Disk 1(of mass m) slides with speed 1.0 m/s across a low friction surface and collides with disk 2( of mass 2m) originally at rest. Disk 1 is observed to bounce off at an angle of 15degrees to its original line of motion, while disk 2 moves at an angle of 55degrees.
a) What is the unknown final speed of each disk?
b) Is this collision elastic? Why?
To find the unknown final speeds of each disk in this collision scenario, we can apply the principles of conservation of momentum and conservation of kinetic energy.
a) Finding the Final Speeds:
First, we need to break down the velocities of both disks into their horizontal and vertical components.
Since disk 1 is observed to bounce off at an angle of 15 degrees to its original line of motion, the horizontal component (V1x) remains the same (1.0 m/s), while the vertical component (V1y) can be found using trigonometry:
V1y = V1 * sin(15°)
= 1.0 m/s * sin(15°)
≈ 0.2588 m/s
On the other hand, disk 2 moves at an angle of 55 degrees. We can calculate its horizontal (V2x) and vertical (V2y) components in a similar way:
V2x = V2 * cos(55°)
≈ 2VM * cos(55°)
≈ 2 * 0 m/s * cos(55°)
= 0 m/s
V2y = V2 * sin(55°)
≈ 2 * 0 m/s * sin(55°)
= 0 m/s
Since the collision is elastic, both momentum and kinetic energy are conserved. We can apply these principles to find the unknown final speeds.
Conservation of Momentum:
In the horizontal direction (x-axis), the total momentum before the collision is given by:
m * V1 = 2m * V2x
Substituting the values we know, we get:
m * 1.0 m/s = 2m * V2x
Simplifying, we find:
V2x = 0.5 m/s
Conservation of Kinetic Energy:
In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Before the collision, the total kinetic energy is given by:
KE1 + KE2 = 0.5 * m * V1^2 + 0.5 * 2m * 0^2
= 0.5 * m * 1.0 m/s * 1.0 m/s
= 0.5 * m * 1.0 m^2/s^2
= 0.5m Joules
After the collision, the total kinetic energy is given by:
KE1' + KE2' = 0.5 * m * V1'^2 + 0.5 * 2m * V2'^2
Since the collision is elastic, the kinetic energy is conserved. So:
0.5m = 0.5 * m * V1'^2 + 0.5 * 2m * V2'^2
Since disk 2 starts from rest, its initial velocity (V2) is zero. The equation becomes:
0.5m = 0.5 * m * V1'^2 + 0.5 * 2m * V2'^2
= 0.5 * m * V1'^2 + 0.5 * 2m * 0^2
= 0.5 * m * V1'^2
Simplifying it further, we find:
1 = V1'^2
Taking the square root, we get:
V1' = 1 m/s
Therefore, the unknown final speed of disk 1 is 1.0 m/s, and the final speed of disk 2 is 0.5 m/s in the horizontal direction (x-axis).
b) Determining Elasticity:
For a collision to be considered elastic, both momentum and kinetic energy must be conserved.
In the horizontal direction (x-axis), where momentum is conserved, we have already verified that the total initial momentum is equal to the total final momentum.
For kinetic energy to be conserved, the total kinetic energy before the collision must be equal to the total kinetic energy after the collision. In this case, considering the kinetic energies of both disks as mentioned earlier, we can see that the kinetic energy remains the same:
Before the collision:
KE1 + KE2 = 0.5m
After the collision:
KE1' + KE2' = 0.5m
Since the total kinetic energy before and after the collision is equal, we can conclude that the collision is indeed elastic.