Three masses connected by a string (momentum conservation) [closed]

Question

Three identical balls each of mass m = 0.5 kg are connected with each other as shown in figure and rest over a smooth horizontal table. At moment t=0, ball B is imparted a horizontal velocity v=9m/s. Calculate velocity of A just before it collides with ball C.

I tried a different method to solve this question where i assumed ball A and B to collide without contact and tension being the internal force providing the impulse, by this we get that in vertical direction, just before collision all balls have same velocity and we can conserve momentum there however this kind of collision is inelastic so we cannot use conservation of kinetic energy, yet conserving it does give the right answer. what is wrong here?

Answer 1

Firstly, the net external force acting on the system (all 3 bodies) is zero, which means that you are free to use the laws of conservation of energy and momentum. You need to understand that the impulse provided by the string only affects the velocities of individual bodies such that the velocity of both the bodies along the string joining them are equal, which seems pretty obvious as the string will break if the velocities of both the balls are unequal. And since we have assumed that the string is inextensible and there is no friction between any surfaces, it is clear that the total energy loss in the system is zero, and hence you can apply the law of conservation of energy. You also need to understand that the loss of energy in an inelastic collision is due to the deformation of bodies and energy thereby getting wasted(in deformation and also heat loss, if any) and also getting stored in the bodies in the form of potential energy. But in your question, all those possibilities are sidelined due to the assumptions we have made (that I have mentioned above).

Answer 2

Note sure why you think A will collide with B, or what "collide without contact" means. My interpretation of the scenario is that B moves in the positive $y$ direction, A and C are pulled after it by strings that connect them to B (I am assuming these are light inextensible strings), so throughout the motion A and C have equal velocities in the $y$ direction (until they collide), but also equal and opposite velocities towards each other in the $x$ direction.

I think we must also assume that the size of the balls is small compared to the distance between them, so that just before A and C collide we can treat both strings as being parallel to the $y$ axis. This means that just before A and C collide the $y$ components of the velocities of A, B and C are all equal. However, we need to remember that the velocities of A and C also have equal and opposite $x$ components.

If we call the $y$ component of the velocities of A, B and C just before A and C collide $v_y$ then by conservation of momentum we have

$mv_0 = 3mv_y$

And if we call the magnitude of the $x$ component of the velocities of A and C just before the collision $v_x$ then by conservation of energy

$\displaystyle \frac 1 2 mv_0^2 = \frac 1 2 mv_y^2 + m(v_x^2+v_y^2)$

So you have two equations in two unknowns $v_x$ and $v_y$. Solve these to find $v_x$ and $v_y$, then the velocity of A just before it collides with C is $\sqrt{v_x^2 + v_y^2}$.

Since we are calculating velocities before any of the balls have collided with one another, the question of whether these collisions are perfectly elastic or not is irrelevant.