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.
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.
a. Momentum conserved. m*225=(5+.06)V
solve for V
I got 2.67 m/s from your equation. But the correct answer is supposed to be 3.6 m/s.
Also, for part C, do I do KE final minus KE initial?
So 1/2(5.06)(75^2) - 1/2(0.06)(225^2) ?
the correct equation is this: (.06)(225) = (5)(v) - (.06)(75)
momentum is conserved. total momentum is equal to the momentum of the block minus the momentum of the bullet. you subtract because the two objects move in opposite directions.
To answer part A, we need to use the principle of conservation of momentum. This principle states that the total momentum of an isolated system remains constant if no external forces act on it.
Let's assume the initial velocity of the block is V. After the collision, the bullet moves backward, so its final velocity is -75 m/s.
The momentum before the collision is given by the sum of the momenta of the bullet and the block:
Initial momentum = (mass of bullet * initial velocity of bullet) + (mass of block * initial velocity of block)
= (0.060 kg * 225 m/s) + (5.0 kg * V)
The momentum after the collision is also the sum of the momenta of the bullet and the block:
Final momentum = (mass of bullet * final velocity of bullet) + (mass of block * final velocity of block)
= (0.060 kg * -75 m/s) + (5.0 kg * velocity of block after collision)
According to the conservation of momentum principle, the initial momentum equals the final momentum:
(0.060 kg * 225 m/s) + (5.0 kg * V) = (0.060 kg * -75 m/s) + (5.0 kg * velocity of block after collision)
Now, we need to solve this equation for the velocity of the block after the collision. Let's calculate it.
(0.060 kg * 225 m/s) + (5.0 kg * V) = (0.060 kg * -75 m/s) + (5.0 kg * velocity of block after collision)
13.5 kgm/s + 5.0 kgV = -4.5 kgm/s + 5.0 kg(velocity of block after collision)
18.0 kgV = -18.0 kgm/s + 5.0 kg(velocity of block after collision)
Simplifying further:
18.0 kgV = -18.0 kgm/s + 5.0 kg(velocity of block after collision)
18.0 kgV + 18.0 kgm/s = 5.0 kg(velocity of block after collision)
18.0 kg(V + 1.0 m/s) = 5.0 kg(velocity of block after collision)
V + 1.0 m/s = (5.0 kg/18.0 kg) * (velocity of block after collision)
V + 1.0 m/s = (5/18) * (velocity of block after collision)
From here, we can solve for the velocity of the block after the collision in terms of V:
velocity of block after collision = (18/5)(V + 1.0) m/s
This gives us the velocity of the block after the collision (part A).
Once you have the velocity of the block after the collision, you can move on to solving part B. The equation you mentioned, 1/2mv^2 = 1/2kx^2, is correct. This equation relates the kinetic energy of the block to the potential energy stored in the compressed spring. By substituting the values of mass (5.0 kg), velocity (from part A), and maximum compression (0.20 m), you can solve for the spring constant (k).
Finally, to answer part C, you need to calculate the initial kinetic energy of the system (bullet and block) before the collision and subtract the final kinetic energy after the collision. The difference between these two energies corresponds to the energy lost during the inelastic collision.