A 2kg ball hits a 3kg ball at rest — find velocities after elastic collision

Question

A ball of mass $m_1 = 2\text{ kg}$ moving at $u_1 = 6\text{ m/s}$ collides head-on with a ball of mass $m_2 = 3\text{ kg}$ at rest ( $u_2 = 0$ ). The collision is perfectly elastic. Find the velocities of both balls after the collision.

Solution — Step by Step

In a perfectly elastic collision, two quantities are conserved:

Linear momentum: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$
Kinetic energy: $\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

Where $v_1, v_2$ are the final velocities.

Since $u_2 = 0$ (second ball at rest), these simplify.

m_1u_1 + m_2 \cdot 0 = m_1v_1 + m_2v_2

2 \times 6 = 2v_1 + 3v_2

12 = 2v_1 + 3v_2 \quad \text{...(1)}

\frac{1}{2}m_1u_1^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

m_1u_1^2 = m_1v_1^2 + m_2v_2^2

2 \times 36 = 2v_1^2 + 3v_2^2

72 = 2v_1^2 + 3v_2^2 \quad \text{...(2)}

Now we have two equations in two unknowns.

For elastic collision with $u_2 = 0$ , we can derive (or use) the standard formulas:

v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 = \frac{2 - 3}{2 + 3} \times 6 = \frac{-1}{5} \times 6 = -1.2\text{ m/s}

v_2 = \frac{2m_1}{m_1 + m_2} u_1 = \frac{2 \times 2}{2 + 3} \times 6 = \frac{4}{5} \times 6 = 4.8\text{ m/s}

Check momentum: $m_1v_1 + m_2v_2 = 2(-1.2) + 3(4.8) = -2.4 + 14.4 = 12 = m_1u_1$ ✓

Check KE: $\frac{1}{2}(2)(1.2)^2 + \frac{1}{2}(3)(4.8)^2 = \frac{1}{2}(2)(1.44) + \frac{1}{2}(3)(23.04) = 1.44 + 34.56 = 36 = \frac{1}{2}(2)(6)^2$ ✓

Result: $v_1 = -1.2\text{ m/s}$ (bounces back), $v_2 = 4.8\text{ m/s}$ (moves forward)

Why This Works

The negative sign for $v_1$ tells us the 2 kg ball bounces backwards after the collision. This makes physical sense: when a lighter object hits a heavier stationary object, the lighter one bounces back.

Intuition check with special cases:

If $m_1 = m_2$ : $v_1 = 0$ , $v_2 = u_1$ — the first ball stops completely and the second ball moves with the first ball’s original velocity. This is the “Newton’s cradle” behavior.
If $m_1 \gg m_2$ (very heavy hits light): $v_1 \approx u_1$ (barely slowed), $v_2 \approx 2u_1$ (light ball shoots forward at twice the original speed).
If $m_1 \ll m_2$ (very light hits heavy): $v_1 \approx -u_1$ (bounces back with nearly the same speed), $v_2 \approx 0$ (heavy ball barely moves). Think of a tennis ball hitting a wall.

Alternative Method

Instead of using the formulas directly, we can solve the system of equations (1) and (2) algebraically. From equation (1): $v_1 = \frac{12 - 3v_2}{2}$ . Substituting into (2):

$2\left(\frac{12 - 3v_2}{2}\right)^2 + 3v_2^2 = 72$

$\frac{(12 - 3v_2)^2}{2} + 3v_2^2 = 72$

$(12 - 3v_2)^2 + 6v_2^2 = 144$

$144 - 72v_2 + 9v_2^2 + 6v_2^2 = 144$

$15v_2^2 - 72v_2 = 0$

$v_2(15v_2 - 72) = 0$

So $v_2 = 0$ (trivial solution — no collision) or $v_2 = 72/15 = 4.8\text{ m/s}$ ✓

Then $v_1 = (12 - 3 \times 4.8)/2 = (12 - 14.4)/2 = -1.2\text{ m/s}$ ✓

In JEE Main, elastic collision problems often come with a choice: use the formulas directly or solve the simultaneous equations. The formulas are faster for head-on collisions with one mass at rest. If one mass is not at rest (both moving), you must set up the equations from scratch or use the relative velocity approach ( $v_2 - v_1 = -(u_2 - u_1)$ for elastic collisions). Memorizing the formulas for “second ball at rest” case saves significant time.

Common Mistake

Students often drop the negative sign for $v_1$ and report both velocities as positive. The direction is crucial — the negative sign means the first ball reverses direction. If asked for “speed” (magnitude), the answer would be 1.2 m/s. If asked for “velocity,” it’s −1.2 m/s (taking the original direction as positive). Always state the direction or use the sign system consistently. Also, some students check only momentum conservation and forget to verify kinetic energy — always check both for elastic collision problems.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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