation of energy.
1. Conservation of momentum:
The total momentum of the system before the collision is equal to the total momentum after the collision. Since the block is at rest, the initial momentum is only due to the bullet, and after the collision, the bullet and block move together as one object.
Let's denote the initial velocity of the bullet as Vb, the final velocity of the bullet and block together as Vf, and the mass of the block as Mb. The mass of the bullet is not needed because it becomes embedded in the block.
So, the momentum before the collision is given by:
Initial momentum = mass of the bullet × initial velocity of the bullet = 11.0 g × Vb
The momentum after the collision is given by:
Final momentum = (mass of the bullet + mass of the block) × final velocity of bullet and block = (11.0 g + 108 g) × Vf
Since momentum is conserved, we can equate the initial and final momenta:
11.0 g × Vb = 119.0 g × Vf ---(Equation 1)
2. Conservation of energy:
The kinetic energy of the bullet before the collision is given by:
Initial kinetic energy = 0.5 × mass of bullet × initial velocity of bullet squared = 0.5 × 11.0 g × Vb^2
The potential energy stored in the spring, when it is compressed by a maximum of 90.0 cm, is given by:
Potential energy = 0.5 × spring constant × compression distance^2
= 0.5 × 151 N/m × (90.0 cm)^2
Since energy is conserved, the initial kinetic energy is equal to the potential energy stored in the spring:
0.5 × 11.0 g × Vb^2 = 0.5 × 151 N/m × (0.9 m)^2 ---(Equation 2)
Now, we have two equations (Equation 1 and Equation 2) with two variables (Vb and Vf). We can solve these equations simultaneously to find the value of Vb.
Using Equation 1, we can rewrite it as:
Vf = (11.0 g × Vb) / 119.0 g
Substituting this value of Vf into Equation 2, we get:
0.5 × 11.0 g × Vb^2 = 0.5 × 151 N/m × (0.9 m)^2
Simplifying the equation and solving for Vb:
Vb^2 = (151 N/m × (0.9 m)^2) / 11.0 g
Vb = √((151 N/m × (0.9 m)^2) / 11.0 g)
Plugging in the given values (g = 9.8 m/s^2), we get:
Vb = √((151 N/m × (0.9 m)^2) / (11.0 g))
= √((151 N/m × (0.9 m)^2) / (11.0 × 9.8 m/s^2))
Evaluating this expression gives the speed of the bullet at impact with the block.