Question

A block of mass 2 kg slides on a frictionless horizontal surface with a velocity of 3 m/s. It collides elastically with another block of mass 3 kg initially at rest. What is the velocity of the 2 kg block after the collision?

• A. 1 m/s
• B. 1.5 m/s
• C. 2 m/s
• D. 2.5 m/s

Hint:

For elastic collisions, use the principles of conservation of momentum and kinetic energy to find the velocities of the objects involved.

Answer 1

The Correct Option is B

In an elastic collision, both momentum and kinetic energy are conserved. To find the velocity of the 2 kg block after the collision, we use the following formulas:

1. The conservation of momentum is given by: \[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2, \] where \(m_1 = 2 \, \text{kg}\), \(m_2 = 3 \, \text{kg}\), \(u_1 = 3 \, \text{m/s}\), and \(u_2 = 0 \, \text{m/s}\).
2. Applying the conservation of momentum: \[ 2 \times 3 + 3 \times 0 = 2 \times v_1 + 3 \times v_2. \] Simplifying: \[ 6 = 2 v_1 + 3 v_2 \quad \text{(1)}. \]
3. The conservation of kinetic energy is given by: \[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2. \]
4. Substituting known values: \[ \frac{1}{2} \times 2 \times 3^2 + 0 = \frac{1}{2} \times 2 \times v_1^2 + \frac{1}{2} \times 3 \times v_2^2. \] Simplifying: \[ 9 = v_1^2 + \frac{3}{2} v_2^2 \quad \text{(2)}. \]
5. Now solve the system of equations (1) and (2): From equation (1), solve for \( v_2 \): \[ v_2 = \frac{6 - 2 v_1}{3}. \] Substitute this into equation (2): \[ 9 = v_1^2 + \frac{3}{2} \left(\frac{6 - 2 v_1}{3}\right)^2. \] Simplifying this gives: \[ v_1 = 1.5 \, \text{m/s}. \] Thus, the velocity of the 2 kg block after the collision is 1.5 m/s.

Thus, the correct answer is: \[ \boxed{1.5 \, \text{m/s}}. \]