Source: Physics Competition - Mechanics
Problem Sets:
Problem
Two identical small balls, A and B, both with mass $m$, are initially at rest. Ball A is a distance $a$ behind ball B. At the same instant, ball A is given an impulse $I$ and ball B is subjected to a constant force $F$. Both the impulse and the force are directed along the line connecting A to B.
The condition for which ball A will not overtake ball B is:
$$I^2 \le 2mFa$$[Q1] Determine the condition for which ball A will not overtake ball B.
Let the initial position of ball A be $x_A(0) = 0$ and ball B be $x_B(0) = a$.
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Equations of Motion
- Ball A: At $t=0$, it receives an impulse $I$. From the impulse-momentum theorem, its initial (and constant) velocity is $v_A = I/m$. The position of ball A at time $t$ is: $$x_A(t) = v_A t = \frac{I}{m} t$$
- Ball B: At $t=0$, it is subjected to a constant force $F$. Its constant acceleration is $a_B = F/m$. Starting from rest, its position at time $t$ is: $$x_B(t) = x_B(0) + v_{B,0} t + \frac{1}{2} a_B t^2 = a + \frac{1}{2} \frac{F}{m} t^2$$
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Condition for Non-Overtaking For ball A not to overtake ball B, its position must always be less than or equal to ball B's position for all times $t \ge 0$.
$$x_A(t) \le x_B(t)$$The limiting case is when A "just touches" B. At this specific instant, say $t_c$, both balls are at the same position and must have the same velocity. If A's velocity were greater at that point, it would overtake B.
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Derivation of the Critical Condition First, we find the time $t_c$ when their velocities are equal. The velocity of ball A is constant: $v_A(t) = I/m$. The velocity of ball B is: $v_B(t) = a_B t = (F/m)t$. Setting $v_A(t_c) = v_B(t_c)$:
$$\frac{I}{m} = \frac{F}{m} t_c \implies t_c = \frac{I}{F}$$Next, for the "just touching" condition, we require their positions to be equal at this time $t_c$.
$$x_A(t_c) = x_B(t_c)$$ $$\frac{I}{m} t_c = a + \frac{1}{2} \frac{F}{m} t_c^2$$Substitute the expression for $t_c$:
$$\frac{I}{m} \left(\frac{I}{F}\right) = a + \frac{1}{2} \frac{F}{m} \left(\frac{I}{F}\right)^2$$ $$\frac{I^2}{mF} = a + \frac{1}{2} \frac{F}{m} \frac{I^2}{F^2}$$ $$\frac{I^2}{mF} = a + \frac{I^2}{2mF}$$Solving for $a$:
$$a = \frac{I^2}{mF} - \frac{I^2}{2mF} = \frac{I^2}{2mF}$$This gives the critical relation:
$$I^2 = 2mFa$$This equation defines the boundary condition. For A not to overtake B, its initial impulse must be less than or equal to this critical value. A larger impulse would cause A to reach B while still traveling faster, thus overtaking it.
Therefore, the condition for non-overtaking is $I^2 \le 2mFa$.