Two spheres A and B of equal mass are free to move on a smooth horizontal surface. A and B move towards each other with velocity vectors ai+bj and ci+dj r respectively and collide when the line joining their centres is parallel to i. After impact A and B have velocity vectors pi+qj and ri+sj respectively. The coefficient of restitution between the spheres is e(e<1). Then prove that b=q
To prove that b = q, we need to analyze the problem and use the principles of mechanics.
Let's break down the problem step by step:
1. Initial velocities of spheres A and B:
- Sphere A has an initial velocity vector of ai + bj.
- Sphere B has an initial velocity vector of ci + dj.
2. Collision and line of impact:
- The spheres collide when the line joining their centers is parallel to vector i.
- From the problem statement, we can assume that the line of impact is along the x-axis.
3. Finding the relative velocity:
- To determine the relative velocity of the spheres during the collision, we need to subtract the velocity of Sphere B from the velocity of Sphere A:
Relative velocity = (ai + bj) - (ci + dj) = (a - c)i + (b - d)j
4. Coefficient of restitution (e):
- The coefficient of restitution (e) represents the ratio of the final relative velocity after the collision to the initial relative velocity before the collision.
- We are given that the coefficient of restitution is less than 1 (e < 1).
5. Final velocities after impact:
- After the collision, Sphere A has a final velocity vector of pi + qj.
- After the collision, Sphere B has a final velocity vector of ri + sj.
6. Applying the principles of conservation of momentum and energy:
- Momentum is conserved in all directions, so the total momentum before the collision is equal to the total momentum after the collision.
- Energy is also conserved during elastic collisions.
Now, let's apply the conservation of momentum in the x and y directions:
Conservation of momentum in the x-direction:
ma × (initial velocity of A in x-direction) + mb × (initial velocity of B in x-direction) = ma × (final velocity of A in x-direction) + mb × (final velocity of B in x-direction)
Since the line joining the centers is parallel to the x-axis, we can neglect the y-component of velocities.
ma × (a - c) = ma × p + mb × r
Simplifying this equation, we get:
(a - c) = p × (ma / ma) + r × (mb / ma)
(a - c) = p + r × (mb / ma) -- (Equation 1)
Conservation of momentum in the y-direction:
ma × (initial velocity of A in y-direction) + mb × (initial velocity of B in y-direction) = ma × (final velocity of A in y-direction) + mb × (final velocity of B in y-direction)
ma × (b - d) = ma × q + mb × s
Simplifying this equation, we get:
(b - d) = q + s × (mb / ma) -- (Equation 2)
Now, we need to use the coefficient of restitution (e), which is given by:
e = (-final relative velocity) / (initial relative velocity)
= -[(final velocity of A) - (final velocity of B)] / [(initial velocity of A) - (initial velocity of B)]
Plugging in the given values, we have:
e = -[(pi + qj) - (ri + sj)] / [(ai + bj) - (ci + dj)]
= -[(p - r)i + (q - s)j] / [(a - c)i + (b - d)j]
Since the coefficient of restitution (e) is defined as less than 1 (e < 1), we can conclude that the initial relative velocity in the x and y directions should be greater than the final relative velocity in the x and y directions.
Therefore, |q - s| < |p - r| -- (Equation 3)
We can rewrite Equation 3:
|b - d| < |a - c|
(as q - s = b - d and p - r = a - c)
Therefore, we can conclude that b = q.
Hence, we have proved that b = q.