To tackle this problem, we need to apply the principles of conservation of momentum and angular momentum. Let's break it down step by step, addressing each part of your question clearly.
Understanding the System Before Collision
Initially, we have two balls A and B, each with mass m, connected by a light rod of length L. The entire system is moving with a velocity v on a frictionless surface. When ball A collides with particle P (also of mass m) at rest, we need to analyze the outcomes of this collision.
(a) Linear Speeds of Balls A and B After the Collision
Since the collision is perfectly inelastic (the particle P sticks to ball A), we can use the conservation of linear momentum. Before the collision, the momentum of the system is:
- Momentum of A: mv
- Momentum of B: mv
- Momentum of P: 0 (since it is at rest)
The total initial momentum of the system is:
P_initial = mv + mv + 0 = 2mv
After the collision, the combined mass of A and P becomes 2m (since P sticks to A). Let the final velocity of A and P together be V, and B continues to have its own velocity, which we will denote as V_B. The total momentum after the collision is:
P_final = (2m)V + mV_B
Setting the initial momentum equal to the final momentum gives us:
2mv = (2m)V + mV_B
We can simplify this equation by dividing through by m:
2v = 2V + V_B
Now, we need another equation to find the relationship between V and V_B. Since ball B is not affected by the collision directly, we can assume it continues moving at its original speed, so:
V_B = v
Substituting this back into our momentum equation:
2v = 2V + v
2V = 2v
V = v
Thus, after the collision, the linear speeds of balls A and B are:
- Speed of A (with P): v
- Speed of B: v
(b) Velocity of the Centre of Mass C of the System A+B+P
The center of mass (CM) velocity can be calculated using the formula:
V_{CM} = \frac{m_1v_1 + m_2v_2 + m_3v_3}{m_1 + m_2 + m_3}
Here, the masses and their velocities are:
- Mass of A (with P): 2m, velocity: v
- Mass of B: m, velocity: v
- Mass of P: m, velocity: v (since it moves with A after collision)
Substituting these values into the CM formula:
V_{CM} = \frac{(2m)v + mv + 0}{2m + m + m} = \frac{3mv}{4m} = \frac{3v}{4}
(c) Angular Speed of the System About C After the Collision
To find the angular speed (ω) of the system about the center of mass C, we first need to determine the moment of inertia (I) of the system about point C. The moment of inertia for point masses is given by:
I = \sum m_i r_i^2
For our system, we have:
- Ball A (with P) is at a distance of L/2 from C.
- Ball B is at a distance of L/2 from C.
Thus, the moment of inertia about C is:
I = 2m\left(\frac{L}{2}\right)^2 + m\left(\frac{L}{2}\right)^2 = 2m\frac{L^2}{4} + m\frac{L^2}{4} = \frac{L^2}{2}m + \frac{L^2}{4}m = \frac{3L^2}{4}m
Now we can use the relationship between linear velocity and angular velocity:
v = ωr
For ball A (with P), the distance r from C is L/2:
v = ω\left(\frac{L}{2}\right)
Substituting for v gives:
ω = \frac{2v}{L}
In summary, after the collision:
- The linear speeds of balls A and B are both v.
- The velocity of the center of mass C is 3v/4.
- The angular speed of the system about C is 2v/L.
