Disk 1 (of inertia m) slides with speed 2.0 m/s across a low-friction surface and collides with disk 2 (of inertia 2m) originally at rest. Disk 1 is observed to turn from its original line of motion by an angle of 15 ∘, while disk 2 moves away from the impact at an angle of 55 ∘.
Calculate the final speeds of disk 1 and disk 2.
Is this collision elastic?
To calculate the final speeds of disk 1 and disk 2, we can use the principle of conservation of linear momentum and the principle of conservation of angular momentum.
Let's start by analyzing the collision in terms of linear momentum. The initial momentum of the system before the collision is zero because disk 2 is at rest. After the collision, the total momentum of the system should still be zero because there are no external forces acting on the system. So we have:
(m × v1) + (2m × 0) = 0
Here, v1 is the final velocity of disk 1, and the initial velocity of disk 2 is zero since it was at rest.
From this equation, we can solve for v1:
m × v1 = 0
v1 = 0
This means that disk 1 comes to a stop after the collision.
Next, let's consider angular momentum. Before the collision, both disks have zero angular momentum because they are not rotating. After the collision, the disks start to rotate and have angular momentum.
Since disk 1 turns from its original line of motion by an angle of 15°, it gains angular momentum. On the other hand, disk 2 moves away from the impact at an angle of 55°, indicating that it also gains angular momentum. These two angular momenta must be equal in magnitude but opposite in direction to ensure conservation of angular momentum.
Let's consider disk 1 first. The initial moment of inertia of disk 1 is m. After the collision, it rotates with a final angular momentum L1. We can calculate L1 as follows:
L1 = I1 × ω1
Here, ω1 is the angular velocity of disk 1 after the collision. Since disk 1 is observed to turn from its original line of motion by an angle of 15°, we can relate the angular velocity to the final speed v1:
v1 = ω1 × r1
where r1 is the radius of disk 1.
Similarly, for disk 2, which has an initial moment of inertia of 2m, the final angular momentum L2 can be calculated as:
L2 = I2 × ω2
where ω2 is the angular velocity of disk 2 after the collision. And we can again relate the angular velocity to the final speed v2 of disk 2:
v2 = ω2 × r2
where r2 is the radius of disk 2.
Now, let's equate the two angular momenta to find relationship between ω1 and ω2:
L1 = L2
m × ω1 × r1 = 2m × ω2 × r2
Simplifying this equation, ω2/ω1 = r1/(2r2).
Next, let's use the relationship between angular velocity and linear velocity to relate ω1 and v1, and ω2 and v2:
v1 = ω1 × r1
v2 = ω2 × r2
Using these equations, we can rewrite ω2/ω1 = r1/(2r2) as v2/v1 = r1/(2r2).
Substituting the value of v1 from the first equation (v1 = 0), we get:
v2/0 = r1/(2r2)
This equation is undefined since division by zero is not possible. Hence, the final speed of disk 2 cannot be calculated.
Therefore, the final velocity of disk 1 is 0 m/s, and the final velocity of disk 2 cannot be determined.
Regarding the elasticity of the collision, in an elastic collision, both kinetic energy and linear momentum are conserved. In this case, only linear momentum is conserved, as the final velocity of disk 1 is 0 m/s. Therefore, this collision is considered inelastic because some kinetic energy is lost in the process.