Question

A 2 kg object is moving with a velocity of 5 m/s on a frictionless surface. It collides elastically with a stationary object of mass 3 kg. Find the velocity of the 3 kg object after the collision.

• A. \(4 \, \text{m/s}\)
• B. \(3 \, \text{m/s}\)
• C. \(5 \, \text{m/s}\)
• D. \(2 \, \text{m/s}\)

Hint:

In perfectly elastic collisions, use standard formulas: \[ v_2 = \frac{2m_1}{m_1 + m_2} u_1 \quad \text{and} \quad v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 \] when second body is at rest.

Answer 1

The Correct Option is A

Step 1: Apply Conservation of Momentum
For any collision, momentum is conserved: \[ m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \] Substituting values: \[ 2 \times 5 + 3 \times 0 = 2v_1 + 3v_2 \] \[ 10 = 2v_1 + 3v_2 \quad \text{(Equation 1)} \] Step 2: Apply Conservation of Kinetic Energy
For elastic collisions, kinetic energy is conserved:
\[ \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 \] Simplifying: \[ 25 + 0 = v_1^2 + 1.5v_2^2 \quad \text{(Equation 2)} \] Step 3: Solve the System of Equations
From Equation 1: \[ v_1 = \frac{10 - 3v_2}{2} = 5 - 1.5v_2 \] Substitute into Equation 2: \[ (5 - 1.5v_2)^2 + 1.5v_2^2 = 25 \] Expanding: \[ 25 - 15v_2 + 2.25v_2^2 + 1.5v_2^2 = 25 \] Simplifying: \[ 3.75v_2^2 - 15v_2 = 0 \] \[ v_2(3.75v_2 - 15) = 0 \] Solutions:

• $v_2 = 0$ (trivial, no collision occurred)
• $3.75v_2 - 15 = 0 \Rightarrow v_2 = 4\ \text{m/s}$

So, the correct answer is \(4 \, \text{m/s}\).