Question

A body of mass $1 kg$ collides head on elastically with a stationary body of mass $3 kg$ After collision, the smaller body reverses its direction of motion and moves with a speed of $2 m / s$ The initial speed of the smaller body before collision is ___ $ms ^{-1}$

Answer 1

Correct Answer: 4

To solve the problem, we apply the principles of conservation of momentum and conservation of kinetic energy, since the collision is elastic.
Step 1: Conservation of Momentum
The momentum before the collision must equal the momentum after the collision. Let \( u \) be the initial speed of the smaller body.
The momentum before collision: \( p_{\text{before}} = (1\,\text{kg}) \cdot u + (3\,\text{kg}) \cdot 0 \)
The momentum after collision: \( p_{\text{after}} = (1\,\text{kg}) \cdot (-2\,\text{m/s}) + (3\,\text{kg}) \cdot v_{\text{3kg}} \)
Equating the two:
\( 1 \cdot u = 1 \cdot (-2) + 3 \cdot v_{\text{3kg}} \)
Step 2: Conservation of Kinetic Energy
The kinetic energy before collision equals the kinetic energy after collision.
Kinetic energy before: \(\frac{1}{2} \cdot 1 \cdot u^2\)
Kinetic energy after: \(\frac{1}{2} \cdot 1 \cdot (-2)^2 + \frac{1}{2} \cdot 3 \cdot v_{\text{3kg}}^2\)
Equating the two:
\( \frac{1}{2}u^2 = \frac{1}{2}(4) + \frac{3}{2}v_{\text{3kg}}^2\)
Step 3: Solve the Equations
Substitute \( v_{\text{3kg}} \) from the momentum equation into the energy equation:
From momentum, \( v_{\text{3kg}} = \frac{u + 2}{3} \)
Substitute this into the energy equation:
\( \frac{1}{2}u^2 = 2 + \frac{3}{2}\left(\frac{u+2}{3}\right)^2\)
Simplify:
\( u^2 = 4 + \frac{(u+2)^2}{3}\)
\( 3u^2 = 12 + (u^2 + 4u + 4)\)
\( 3u^2 = u^2 + 4u + 16\)
\( 2u^2 - 4u - 16 = 0\)
Divide by 2:
\( u^2 - 2u - 8 = 0\)
Factorize:
\((u - 4)(u + 2) = 0\)
Solutions: \( u = 4 \) or \( u = -2 \)
Since speed cannot be negative, \( u = 4 \,\text{m/s} \).
Conclusion: The initial speed of the smaller body before collision was \( 4 \,\text{m/s} \), which falls within the given range of 4 to 4.