A mass of 9.8 kg is moving in the positive x direction at 16.6 m/s and collides in a perfectly elastic, head on collision with a mass of 12.7 kg at rest. The same mass of 9.8 kg in the positive direction at 16.6 m/s also collides with the mass of 12.7 kg at rest in an inelastic collision. If both collisions occur over a time of 0.13 seconds, what is the difference between the magnitude of the average force on the 9.8 kg mass in the elastic collision, and magnitude of the the average force on the 9.8 kg mass in the inelastic collision? That is, how much greater is magnitude of the the average force in the elastic collision compared to the inelastic collision? Answer in Newtons.
To solve this problem, we can use the laws of conservation of momentum and kinetic energy.
In the elastic collision, the momentum and the kinetic energy of the system are conserved. We can first calculate the final velocities of both masses using the conservation of momentum:
9.8 kg mass before collision:
Initial momentum = mass * velocity = 9.8 kg * 16.6 m/s = 162.68 kg·m/s
12.7 kg mass at rest:
Initial momentum = mass * velocity = 12.7 kg * 0 m/s = 0 kg·m/s
Total initial momentum = 162.68 kg·m/s
Total final momentum = 9.8 kg mass final momentum + 12.7 kg mass final momentum
Let the final velocity of the 9.8 kg mass be v1 and the final velocity of the 12.7 kg mass be v2.
Total final momentum = 9.8 kg * v1 + 12.7 kg * v2
Applying the conservation of momentum:
Total initial momentum = Total final momentum
162.68 kg·m/s = 9.8 kg * v1 + 12.7 kg * v2
In the inelastic collision, only momentum is conserved, and kinetic energy is not conserved. The two masses will stick together and move with the same final velocity. Let v be the final velocity of the masses after collision.
Total initial momentum = Total final momentum
162.68 kg·m/s = (9.8 kg + 12.7 kg) * v
Now we have two equations:
1) 162.68 kg·m/s = 9.8 kg * v1 + 12.7 kg * v2
2) 162.68 kg·m/s = (9.8 kg + 12.7 kg) * v
Solving these two equations will give us the final velocities for both the elastic and inelastic collisions.
After finding the final velocities, we can calculate the change in momentum for both the elastic and inelastic collisions. The change in momentum is simply the final momentum minus the initial momentum.
Finally, we can calculate the average force using the formula:
Average force = change in momentum / time
By comparing the magnitudes of the average forces in the elastic and inelastic collisions, we can determine the difference in force.
To find the difference in magnitude of the average forces in the two collisions, we need to calculate the average force in both the elastic and inelastic collisions.
In an elastic collision, both kinetic energy and momentum are conserved. We can use these principles to find the average force on the 9.8 kg mass.
First, let's find the final velocities after the elastic collision. Using the conservation of momentum:
(mass1 * velocity1) + (mass2 * velocity2) = (mass1 * final_velocity1) + (mass2 * final_velocity2)
where:
mass1 = 9.8 kg (mass initially moving)
mass2 = 12.7 kg (mass at rest initially)
velocity1 = 16.6 m/s (initial velocity of mass 1)
velocity2 = 0 m/s (initial velocity of mass 2)
Plugging in the values:
(9.8 kg * 16.6 m/s) + (12.7 kg * 0 m/s) = (9.8 kg * final_velocity1) + (12.7 kg * final_velocity2)
(162.68 kg*m/s) = (9.8 kg * final_velocity1) + (12.7 kg * final_velocity2)
Since both masses have the same final velocity due to conservation of kinetic energy, final_velocity1 = final_velocity2 = final_velocity.
(162.68 kg*m/s) = (9.8 kg * final_velocity) + (12.7 kg * final_velocity)
162.68 kg*m/s = 22.5 kg * final_velocity
final_velocity = 162.68 kg*m/s / 22.5 kg ≈ 7.23 m/s
Now, we can find the change in velocity for mass 1:
delta_v = final_velocity - velocity1
delta_v = 7.23 m/s - 16.6 m/s ≈ -9.37 m/s
The time taken for the elastic collision is given as 0.13 seconds. Now, we can calculate the acceleration and thus the force using Newton's second law:
acceleration = delta_v / time
acceleration = -9.37 m/s / 0.13 s ≈ -72.08 m/s^2
Force = mass1 * acceleration
Force = 9.8 kg * -72.08 m/s^2 ≈ -705.78 N (negative because it is acting in the opposite direction)
For the inelastic collision, assume both masses stick together after the collision. We can use the principle of conservation of momentum to find the final velocity:
(mass1 * velocity1) + (mass2 * velocity2) = (mass1 + mass2) * final_velocity
(9.8 kg * 16.6 m/s) + (12.7 kg * 0 m/s) = (9.8 kg + 12.7 kg) * final_velocity
162.68 kg*m/s = 22.5 kg * final_velocity
final_velocity = 162.68 kg*m/s / 22.5 kg ≈ 7.23 m/s
Since the masses stick together, the final velocity is the same as in the elastic collision.
Using the same time of 0.13 seconds, we can calculate the acceleration and force:
acceleration = delta_v / time
acceleration = (final_velocity - velocity1) / time
acceleration = (7.23 m/s - 16.6 m/s) / 0.13 s ≈ -82.31 m/s^2
Force = mass1 * acceleration
Force = 9.8 kg * -82.31 m/s^2 ≈ -806 N (negative because it is acting in the opposite direction)
Therefore, the difference in magnitude of the average forces is:
|average force in elastic collision| - |average force in inelastic collision|
= |-705.78 N| - |-806 N|
= 705.78 N - 806 N
≈ 100.22 N
So, the magnitude of the average force in the elastic collision is approximately 100.22 N greater than the magnitude of the average force in the inelastic collision.