COM O11

Solution to Question 11

Physics Solution: Collision Conditions

1. Initial Collision Analysis

Block \( m \) moves with velocity \( u \) towards stationary block \( M \). The collision is elastic.

Let \( v_1 \) be the velocity of \( m \) and \( v_2 \) be the velocity of \( M \) after the first collision. Using standard elastic collision formulas:

$$ v_1 = \left( \frac{m – M}{m + M} \right) u $$ $$ v_2 = \left( \frac{2m}{m + M} \right) u $$

2. Analyzing the “Collide Only Once” Condition

For the blocks to collide only once, block \( m \) must never catch up to block \( M \) after the first interaction.

Case A: \( m \ge M \)

If \( m \ge M \), then \( v_1 \ge 0 \). Both blocks move to the right. Since \( v_2 > v_1 \) (as the target in an elastic collision always moves faster than the projectile if \( m \ge M \)), block \( M \) moves away faster. They will not collide again. This works for \( m \ge M \).

Case B: \( m < M \)

If \( m < M \), then \( v_1 < 0 \). Block \( m \) rebounds (moves left), hits the wall, and reflects elastically. Its new velocity is \( |v_1| \) towards the right.

Now, both blocks are moving right. A second collision will occur if \( m \) is faster than \( M \): \( |v_1| > v_2 \). To ensure they collide only once, we need \( m \) to be slower or equal speed:

$$ |v_1| \le v_2 $$

Substituting the magnitudes:

$$ \frac{M – m}{m + M} u \le \frac{2m}{m + M} u $$ $$ M – m \le 2m $$ $$ M \le 3m $$

3. Combining Conditions

We found that single collision occurs if:

  • \( m \ge M \) (which implies \( M \le m \), and naturally \( M \le 3m \))
  • \( m < M \) provided that \( M \le 3m \)

Both cases are covered by the single condition: \( M \le 3m \).

Final Answer

The condition under which the blocks will collide only once is (b) \( M \le 3m \).

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