Question

A particle A of mass m moving on smooth horizontal surface collides with stationary particle B of mass 2m directly, situated at a distance d=1.3m from wall. The coefficient of restitution between A and B and between B and wall is e=$\frac{1}{4}$. If the entire motion takes place along a straight line perpendicular to wall, then calculate distance (in m) from wall where they collide again.

Answer 1

0.3

Answer 2

The problem involves a series of one-dimensional collisions. Let's define the wall's position as $x=0$. The initial position of particle B is $x=d=1.3$ m. Let particle A approach B from $x>d$. We take the direction towards the wall (left) as the negative direction, and away from the wall (right) as the positive direction. Let the initial speed of particle A be $u_0$. So, its initial velocity is $u_A = -u_0$. Particle B is initially stationary, $u_B = 0$.

1. First collision (A and B): Mass of A = m, Mass of B = 2m. Initial velocities: $u_A = -u_0$, $u_B = 0$. Coefficient of restitution $e = 1/4$. Let the velocities after the first collision be $v_{A1}$ and $v_{B1}$.

Using conservation of momentum: $m u_A + 2m u_B = m v_{A1} + 2m v_{B1}$ $m(-u_0) + 2m(0) = m v_{A1} + 2m v_{B1}$ $-u_0 = v_{A1} + 2v_{B1}$ (Equation 1)

Using the definition of coefficient of restitution: $e = \frac{v_{B1} - v_{A1}}{u_A - u_B}$ $\frac{1}{4} = \frac{v_{B1} - v_{A1}}{-u_0 - 0}$ $-\frac{1}{4} u_0 = v_{B1} - v_{A1}$ (Equation 2)

Adding Equation 1 and Equation 2: $(-u_0) + (-\frac{1}{4} u_0) = (v_{A1} + 2v_{B1}) + (v_{B1} - v_{A1})$ $-\frac{5}{4} u_0 = 3v_{B1}$ $v_{B1} = -\frac{5}{12} u_0$

Substitute $v_{B1}$ into Equation 2: $-\frac{1}{4} u_0 = -\frac{5}{12} u_0 - v_{A1}$ $v_{A1} = -\frac{5}{12} u_0 + \frac{1}{4} u_0 = \frac{-5+3}{12} u_0 = -\frac{2}{12} u_0 = -\frac{1}{6} u_0$

So, after the first collision, both particles A and B move towards the wall (in the negative x-direction): $v_{A1} = -\frac{1}{6} u_0$ $v_{B1} = -\frac{5}{12} u_0$

Since $|v_{B1}| = \frac{5}{12} u_0$ is greater than $|v_{A1}| = \frac{1}{6} u_0 = \frac{2}{12} u_0$, particle B will reach the wall first.

2. Collision of B with the wall: Particle B starts at $x=d$ and moves towards the wall at $x=0$ with velocity $v_{B1} = -\frac{5}{12} u_0$. Time taken for B to reach the wall: $t_1 = \frac{\text{distance}}{\text{speed}} = \frac{d}{|v_{B1}|} = \frac{d}{\frac{5}{12} u_0} = \frac{12d}{5u_0}$

During this time $t_1$, particle A also moves towards the wall. Its initial position was $x=d$. Position of A when B hits the wall: $x_A(t_1) = d + v_{A1} t_1 = d + (-\frac{1}{6} u_0) \left(\frac{12d}{5u_0}\right) = d - \frac{2d}{5} = \frac{3d}{5}$

Now, particle B collides with the wall. The wall is stationary. Velocity of B before collision with wall = $v_{B1} = -\frac{5}{12} u_0$. Coefficient of restitution between B and wall $e = 1/4$. Let $v_{B2}$ be the velocity of B after reflection from the wall. $e = -\frac{v_{B2} - v_{wall}}{v_{B1} - v_{wall}} \implies e = -\frac{v_{B2}}{v_{B1}}$ $v_{B2} = -e v_{B1} = -\frac{1}{4} \left(-\frac{5}{12} u_0\right) = \frac{5}{48} u_0$ So, after hitting the wall, B moves away from the wall (in the positive x-direction) with velocity $v_{B2} = \frac{5}{48} u_0$.

3. Second collision (A and B): At time $t_1$ (let's reset our clock to $t=0$ for this phase): Particle A: position $x_A = \frac{3d}{5}$, velocity $v_A = -\frac{1}{6} u_0$. Particle B: position $x_B = 0$, velocity $v_B = \frac{5}{48} u_0$.

A is moving towards the wall (left), and B is moving away from the wall (right). They are moving towards each other. Let them collide at time $t'$ (after $t_1$) at position $x_f$. Position of A at time $t'$: $x_A(t') = x_A + v_A t' = \frac{3d}{5} - \frac{1}{6} u_0 t'$ Position of B at time $t'$: $x_B(t') = x_B + v_B t' = 0 + \frac{5}{48} u_0 t'$

For collision, $x_A(t') = x_B(t')$: $\frac{3d}{5} - \frac{1}{6} u_0 t' = \frac{5}{48} u_0 t'$ $\frac{3d}{5} = \frac{1}{6} u_0 t' + \frac{5}{48} u_0 t'$ $\frac{3d}{5} = \left(\frac{8}{48} + \frac{5}{48}\right) u_0 t'$ $\frac{3d}{5} = \frac{13}{48} u_0 t'$ $t' = \frac{3d}{5} \cdot \frac{48}{13 u_0} = \frac{144d}{65 u_0}$

The distance from the wall where they collide again is $x_f$. We can find it using B's position equation: $x_f = x_B(t') = \frac{5}{48} u_0 t'$ $x_f = \frac{5}{48} u_0 \left(\frac{144d}{65 u_0}\right)$ $x_f = \frac{5 \cdot 144 d}{48 \cdot 65}$ $x_f = \frac{5 \cdot (3 \cdot 48) d}{48 \cdot (13 \cdot 5)}$ $x_f = \frac{3d}{13}$

Given $d = 1.3$ m. $x_f = \frac{3 \cdot 1.3}{13} = \frac{3 \cdot (13/10)}{13} = \frac{3}{10} = 0.3$ m.

The distance from the wall where they collide again is 0.3 m.