A bullet of mass m=0.060 kg hits a 5.0 kg block with an initial speed of 225 m/s. The block is connected to a spring that is attached to a wall. The friction between the block and the table is negligible. Upon impact, the bullet bounces back from the box with a speed of 75 m/s.
A) Calculate the speed of the block right after the collision.
I thought the equation to use was mv of bullet = (M+m)V of block. I came up with 2.67 but the answer is supposed to be 3.6 m/s.
B) As a result of the collision, the spring compresses to a mazimum of 0.20m. Find the spring constant
--The equation to this is 1/2mv^2 = 1/2kx^2 right? I can't get the answer to this question without the answer to part A.
C) Find the inelastic energy loss during the collision
To calculate the speed of the block right after the collision, you can use the principle of conservation of linear momentum. According to this principle, the initial momentum of the system (bullet + block) before the collision is equal to the final momentum of the system after the collision.
The initial momentum of the bullet is given by:
Initial momentum of bullet = mass of bullet × initial speed of bullet
= 0.060 kg × 225 m/s
The initial momentum of the block is zero since it is initially at rest.
The final momentum of the bullet after the collision is given by:
Final momentum of bullet = mass of bullet × final speed of bullet
= 0.060 kg × (-75 m/s) (letting the direction be negative since it's opposite to initial direction)
According to the principle of conservation of linear momentum:
Initial momentum of system = Final momentum of system
Therefore, we can write:
(mass of bullet × initial speed of bullet) + (mass of block × 0) = (mass of bullet × final speed of bullet) + (mass of block × final speed of block)
Simplifying the equation with given values:
(0.060 kg × 225 m/s) = (0.060 kg × (-75 m/s)) + (5.0 kg × final speed of block)
Solving for the final speed of the block, we get:
Final speed of block = [(0.060 kg × 225 m/s) - (0.060 kg × (-75 m/s))] / 5.0 kg
Final speed of block = (13.5 - (-4.5)) / 5.0 m/s
Final speed of block = 18 / 5.0 m/s
Final speed of block = 3.6 m/s
So, the speed of the block right after the collision is 3.6 m/s, which matches the correct answer.
Now, moving on to part B, to find the spring constant, we can use the equation for potential energy of a spring:
Potential energy of spring = 1/2 × spring constant × (maximum compression)^2
Given that the maximum compression is 0.20 m, and the potential energy of the spring is equal to the kinetic energy of bullet before the collision, which can be calculated using the equation:
Kinetic energy of bullet = 1/2 × mass of bullet × (initial speed of bullet)^2
Set the kinetic energy of the bullet equal to the potential energy of the spring:
1/2 × mass of bullet × (initial speed of bullet)^2 = 1/2 × spring constant × (maximum compression)^2
Now, rearrange the equation to solve for the spring constant:
spring constant = (mass of bullet × (initial speed of bullet)^2) / (maximum compression)^2
Plug in the given values and solve the equation to find the spring constant.
For part C, inelastic energy loss during the collision can be calculated by finding the difference between the initial kinetic energy and the final kinetic energy of the system. The initial kinetic energy is given by:
Initial kinetic energy = 1/2 × mass of bullet × (initial speed of bullet)^2
The final kinetic energy can be calculated by considering the final kinetic energy of the bullet and the block. The final kinetic energy of the bullet is given by:
Final kinetic energy of bullet = 1/2 × mass of bullet × (final speed of bullet)^2
The final kinetic energy of the block is given by:
Final kinetic energy of block = 1/2 × mass of block × (final speed of block)^2
The inelastic energy loss is the difference between the initial kinetic energy and the sum of final kinetic energies of the bullet and the block:
Inelastic energy loss = Initial kinetic energy - (Final kinetic energy of bullet + Final kinetic energy of block)
Evaluate each term using the given values to find the inelastic energy loss during the collision.